Algebra I math worksheets

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Write linear equations

Learning goals

Identifies slope and an appropriate point or intercept from given information.
Writes a linear equation in a suitable form.
Converts among equivalent linear forms when useful.
Checks the equation against given points or context.

Difficulty: 1 – 5

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Write linear equations 1

Write a linear equation in \(x\) and \(y\) for each verbal relationship. a) The sum of \(x\) and \(y\) is \(18\). b) \(y\) is \(7\) less than \(x\). c) \(x\) is three times \(y\). d) \(y\) is the opposite of \(x\).

Hints

- Identify the operation named in each statement. - Translate “less than,” “times,” and “opposite” carefully. - Check that the variable named first in the statement is on the correct side of the equation.

Solution

1. a) “The sum of \(x\) and \(y\)” is \(x+y\), so the equation is \(x+y=18\). 2. b) “\(7\) less than \(x\)” is \(x-7\), so the equation is \(y=x-7\). 3. c) “Three times \(y\)” is \(3y\), so the equation is \(x=3y\). 4. d) The opposite of \(x\) is \(-x\), so the equation is \(y=-x\).

Answer

a) \(x+y=18\) b) \(y=x-7\) c) \(x=3y\) d) \(y=-x\)

Write linear equations 2

A tank starts with \(40\,\text{gal}\) of water. Water is added at a constant rate of \(6\,\text{gal}\) per minute. Write a linear function \(V(t)\) for the volume of water, in gallons, after \(t\) minutes.

Hints

- Identify the amount present before any time passes. - Identify how much the output changes for each one-unit increase in time. - Put those two quantities into a linear rule with the initial amount as the constant term.

Solution

1. The initial volume is the value when \(t=0\), so the y-intercept is \(40\). 2. The volume increases by \(6\,\text{gal}\) each minute, so the slope is \(6\). 3. Therefore, \(V(t)=40+6t\).

Answer

\(V(t)=40+6t\)

Write linear equations 3

Write the equation of a line with slope \(-3\) and y-intercept \(4\). Use slope-intercept form.

Hints

- Which number is the coefficient of \(x\) in slope-intercept form? - Which number is added as a constant?

Solution

1. Slope-intercept form is \(y=mx+b\), where \(m\) is the slope and \(b\) is the y-intercept. 2. Substitute \(m=-3\) and \(b=4\), giving \(y=-3x+4\).

Answer

\(y=-3x+4\).

Write linear equations 4

Point-slope form is \(y-y_1=m(x-x_1)\). Write a point-slope equation for a line with slope \(3\) passing through \((2,5)\).

Hints

- Match each part of the given point with \(x_1\) and \(y_1\). - The slope belongs in front of the parentheses.

Solution

1. Use \(m=3\), \(x_1=2\), and \(y_1=5\). 2. Substitute into point-slope form to obtain \(y-5=3(x-2)\).

Answer

\(y-5=3(x-2)\).

Write linear equations 5

A line has the equation \(y+2=-3(x-1)\). Without expanding, give its slope and one point on the line.

Hints

- Which number gives the slope in the form \(y-y_1=m(x-x_1)\)? - Check a possible point by putting its coordinates into the equation.

Solution

1. In point-slope form, \(y-y_1=m(x-x_1)\). Here the coefficient is \(m=-3\). 2. Since \(y+2=y-(-2)\) and \(x-1\) uses \(x_1=1\), the point is \((1,-2)\).

Answer

Slope: \(-3\). Point: \((1,-2)\).

Write linear equations 6

A bike rental costs \(\$5\) to start, plus \(\$3\) for each hour rented. Write a linear function \(C(h)\) for the total cost in dollars after \(h\) hours.

Hints

- Which cost is paid even before any hours have passed? - Which part grows with the number of rental hours?

Solution

1. The starting cost is \(5\) dollars. 2. Each hour adds \(3\) dollars, so the total is \(C(h)=3h+5\), for \(h\ge0\).

Answer

\(C(h)=3h+5\), with \(h\ge0\).

Write linear equations 7

A museum charges \(\$7\) for a day ticket. The price stays the same no matter how many hours \(h\) a visitor stays. Write the cost function \(C(h)\). What are its slope and y-intercept?

Hints

- What happens to the price when the visit lasts another hour? - How could a constant price be written in slope-intercept form? - Check what the rule gives at zero hours.

Solution

1. The cost does not change with time, so \(C(h)=7\). 2. In \(C(h)=0h+7\), the slope is \(0\) and the y-intercept is \((0,7)\).

Answer

\(C(h)=7\); slope \(0\); y-intercept \((0,7)\).

Write linear equations 8

A line passes through the points \(A(-1,-2)\) and \(B(3,4)\). Write the equation of the line in slope-intercept form.

Hints

- Find the slope from the change in y divided by the change in x. - After finding the slope, substitute either given point into \(y=mx+b\). - Check that your final equation contains both given points.

Solution

1. Find the slope: \(m=\frac{4-(-2)}{3-(-1)}=\frac{6}{4}=1.5\). 2. Use \(B(3,4)\) to find the y-intercept: \(4=1.5\cdot3+b\), so \(b=-0.5\). 3. Therefore, the equation is \(y=1.5x-0.5\).

Answer

\(y=1.5x-0.5\)

Write linear equations 9

Linear functions can be written in the form \(f(x) = mx + b\). a) Find the equation of a line with y-intercept \(-3\) and x-intercept \(6\). b) A line has no x-intercept and passes through \(P(2, 5)\). Find its equation and justify your answer using the slope \(m\).

Hints

- What does the y-intercept tell you about \(b\)? - What must be true about the slope of a line that never crosses the x-axis? - What ordered pair represents the given x-intercept in part a?

Solution

1. For part a, the y-intercept gives \(b = -3\). Since \((6, 0)\) is on the line, \(0 = 6m - 3\), so \(m = 0.5\). Thus \(f(x) = 0.5x - 3\). 2. For part b, any nonhorizontal line has an x-intercept, so a line with no x-intercept must have slope \(m = 0\) and a nonzero y-intercept. 3. Because the horizontal line passes through \((2, 5)\), its equation is \(f(x) = 5\).

Answer

a) \(f(x) = 0.5x - 3\) b) \(f(x) = 5\). Its slope is \(m = 0\), so the line is horizontal and never crosses the x-axis.

Write linear equations 10

Find the equation of a linear function \(f\). Its graph has y-intercept \((0, 4)\) and x-intercept \((-2, 0)\).

Hints

- What does the y-intercept tell you about \(b\) in \(y = mx + b\)? - What is the y-coordinate of every point on the x-axis? - Substitute a known point into the linear equation to find the unknown slope.

Solution

1. The y-intercept gives \(b = 4\), so \(f(x) = mx + 4\). 2. Since \((-2, 0)\) is on the graph, \(0 = -2m + 4\). 3. Solving gives \(m = 2\), so \(f(x) = 2x + 4\).

Answer

\(f(x) = 2x + 4\)

Write linear equations 11

A line \(g\) has slope \(m = -0.75\) and x-intercept \((4, 0)\). a) Find the y-intercept \(b\). b) Write the equation of \(g\).

Hints

- What is the slope-intercept form of a linear equation? - What is the y-coordinate of a point on the x-axis? - Substitute the known slope and point into the equation to find the missing intercept.

Solution

1. Start with \(g(x) = -0.75x + b\). 2. Since \((4, 0)\) is on the line, \(0 = -0.75 \cdot 4 + b\). 3. Thus \(0 = -3 + b\), so \(b = 3\). 4. Therefore, \(g(x) = -0.75x + 3\).

Answer

a) \(b = 3\) b) \(g(x) = -0.75x + 3\)

Write linear equations 12

A linear function has the form \(f(x) = 0.5x + b\) and has an x-intercept at \(x = 6\). a) Find \(b\). b) What is the y-intercept of the graph? c) A second function is \(g(x) = 0.5x\). Describe the translation that maps the graph of \(g\) to the graph of \(f\).

Hints

- What is the y-coordinate of a point on the x-axis? - Substitute the x-intercept into the given function form. - Compare the two equations. Which term changes the vertical position of the graph?

Solution

1. Since \(f(6) = 0\), \(0 = 0.5 \cdot 6 + b\). Thus \(b = -3\). 2. The y-intercept is \((0, b) = (0, -3)\). 3. Since \(f(x) = g(x) - 3\), the graph of \(g\) is translated down \(3\) units to obtain the graph of \(f\).

Answer

a) \(b = -3\) b) \((0, -3)\) c) Translate the graph of \(g\) down \(3\) units.

Write linear equations 13

A linear function \(f\) has y-intercept \(-3\). To graph the line from the y-intercept, move \(4\) units to the right and \(3\) units down to reach another point. Find the equation of \(f\).

Hints

- What sign should the rise have when the line moves downward from left to right? - How do rise and run determine the slope? - Which two values determine an equation in slope-intercept form?

Solution

1. The y-intercept is \(b = -3\). 2. Moving \(4\) units right gives a run of \(4\), and moving \(3\) units down gives a rise of \(-3\). 3. The slope is \(m = \frac{-3}{4}\). 4. Therefore, \(f(x) = -\frac{3}{4}x - 3\).

Answer

\(f(x) = -\frac{3}{4}x - 3\)

Write linear equations 14

The point \(S(0, 5)\) lies on the y-axis. a) Write three different linear equations of the form \(y = mx + b\) whose graphs all pass through \(S\). b) A line through \(S\) also passes through \(Q(4, 2)\). Find its slope \(m\).

Hints

- What does the point \((0, 5)\) tell you about the y-intercept? - How many points are needed to determine a unique slope? - In slope-intercept form, which value can change while the line still passes through \((0, 5)\)? - For part b, use the two given points to calculate slope.

Solution

1. Any line through \((0, 5)\) has y-intercept \(b = 5\). 2. For part a, choose any three different slopes. For example, \(y = x + 5\), \(y = 2x + 5\), and \(y = -x + 5\) all pass through \(S\). 3. For part b, use \(S(0, 5)\) and \(Q(4, 2)\): \(m = \frac{2 - 5}{4 - 0} = -\frac{3}{4}\).

Answer

a) Answers will vary. One possible set is \(y = x + 5\), \(y = 2x + 5\), and \(y = -x + 5\). b) \(m = -\frac{3}{4}\)

Write linear equations 15

Find the equation of the line that passes through \(A(-2, -5)\) and \(B(3, 5)\). Show your calculations.

Hints

- What form can you use for a linear equation with slope and y-intercept? - How do you calculate slope from two points? - After finding the slope, substitute either point to find the y-intercept.

Solution

1. Find the slope: \(m = \frac{5 - (-5)}{3 - (-2)} = \frac{10}{5} = 2\). 2. Substitute \((3, 5)\) into \(y = mx + b\): \(5 = 2 \cdot 3 + b\), so \(b = -1\). 3. Therefore, the equation is \(y = 2x - 1\).

Answer

\(y = 2x - 1\)

Write linear equations 16

Line \(h\) has equation \(y = -1.5x + 4\). A line \(k\) is parallel to \(h\) and has x-intercept \((4, 0)\). Find the equation of \(k\).

Hints

- What does parallelism tell you about the slopes of the two lines? - What is the y-coordinate of a point on the x-axis? - Substitute the known slope and point into slope-intercept form.

Solution

1. Since \(k\) is parallel to \(h\), its slope is \(m = -1.5\). 2. Substitute \((4, 0)\) into \(y = -1.5x + b\): \(0 = -1.5 \cdot 4 + b\). 3. Thus \(b = 6\), so the equation is \(y = -1.5x + 6\).

Answer

\(y = -1.5x + 6\)

Write linear equations 17

Line \(g\) is parallel to the line \(h\), whose equation is \(y = -2x + 7\). The graph of \(g\) passes through \(P(3, 4)\). Find the equation of \(g\).

Hints

- What do parallel lines have in common about their slopes? - Write a linear equation using the known slope and an unknown y-intercept. - Substitute the given point to find the y-intercept.

Solution

1. Since \(g\) is parallel to \(h\), its slope is \(m = -2\). 2. Substitute \((3, 4)\) into \(y = -2x + b\): \(4 = -2 \cdot 3 + b\). 3. Solving gives \(b = 10\), so \(g\) has equation \(y = -2x + 10\).

Answer

\(y = -2x + 10\)

Write linear equations 18

A linear function has y-intercept \(-3\) and slope \(1.5\). Its output is \(1.5\) at one x-value. a) Starting from the y-intercept, what vertical change is needed to reach an output of \(1.5\)? b) Use the slope as rise divided by run to find the corresponding horizontal change and x-value. c) Write the linear equation for this situation and solve it using equivalent transformations. Confirm that it gives the same x-value as part b).

Hints

- Compare the target output with the y-intercept before using the slope. - Interpret the slope as a ratio of vertical change to horizontal change. - After the graphical-rate reasoning, translate the same information into an equation.

Solution

1. The vertical change from \(-3\) to \(1.5\) is \(1.5-(-3)=4.5\). 2. The slope is \(1.5=\frac{4.5}{3}\), so a rise of \(4.5\) corresponds to a run of \(3\). Starting at \(x=0\), the required x-value is \(x=3\). 3. The function is \(f(x)=1.5x-3\), so the required equation is \(1.5x-3=1.5\). 4. Add \(3\): \(1.5x=4.5\). Divide by \(1.5\): \(x=3\), agreeing with the slope reasoning.

Answer

a) A vertical change of \(4.5\) b) A horizontal change of \(3\), so \(x=3\) c) \(1.5x-3=1.5\), which gives \(x=3\)

Write linear equations 19

A linear function passes through \(P(0, -3)\) and \(Q(4, 0)\). a) Find its equation in the form \(f(x) = mx + b\). b) Interpret \(f(x) = 0\) graphically using the given points. Which point gives the solution directly?

Hints

- How do you find the slope of a line from two points? - What is true about the coordinates of a point on the x-axis? - How is the equation \(f(x) = 0\) connected to a zero of the function?

Solution

1. Since \(P(0, -3)\) lies on the y-axis, the y-intercept is \(b = -3\). 2. The slope is \(m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4}\). 3. Therefore, \(f(x) = \frac{3}{4}x - 3\). 4. The equation \(f(x) = 0\) asks for the x-value where the graph crosses the x-axis. 5. Point \(Q(4, 0)\) is on the x-axis, so it shows directly that the solution is \(x = 4\).

Answer

a) \(f(x) = \frac{3}{4}x - 3\) b) \(Q(4, 0)\) gives the solution directly, so \(x = 4\).

Write linear equations 20

A linear function has the form \(f(x) = mx - 1\), and its graph passes through \(P(-2, 5)\). a) Find the slope \(m\). b) Use a calculation to determine whether \(Q(1, -4)\) also lies on the graph of \(f\).

Hints

- Substitute the known point into the function to solve for the slope. - To test \(Q\), substitute its x-coordinate and compare the output with its y-coordinate.

Solution

1. Substitute \(P(-2, 5)\): \(5 = -2m - 1\). Thus \(6 = -2m\), so \(m = -3\). The function is \(f(x) = -3x - 1\). 2. Check \(Q\): \(f(1) = -3 \cdot 1 - 1 = -4\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the graph.

Answer

a) \(m = -3\) b) Yes. Since \(f(1) = -4\), \(Q(1, -4)\) lies on the graph.

Write linear equations 21

For each description, write the equation of the line in the form \(f(x) = mx + b\). a) The line has y-intercept \(2\). From one point on the line, moving \(3\) units right and \(1\) unit down reaches another point on the line. b) The line passes through \(P(0, -1.5)\). Starting at the y-intercept, moving \(1\) unit right and \(2.5\) units up reaches another point on the line. c) The line passes through the origin. A move of \(4\) units right corresponds to a move of \(3\) units up along the line.

Hints

- Which value in slope-intercept form tells where the line crosses the y-axis? - How do rise and run determine the slope? - What is the y-intercept of a line through the origin? - Use the direction of the vertical change to determine the sign of the slope.

Solution

1. For part a, the y-intercept is \(2\), and the slope is \(m = \frac{-1}{3}\). Thus \(f(x) = -\frac{1}{3}x + 2\). 2. For part b, the y-intercept is \(-1.5\), and the slope is \(m = \frac{2.5}{1} = 2.5\). Thus \(f(x) = 2.5x - 1.5\). 3. For part c, passing through the origin gives \(b = 0\). The slope is \(m = \frac{3}{4}\), so \(f(x) = \frac{3}{4}x\).

Answer

a) \(f(x) = -\frac{1}{3}x + 2\) b) \(f(x) = 2.5x - 1.5\) c) \(f(x) = \frac{3}{4}x\)

Write linear equations 22

A line passes through \((-2, 5)\). Whenever \(x\) increases by \(3\), \(y\) decreases by \(6\). a) Write a linear function \(f(x)\) whose graph is this line. b) Find the x-intercept.

Hints

- Convert the stated horizontal and vertical changes into a slope. - Use the given point to determine the constant term in \(y=mx+b\). - At the x-intercept, the y-coordinate is zero.

Solution

1. The slope is \(m=\frac{-6}{3}=-2\). 2. Substitute \((-2, 5)\) into \(y=-2x+b\): \(5=-2(-2)+b\), so \(b=1\). Thus, \(f(x)=-2x+1\). 3. At the x-intercept, \(0=-2x+1\), so \(x=\frac{1}{2}\). The intercept is \(\left(\frac{1}{2}, 0\right)\).

Answer

a) \(f(x)=-2x+1\) b) \(\left(\frac{1}{2}, 0\right)\)

Write linear equations 23

Write an equation for each linear function. a) The graph of \(f\) passes through \(A(-2, 7)\) and \(B(4, -5)\). b) The graph of \(g\) has slope \(-1.5\) and passes through \(C(2, 1)\).

Hints

- Use the slope formula when two points are given. - Write the line in the form \(y=mx+b\). - Substitute a known point to find \(b\).

Solution

1. For \(f\), the slope is \(m=\frac{-5-7}{4-(-2)}=-2\). 2. Substitute \(A(-2, 7)\) into \(y=-2x+b\): \(7=4+b\), so \(b=3\). Therefore, \(f(x)=-2x+3\). 3. For \(g\), substitute \(C(2, 1)\) into \(y=-1.5x+b\): \(1=-3+b\), so \(b=4\). Therefore, \(g(x)=-1.5x+4\).

Answer

a) \(f(x)=-2x+3\) b) \(g(x)=-1.5x+4\)

Write linear equations 24

The graph shows two linear functions, \(f\) and \(g\). a) For \(f\), find the y-intercept \(b_1\) and slope \(m_1\). Then write the function equation. b) For \(g\), find the y-intercept \(b_2\) and slope \(m_2\). Then write the function equation.

Figure for problem 532209

Hints

- Where does each line cross the y-axis? - Choose two grid points on each line to form a slope triangle. - Find slope as vertical change divided by horizontal change, including the correct sign. - Substitute the slope and y-intercept into \(y = mx + b\).

Solution

1. The graph of \(f\) crosses the y-axis at \((0, -1)\), so \(b_1 = -1\). Using \((0, -1)\) and \((2, 0)\), the slope is \(m_1 = \frac{0 - (-1)}{2 - 0} = \frac{1}{2}\). Therefore, \(f(x) = \frac{1}{2}x - 1\). 2. The graph of \(g\) crosses the y-axis at \((0, 3)\), so \(b_2 = 3\). Using \((0, 3)\) and \((2, 0)\), the slope is \(m_2 = \frac{0 - 3}{2 - 0} = -\frac{3}{2}\). Therefore, \(g(x) = -\frac{3}{2}x + 3\).

Answer

a) \(b_1 = -1\), \(m_1 = \frac{1}{2}\), and \(f(x) = \frac{1}{2}x - 1\) b) \(b_2 = 3\), \(m_2 = -\frac{3}{2}\), and \(g(x) = -\frac{3}{2}x + 3\)

Write linear equations 25

The figure shows two linear functions, \(f\) and \(g\). Two points are marked on each line. For each line, find the slope \(m\), the y-intercept \(b\), and the complete equation in the form \(y = mx + b\).

Figure for problem 532406

Hints

- Read the coordinates of the two marked points on each line. - Use change in y divided by change in x to calculate the slope. - Where does each line cross the y-axis? - Substitute the slope and y-intercept into \(y = mx + b\).

Solution

1. For \(f\), the marked points are \((0, -1)\) and \((3, 1)\). The slope is \(m = \frac{1 - (-1)}{3 - 0} = \frac{2}{3}\). Since \((0, -1)\) is on the y-axis, \(b = -1\). Therefore, \(f(x) = \frac{2}{3}x - 1\). 2. For \(g\), the marked points are \((-1, 3)\) and \((1, -1)\). The slope is \(m = \frac{-1 - 3}{1 - (-1)} = -2\). Using \((1, -1)\), \(-1 = -2 \cdot 1 + b\), so \(b = 1\). Therefore, \(g(x) = -2x + 1\).

Answer

\(f\): \(m = \frac{2}{3}\), \(b = -1\), \(f(x) = \frac{2}{3}x - 1\) \(g\): \(m = -2\), \(b = 1\), \(g(x) = -2x + 1\)

Write linear equations 26

Find the equation of each line \(f\), \(g\), and \(h\) shown in the coordinate plane. Write each equation in the form \(y = mx + b\).

Figure for problem 532410

Hints

- First read the y-intercept of each line. - Choose two grid points on each line to form a slope triangle. - Compute slope as \(\frac{\Delta y}{\Delta x}\). - A line that falls from left to right has a negative slope.

Solution

1. Line \(f\) crosses the y-axis at \((0, 1)\). Using \((0, 1)\) and \((2, 4)\), the slope is \(\frac{4 - 1}{2 - 0} = \frac{3}{2}\). Therefore, \(f(x) = \frac{3}{2}x + 1\). 2. Line \(g\) crosses the y-axis at \((0, -2)\). Using \((0, -2)\) and \((2, -3)\), the slope is \(\frac{-3 - (-2)}{2 - 0} = -\frac{1}{2}\). Therefore, \(g(x) = -\frac{1}{2}x - 2\). 3. Line \(h\) crosses the y-axis at \((0, 3)\) and the x-axis at \((3, 0)\). Its slope is \(\frac{0 - 3}{3 - 0} = -1\). Therefore, \(h(x) = -x + 3\).

Answer

\(f(x) = \frac{3}{2}x + 1\), \(g(x) = -\frac{1}{2}x - 2\), and \(h(x) = -x + 3\).

Write linear equations 27

The line \(g\) shown in the coordinate plane passes through points \(P_1\) and \(P_2\). a) Use the coordinates of \(P_1\) and \(P_2\) to calculate the slope \(m\). b) Calculate the y-intercept \(b\). c) Write the complete function equation in the form \(g(x) = mx + b\).

Figure for problem 533201

Hints

- How can two points be used to calculate a slope? - Once you know the slope, substitute one point into \(y = mx + b\). - Pay attention to the scale when reading the point coordinates. - Check that your equation gives the correct y-value for both points.

Solution

1. The points are \(P_1(1, 5)\) and \(P_2(4, -1)\). 2. The slope is \(m = \frac{-1 - 5}{4 - 1} = -2\). 3. Substitute \(P_1(1, 5)\) into \(y = -2x + b\): \(5 = -2 \cdot 1 + b\), so \(b = 7\). 4. Therefore, \(g(x) = -2x + 7\).

Answer

a) \(m = -2\) b) \(b = 7\) c) \(g(x) = -2x + 7\)

Write linear equations 28

The coordinate plane shows three lines, \(f\), \(g\), and \(h\). Find the equation of each line.

Figure for problem 533212

Hints

- A horizontal line has slope \(0\) and an equation of the form \(y = c\). - For the other lines, notice whether each line rises or falls from left to right.

Solution

1. Line \(f\) is horizontal at \(y = 2\), so its slope is \(0\) and its equation is \(f(x) = 2\). 2. Line \(g\) has y-intercept \(1\). Moving \(1\) unit right corresponds to moving \(1\) unit down, so the slope is \(-1\). Thus \(g(x) = -x + 1\). 3. Line \(h\) has y-intercept \(-3\). Moving \(1\) unit right corresponds to moving \(1\) unit up, so the slope is \(1\). Thus \(h(x) = x - 3\).

Answer

\(f(x) = 2\), \(g(x) = -x + 1\), and \(h(x) = x - 3\).

Write linear equations 29

Consider the line \(g\) shown in the coordinate plane. a) Find the y-intercept \(b\). b) Find the slope \(m\). c) Write the equation in the form \(y = mx + b\).

Figure for problem 533257

Hints

- Find the point where the line crosses the y-axis. - Choose two grid points on the line. - Compare the vertical change with the horizontal change between those points.

Solution

1. The line crosses the y-axis at \((0, 1)\), so \(b = 1\). 2. Using \((0, 1)\) and \((2, 2)\), the slope is \(m = \frac{2 - 1}{2 - 0} = \frac{1}{2}\). 3. Therefore, the equation is \(g(x) = \frac{1}{2}x + 1\).

Answer

a) \(b = 1\) b) \(m = \frac{1}{2}\) c) \(g(x) = \frac{1}{2}x + 1\)

Write linear equations 30

Find the equation of each line \(f\) and \(h\) shown in the coordinate plane. Use the y-intercept and a slope triangle.

Figure for problem 533258

Hints

- Start by finding where each line crosses the y-axis. - Choose grid points with easy-to-read coordinates for the slope triangle. - Compare the vertical change with the horizontal change. - Use the direction of the line to determine the sign of the slope.

Solution

1. Line \(f\) crosses the y-axis at \((0, 4)\). Using \((0, 4)\) and \((1, 1)\), the slope is \(m = \frac{1 - 4}{1 - 0} = -3\). Therefore, \(f(x) = -3x + 4\). 2. Line \(h\) crosses the y-axis at \((0, -3)\). Using \((0, -3)\) and \((5, -1)\), the slope is \(m = \frac{-1 - (-3)}{5 - 0} = \frac{2}{5}\). Therefore, \(h(x) = \frac{2}{5}x - 3\).

Answer

\(f(x) = -3x + 4\) and \(h(x) = \frac{2}{5}x - 3\).

Write linear equations 31

Each line passes through the given point \(P\) and has the given slope \(m\). Find the y-intercept \(b\) and write the equation of each line. a) \(P(2, 5)\), \(m=-2\) b) \(P(-4, -1)\), \(m=\frac{1}{2}\) c) \(P(3, 0)\), \(m=\frac{4}{3}\) d) \(P(0, -2.5)\), \(m=-1.5\)

Hints

- Start with the form \(y=mx+b\). - Substitute the known slope and the coordinates of the given point. - Solve the resulting equation for \(b\). - When the x-coordinate is \(0\), the point's y-coordinate is the y-intercept.

Solution

1. For a), substitute \(P(2, 5)\) into \(y=-2x+b\): \(5=-2\cdot2+b\), so \(b=9\). The equation is \(y=-2x+9\). 2. For b), substitute \(P(-4, -1)\) into \(y=\frac{1}{2}x+b\): \(-1=\frac{1}{2}\cdot(-4)+b\), so \(b=1\). The equation is \(y=\frac{1}{2}x+1\). 3. For c), substitute \(P(3, 0)\) into \(y=\frac{4}{3}x+b\): \(0=\frac{4}{3}\cdot3+b\), so \(b=-4\). The equation is \(y=\frac{4}{3}x-4\). 4. For d), the point \(P(0, -2.5)\) is the y-intercept, so \(b=-2.5\). The equation is \(y=-1.5x-2.5\).

Answer

a) \(b=9\); \(y=-2x+9\) b) \(b=1\); \(y=\frac{1}{2}x+1\) c) \(b=-4\); \(y=\frac{4}{3}x-4\) d) \(b=-2.5\); \(y=-1.5x-2.5\)

Write linear equations 32

The graph shows four lines, \(h_1\) through \(h_4\). Find the equation of each line.

Figure for problem 533272

Hints

- Find a point where each line crosses the y-axis. - What is special about the equation of a line through the origin? - What is the slope of a horizontal line? - Use two points to calculate slope as change in y divided by change in x.

Solution

1. Line \(h_1\) passes through \((0, 0)\) and \((4, 1)\), so its slope is \(\frac{1}{4}\). Thus \(h_1(x) = \frac{1}{4}x\). 2. Line \(h_2\) has y-intercept \(4\) and passes through \((4, 2)\). Its slope is \(\frac{2 - 4}{4 - 0} = -\frac{1}{2}\). Thus \(h_2(x) = -\frac{1}{2}x + 4\). 3. Line \(h_3\) is horizontal at \(y = -3\), so \(h_3(x) = -3\). 4. Line \(h_4\) has y-intercept \(4\) and passes through \((-2, 0)\). Its slope is \(\frac{4 - 0}{0 - (-2)} = 2\). Thus \(h_4(x) = 2x + 4\).

Answer

\(h_1(x) = \frac{1}{4}x\), \(h_2(x) = -\frac{1}{2}x + 4\), \(h_3(x) = -3\), and \(h_4(x) = 2x + 4\).

Write linear equations 33

Two hikers, Alex and Bea, travel along a long straight road. Alex starts at mile marker \(0\) and walks at a constant speed toward higher mile markers. Bea starts at the same time at mile marker \(10\) and walks at a constant speed toward mile marker \(0\). On the graph, curve \(a\) represents Alex and curve \(b\) represents Bea. Write a function of the form \(s(t)=mt+b\) for each hiker on the displayed interval \(0\le t\le4\).

Figure for problem 533288

Hints

- The y-intercept is each hiker's starting position. - A positive slope means moving toward higher mile markers; a negative slope means moving toward lower mile markers. - Use another point on each line to calculate slope.

Solution

1. Alex's graph starts at \(0\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-0}{2-0}=3\), so \(s_A(t)=3t\), with \(0\le t\le4\). 2. Bea's graph starts at \(10\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-10}{2-0}=-2\), so \(s_B(t)=-2t+10\), with \(0\le t\le4\).

Answer

\(s_A(t)=3t\) and \(s_B(t)=-2t+10\), with \(0\le t\le4\) for both functions

Write linear equations 34

Find the equations of lines \(a\) and \(b\). Pay special attention to the direction of line \(a\).

Figure for problem 533525

Hints

- What is the slope of a line parallel to the x-axis? - For line \(b\), compare how the y-value changes as the x-value changes.

Solution

1. Line \(a\) is horizontal, so its slope is \(0\). It crosses the y-axis at \(-3\), so \(a(x) = -3\). 2. Line \(b\) crosses the y-axis at \(2\). Using \((0, 2)\) and \((-1, 5)\), the slope is \(m = \frac{5 - 2}{-1 - 0} = -3\). Therefore, \(b(x) = -3x + 2\).

Answer

Line \(a\): \(a(x) = -3\); line \(b\): \(b(x) = -3x + 2\)

Write linear equations 35

Two water tanks are observed at the same time. The graph shows volume \(V\), in gallons, as a function of time \(t\), in minutes. Curve \(a\) represents Tank 1, and curve \(b\) represents Tank 2. Write an equation of the form \(V(t)=mt+b\) for each tank, including the displayed domain.

Figure for problem 533528

Hints

- The y-intercept is the initial volume. - The slope is the change in gallons per minute. - Use two clear points on each graph. - Read each displayed time interval from the graph.

Solution

1. Tank 1 starts with \(10\) gallons, so \(b=10\). After \(5\) minutes it contains \(30\) gallons, so \(m=\frac{30-10}{5-0}=4\). Therefore, \(V_1(t)=4t+10\), with \(0\le t\le12\). 2. Tank 2 starts with \(50\) gallons, so \(b=50\). It is empty after \(10\) minutes, so \(m=\frac{0-50}{10-0}=-5\). Therefore, \(V_2(t)=-5t+50\), with \(0\le t\le10\).

Answer

Tank 1: \(V_1(t)=4t+10\), with \(0\le t\le12\); Tank 2: \(V_2(t)=-5t+50\), with \(0\le t\le10\)

Write linear equations 36

A taxi company charges a fixed starting fee plus a constant price per mile. The graph shows the total cost \(y\), in dollars, for a trip of \(x\) miles. Find the equation of the line.

Figure for problem 533877

Hints

- Where does the line cross the vertical axis? That value represents the starting fee. - How much does the cost increase for each additional mile? - Use the form \(y = mx + b\).

Solution

1. The line crosses the y-axis at \(4\), so the starting fee is \(\$4\). 2. Using \((0, 4)\) and \((3, 10)\), the slope is \(m = \frac{10 - 4}{3 - 0} = 2\). This means the taxi charges \(\$2\) per mile. 3. Therefore, the equation is \(y = 2x + 4\).

Answer

\(y = 2x + 4\)

Write linear equations 37

A scooter rental costs \(y=6x+4\) dollars for \(x\) hours of riding. Let \(m\) be the riding time in minutes. a) Write the cost formula using \(m\) instead of \(x\). b) Check that both formulas give the same cost for a \(90\)-minute ride.

Hints

- How do you express the number of hours using a time in minutes? - Substitute that expression for the original input letter. - Compare the same trip duration in both units when checking.

Solution

1. One hour is \(60\) minutes, so \(x=\frac{m}{60}\). 2. Substitute: \(y=6\left(\frac{m}{60}\right)+4=\frac{m}{10}+4\). 3. For \(90\) minutes, \(x=1.5\) hours. The first formula gives \(6(1.5)+4=13\) dollars, and the new one gives \(\frac{90}{10}+4=13\) dollars.

Answer

a) \(y=\frac{m}{10}+4\) b) Both formulas give \(\$13\).

Write linear equations 38

An art studio charges a sign-up fee plus the same price for each class. Three classes cost \(\$22\) in total, and seven classes cost \(\$34\). Write a linear function \(C(n)\) for the cost of \(n\) classes. Find the cost per class and the sign-up fee.

Hints

- How many extra classes account for the increase in total cost? - Once you know the price per class, which part of the cost remains fixed? - Test your equation with both class counts.

Solution

1. The total goes up by \(34-22=12\) dollars for \(7-3=4\) more classes, so the rate is \(12\div4=3\) dollars per class. 2. Write \(C(n)=3n+b\). From the three-class total, \(22=3\cdot3+b\), so \(b=13\). 3. Therefore \(C(n)=3n+13\). It gives \(C(7)=34\), as required.

Answer

\(C(n)=3n+13\). Cost per class: \(\$3\). Sign-up fee: \(\$13\).

Write linear equations 39

A candle burns at a constant rate. After \(2\) hours, it is \(25\,\text{cm}\) tall. After \(6\) hours, it is \(17\,\text{cm}\) tall. Write a linear function \(L(t)\) for the candle's height, in centimeters, after \(t\) hours.

Hints

- Treat the two measurements as points whose coordinates include time and height. - First find the constant change in height per hour. - Then use either measurement to determine the starting-value term of the model.

Solution

1. The slope is \(\frac{17-25}{6-2}=-2\,\text{cm/h}\). 2. Use \(L(t)=-2t+b\) and the point \((2,25)\): \(25=-2(2)+b\), so \(b=29\). 3. Therefore, \(L(t)=-2t+29\).

Answer

\(L(t)=-2t+29\)

Write linear equations 40

A savings account is modeled over a short period by the table below, where \(w\) is the number of weeks and \(B\) is the balance in dollars. <table><tr><td>\(w\)</td><td>\(0\)</td><td>\(2\)</td><td>\(5\)</td><td>\(7\)</td></tr><tr><td>\(B\)</td><td>\(18\)</td><td>\(24\)</td><td>\(33\)</td><td>\(39\)</td></tr></table> Determine whether the data are linear. If they are, write a linear function \(B(w)\) that matches the table.

Hints

- Because the input gaps are not all the same, compare change in output per unit of input rather than raw output differences. - A linear table has the same rate of change across every interval. - If the data are linear, use the row with input zero to identify the constant term directly.

Solution

1. From week \(0\) to week \(2\), the balance changes by \(6\) dollars in \(2\) weeks, or \(3\) dollars per week. 2. From week \(2\) to week \(5\), the balance changes by \(9\) dollars in \(3\) weeks, also \(3\) dollars per week. From week \(5\) to week \(7\), it changes by \(6\) dollars in \(2\) weeks, again \(3\) dollars per week. 3. The constant rate shows the data are linear. Since \(B(0)=18\), the model is \(B(w)=3w+18\).

Answer

The data are linear, and \(B(w)=3w+18\).

Write linear equations 41

Write the equation of the line in slope-intercept form. Use its crossing of the y-axis and the marked point to find the slope.

Figure for problem 560230

Hints

- Where does the line cross the y-axis? - Use the two marked points to calculate rise divided by run.

Solution

1. The line crosses the y-axis at \((0,-1)\), so \(b=-1\). 2. From \((0,-1)\) to \((2,3)\), the rise is \(4\) and the run is \(2\), so \(m=2\). 3. The equation is \(y=2x-1\).

Answer

\(y=2x-1\).

Write linear equations 42

Write the equation of the line from the graph in slope-intercept form. Use the points where it crosses the two axes.

Figure for problem 560231

Hints

- Read each coordinate where the line meets an axis. - One intercept gives the constant term; both intercepts give the slope.

Solution

1. The line crosses the axes at \((0,4)\) and \((3,0)\). 2. Its slope is \(m=\frac{0-4}{3-0}=-\frac{4}{3}\). 3. Its y-intercept is \(4\), so the equation is \(y=-\frac{4}{3}x+4\).

Answer

\(y=-\frac{4}{3}x+4\).

Write linear equations 43

A line has slope \(-2\) and passes through \((-1,4)\). Write its equation first in point-slope form and then in slope-intercept form.

Hints

- Be careful when subtracting a negative x-coordinate in point-slope form. - Expand the product before adding the fixed number to both sides. - Check your result with the original point.

Solution

1. Substitute the point and slope into point-slope form: \(y-4=-2(x+1)\). 2. Expand and isolate \(y\): \(y-4=-2x-2\), so \(y=-2x+2\). 3. Check the point: \(-2(-1)+2=4\).

Answer

Point-slope: \(y-4=-2(x+1)\). Slope-intercept: \(y=-2x+2\).

Write linear equations 44

A student writes \(y=2x+1\) for a line with slope \(2\) that passes through \((3,1)\). Explain why the equation is wrong, then write the correct equation.

Hints

- Does the student's equation give the stated y-coordinate when \(x=3\)? - Keep the stated slope and find a different constant term.

Solution

1. The student's equation has slope \(2\), but substituting \(x=3\) gives \(y=7\), not \(1\). 2. Write \(y=2x+b\). Substituting \((3,1)\) gives \(1=6+b\), so \(b=-5\). 3. The correct equation is \(y=2x-5\).

Answer

The student's line does not pass through \((3,1)\). The correct equation is \(y=2x-5\).

Write linear equations 45

The points in this table lie on a line. <table><tr><th>\(x\)</th><td>\(-2\)</td><td>\(0\)</td><td>\(3\)</td></tr><tr><th>\(y\)</th><td>\(9\)</td><td>\(5\)</td><td>\(-1\)</td></tr></table> Find the slope, write the equation in slope-intercept form, and check it with the third point.

Hints

- Look for two columns that make it easy to find vertical change over horizontal change. - Which row of the table tells you the value when the input is zero? - Check the rule with the column you did not use to find the slope.

Solution

1. Between \((-2,9)\) and \((0,5)\), the slope is \(\frac{5-9}{0-(-2)}=-2\). 2. The table shows \((0,5)\), so the y-intercept is \(5\) and the equation is \(y=-2x+5\). 3. At \(x=3\), the rule gives \(y=-2(3)+5=-1\), matching the third point.

Answer

Slope \(-2\); equation \(y=-2x+5\). At \(x=3\), \(y=-1\), as in the table.

Write linear equations 46

A linear function \(p\) has the form \(p(x) = 3x + b\). Its graph has an x-intercept at \(x = -2\). a) Find the value of \(b\). b) Let \(P(1, p(1))\). Compared with the graph of \(q(x) = -2x + 5\), is \(P\) above the graph, below the graph, or on the graph?

Hints

- What is the function value at an x-intercept? - Substitute the given x-value into \(p\) to find the y-coordinate of \(P\). - Compare the two functions at the same x-value.

Solution

1. Since the x-intercept is at \(x = -2\), \(p(-2) = 0\). Thus \(3 \cdot (-2) + b = 0\), so \(b = 6\). 2. Then \(p(1) = 3 \cdot 1 + 6 = 9\), so \(P = (1, 9)\). 3. Evaluate the other function at the same x-value: \(q(1) = -2 \cdot 1 + 5 = 3\). 4. Because \(9 > 3\), \(P\) is above the graph of \(q\).

Answer

a) \(b = 6\) b) \(P(1, 9)\) is above the graph of \(q\).

Write linear equations 47

A linear function \(g\) has the form \(g(x) = mx + b\) and passes through \(P(0, 6)\). In the first quadrant, the graph of \(g\) and the coordinate axes enclose a triangle with area \(15\) square units. a) Find the x-intercept of \(g\). b) Find the equation of \(g\).

Hints

- What information does the point on the y-axis give you about the linear equation? - Use the area formula for a right triangle to relate the two intercepts. - Once you know both intercepts, how can you find the slope?

Solution

1. Since \(P(0, 6)\) is on the y-axis, the y-intercept is \(b = 6\). 2. Let the x-intercept be \((x_0, 0)\). The triangle has base \(x_0\) and height \(6\), so \(15 = \frac{1}{2} \cdot x_0 \cdot 6\). Thus \(x_0 = 5\). 3. Use the points \((0, 6)\) and \((5, 0)\) to find the slope: \(m = \frac{0 - 6}{5 - 0} = -1.2\). 4. Therefore, \(g(x) = -1.2x + 6\).

Answer

a) The x-intercept is \((5, 0)\). b) \(g(x) = -1.2x + 6\)

Write linear equations 48

The graph of a linear function \(h\) passes through \(P(2, 1)\). From any point on the line, moving \(5\) units left and \(2\) units up reaches another point on the line. Find the slope \(m\), the y-intercept \(b\), and the equation of \(h\).

Hints

- Moving left means the change in \(x\) is negative. - Once you know the slope, substitute the coordinates of the given point into slope-intercept form. - Solve the resulting equation for the y-intercept.

Solution

1. Moving \(5\) units left gives \(\Delta x = -5\), and moving \(2\) units up gives \(\Delta y = 2\). Thus \(m = \frac{2}{-5} = -0.4\). 2. Substitute \((2, 1)\) into \(y = mx + b\): \(1 = -0.4 \cdot 2 + b\). 3. Solving gives \(b = 1.8\). 4. Therefore, \(h(x) = -0.4x + 1.8\).

Answer

The slope is \(m = -0.4\), the y-intercept is \(b = 1.8\), and \(h(x) = -0.4x + 1.8\).

Write linear equations 49

For each condition, write the requested linear equation. a) The graph of \(h\) is parallel to \(p(x) = -4x + 10\) and has the same y-intercept as \(q(x) = 0.5x - 2\). Find \(h(x)\). b) Choose \(m\) and \(b\) so that \(h(x) = mx + b\) is a horizontal line through \(P(5, 3)\). Then write \(h(x)\). c) Start with \(f(x) = x\). Double its slope, then translate the graph down \(3\) units. Write the new function \(h(x)\).

Hints

- What must be equal when two lines are parallel? - Does the y-value change along a horizontal line? - Apply the two changes in part c one at a time to the slope and y-intercept.

Solution

1. Parallel lines have the same slope, so part a has slope \(-4\). The y-intercept of \(q\) is \(-2\). Therefore, \(h(x) = -4x - 2\). 2. A horizontal line has slope \(m = 0\). Passing through \((5, 3)\) makes its constant value \(3\), so \(b = 3\) and \(h(x) = 3\). 3. The slope of \(f(x) = x\) is \(1\). Doubling it gives slope \(2\). Translating down \(3\) units gives y-intercept \(-3\), so \(h(x) = 2x - 3\).

Answer

a) \(h(x) = -4x - 2\) b) \(m = 0\), \(b = 3\), and \(h(x) = 3\) c) \(h(x) = 2x - 3\)

Write linear equations 50

The line \(g\) has equation \(y = 1.5x - 2\). Write equations for lines \(h\), \(i\), and \(j\) that satisfy these conditions: - \(h\) is parallel to \(g\) and has y-intercept \(4\). - \(i\) has the same slope as \(g\) and passes through \(P(2, 1)\). - \(j\) has the same y-intercept as \(g\) and is parallel to the x-axis. Which of the lines \(g\), \(h\), \(i\), and \(j\) are distinct parallel lines, and which are the same line?

Hints

- What is the slope of a line parallel to the x-axis? - How can you use a point and a known slope to find the y-intercept? - What must match for two linear equations to represent the same line?

Solution

1. Line \(h\) has slope \(1.5\) and y-intercept \(4\), so \(h: y = 1.5x + 4\). 2. For \(i\), use slope \(1.5\) and point \((2, 1)\): \(1 = 1.5 \cdot 2 + b\), so \(b = -2\). Thus \(i: y = 1.5x - 2\). 3. Line \(j\) is horizontal, so its slope is \(0\). Its y-intercept is \(-2\), so \(j: y = -2\). 4. Lines \(g\) and \(i\) have the same equation, so they are the same line. Line \(h\) has the same slope but a different y-intercept, so it is parallel to both \(g\) and \(i\). Line \(j\) has a different slope and is not parallel to them.

Answer

\(h: y = 1.5x + 4\), \(i: y = 1.5x - 2\), and \(j: y = -2\). Lines \(g\) and \(i\) are the same line. Line \(h\) is parallel to \(g\) and \(i\). Line \(j\) is not parallel to any of the other three.

Write linear equations 51

Consider the four line equations: 1) \(y = \frac{1}{3}x + 2\) 2) \(y = 0.3x + 2\) 3) \(3y = x + 12\) 4) \(y - 2 = \frac{1}{3}x\). Decide whether each statement is true or false, and briefly justify your answer. a) All four lines have y-intercept \((0, 2)\). b) Exactly two of the equations represent the same line. c) Lines (1), (3), and (4) have the same slope.

Hints

- Rewrite every equation in slope-intercept form before comparing them. - Be careful: \(0.3\) is not equal to \(\frac{1}{3}\). - Equal slopes make lines parallel; equal slopes and equal y-intercepts make them identical.

Solution

1. Rewrite each equation in slope-intercept form. Equation (1) is \(y = \frac{1}{3}x + 2\). Equation (2) is \(y = 0.3x + 2\), and \(0.3 \neq \frac{1}{3}\). Equation (3) becomes \(y = \frac{1}{3}x + 4\). Equation (4) becomes \(y = \frac{1}{3}x + 2\). 2. Statement a) is false. Equations (1), (2), and (4) have y-intercept \(2\), while equation (3) has y-intercept \(4\). 3. Statement b) is true. Equations (1) and (4) represent the same line. 4. Statement c) is true. Equations (1), (3), and (4) all have slope \(\frac{1}{3}\). Equations (1) and (4) are identical, while line (3) is parallel to them.

Answer

a) False. Line (3) has y-intercept \((0, 4)\), not \((0, 2)\). b) True. Equations (1) and (4) represent the same line. c) True. Lines (1), (3), and (4) all have slope \(\frac{1}{3}\).

Write linear equations 52

A linear function \(g\) has graph passing through \(P(-1,7)\) and \(Q(2,-2)\). a) Construct the function in the form \(g(x)=mx+b\). b) Use your function to determine whether \(R(0,5)\) and \(S(1,1)\) are input-output pairs of \(g\).

Hints

- A linear function through two points is determined by its slope and one point. - After finding the slope, use one given point to determine the initial value \(b\). - Only after the function is constructed should you test the proposed ordered pairs.

Solution

1. The slope through \(P\) and \(Q\) is \(m=\frac{-2-7}{2-(-1)}=-3\). 2. Substitute \(P(-1,7)\) into \(g(x)=-3x+b\): \(7=-3(-1)+b\), so \(b=4\). 3. Therefore, \(g(x)=-3x+4\). 4. \(g(0)=4\), not \(5\), so \(R(0,5)\) is not an input-output pair. 5. \(g(1)=1\), so \(S(1,1)\) is an input-output pair.

Answer

a) \(g(x)=-3x+4\) b) \(R(0,5)\) is not an input-output pair; \(S(1,1)\) is.

Write linear equations 53

A line \(g\) passes through \(P(4, -1)\) and has slope \(m = -0.75\). Temuulen wants to find more points on the line using slope triangles. a) Starting at \(P\), Temuulen moves \(8\,\text{units}\) left. How many units must Temuulen move up or down to land on the line again? b) What is the vertical change when Temuulen moves exactly \(1\,\text{unit}\) right? c) Find an equation for line \(g\).

Hints

- Relate slope, horizontal change, and vertical change. - Use the sign of the slope together with the direction of the horizontal move. - Use the given point to determine the intercept in a linear equation with the known slope.

Solution

1. For part a, \(\Delta x = -8\). Since \(\Delta y = m\Delta x\), \(\Delta y = -0.75 \cdot (-8) = 6\). Temuulen moves \(6\,\text{units}\) up. 2. For part b, \(\Delta x = 1\), so \(\Delta y = -0.75 \cdot 1 = -0.75\). Temuulen moves \(0.75\,\text{unit}\) down. 3. Write \(g(x) = -0.75x + b\). Using \(P(4, -1)\), \(-1 = -0.75(4) + b\), so \(b = 2\). Therefore, \(g(x) = -0.75x + 2\).

Answer

a) \(6\,\text{units}\) up b) \(\Delta y = -0.75\), so \(0.75\,\text{unit}\) down c) \(g(x) = -0.75x + 2\)

Write linear equations 54

A linear function \(p\) has slope \(m = -\frac{2}{3}\), and its graph passes through \(A(3, 1)\). a) Find the equation of \(p\). b) Give the coordinates of another point \(B\) on the graph whose coordinates are both integers. c) Use a calculation to determine whether \(C(-6, 7)\) lies on the graph of \(p\).

Hints

- Substitute the given point into slope-intercept form to find the y-intercept. - Which x-values make a fraction with denominator \(3\) especially convenient? - To test a point, substitute its x-coordinate and compare the result with its y-coordinate.

Solution

1. Write \(p(x) = -\frac{2}{3}x + b\). Substitute \((3, 1)\): \(1 = -\frac{2}{3} \cdot 3 + b = -2 + b\), so \(b = 3\). Therefore, \(p(x) = -\frac{2}{3}x + 3\). 2. Choose an x-value that makes \(-\frac{2}{3}x\) an integer. For example, \(p(0) = 3\), so \(B(0, 3)\) is one possible point. 3. Check \(C\): \(p(-6) = -\frac{2}{3} \cdot (-6) + 3 = 7\). This matches the y-coordinate of \(C\), so \(C\) lies on the graph.

Answer

a) \(p(x) = -\frac{2}{3}x + 3\) b) Answers will vary. One possible point is \(B(0, 3)\). c) Yes, because \(p(-6) = 7\).

Write linear equations 55

A line passes through \(C(0, 4)\) and \(D(5, 2)\). a) Find the equation of the line. b) Find the x-intercept.

Hints

- What does a point of the form \((0, y)\) tell you about the equation? - How can you use the two points to find the slope? - What is the y-coordinate at the x-intercept?

Solution

1. Since \(C(0, 4)\) is on the y-axis, the y-intercept is \(b = 4\). 2. The slope is \(m = \frac{2 - 4}{5 - 0} = -\frac{2}{5}\). 3. The equation is \(y = -\frac{2}{5}x + 4\). 4. For the x-intercept, set \(y = 0\): \(0 = -\frac{2}{5}x + 4\), so \(x = 10\).

Answer

a) \(y = -\frac{2}{5}x + 4\) b) \((10, 0)\)

Write linear equations 56

The points \(P(1, 2)\) and \(Q(3, 3)\) lie in the coordinate plane. a) Write the equation of the line through \(P\) and \(Q\). b) Use a calculation to determine whether \(R(7, 5)\) also lies on the line. Explain your answer.

Hints

- Find the slope from \(P\) and \(Q\), then use one point to find the y-intercept. - To test \(R\), substitute its x-coordinate into the equation and compare the result with its y-coordinate.

Solution

1. Find the slope: \(m = \frac{3 - 2}{3 - 1} = \frac{1}{2}\). 2. Substitute \((1, 2)\) into \(y = mx + b\): \(2 = \frac{1}{2} \cdot 1 + b\), so \(b = \frac{3}{2}\). 3. The equation is \(y = \frac{1}{2}x + \frac{3}{2}\). 4. Check \(R\): when \(x = 7\), \(y = \frac{1}{2} \cdot 7 + \frac{3}{2} = 5\). This matches the y-coordinate of \(R\), so \(R\) lies on the line.

Answer

a) \(y = \frac{1}{2}x + \frac{3}{2}\) b) Yes. Substituting \(x = 7\) gives \(y = 5\), so \(R(7, 5)\) lies on the line.

Write linear equations 57

Line \(f\) passes through \(A(-4, 1)\) and \(B(2, 4)\). Find the equation of a line \(g\) that is parallel to \(f\) and passes through \(P(2, -2)\).

Hints

- What must be true about the slopes of two parallel lines? - How do you calculate a slope when two points on a line are known? - What is the general slope-intercept form of a linear equation? - How can a given point be used to determine the missing part of the equation?

Solution

1. Find the slope of \(f\): \(m = \frac{4 - 1}{2 - (-4)} = \frac{3}{6} = \frac{1}{2}\). 2. Since \(g\) is parallel to \(f\), it also has slope \(\frac{1}{2}\). 3. Substitute \((2, -2)\) into \(y = \frac{1}{2}x + b\): \(-2 = \frac{1}{2} \cdot 2 + b\), so \(b = -3\). 4. Therefore, \(g\) has equation \(y = \frac{1}{2}x - 3\).

Answer

\(y = \frac{1}{2}x - 3\)

Write linear equations 58

Lines \(g_1\) and \(g_2\) are parallel. Line \(g_1\) passes through \(C(-2, 7)\) and \(D(1, -2)\). Line \(g_2\) has y-intercept \(1\) and passes through \(E(3, y_E)\). a) Find the equation of \(g_2\). b) Find \(y_E\).

Hints

- First find the slope of \(g_1\). - Use the fact that parallel lines have the same slope. - Once you know the slope and y-intercept of \(g_2\), write its equation. - Substitute the x-coordinate of \(E\) to find its y-coordinate.

Solution

1. Find the slope of \(g_1\): \(m = \frac{-2 - 7}{1 - (-2)} = \frac{-9}{3} = -3\). 2. Since \(g_2\) is parallel to \(g_1\), it also has slope \(-3\). 3. With y-intercept \(1\), \(g_2\) has equation \(y = -3x + 1\). 4. Substitute \(x = 3\): \(y_E = -3 \cdot 3 + 1 = -8\).

Answer

a) \(y = -3x + 1\) b) \(y_E = -8\)

Write linear equations 59

A linear function \(f\) has slope \(m = -2\), and its graph passes through \(A(4, 6)\). a) Find the equation of \(f\). b) Find the x-intercept of the graph.

Hints

- How can you find the y-intercept when you know the slope and one point? - What is the y-coordinate at the x-intercept?

Solution

1. Substitute \(m = -2\) and \((4, 6)\) into \(y = mx + b\): \(6 = -2 \cdot 4 + b\), so \(b = 14\). 2. Therefore, \(f(x) = -2x + 14\). 3. For the x-intercept, set \(f(x) = 0\): \(0 = -2x + 14\), so \(x = 7\).

Answer

a) \(f(x) = -2x + 14\) b) \((7, 0)\)

Write linear equations 60

Line \(g\) passes through the origin and has slope \(3\). Line \(h\) has the same slope but is shifted 4 units down from \(g\). 1. Write equations for \(g\) and \(h\). 2. A third line \(k\) keeps the y-intercept of \(h\), but its slope is the slope of \(h\) multiplied by \(-\frac{1}{3}\). Find an equation for \(k\). 3. Find the coordinates of the intersection of \(g\) and \(k\).

Hints

- What is the slope-intercept equation of a line through the origin? - How does shifting a graph downward change its y-intercept? - To find the intersection, set the two function expressions equal.

Solution

1. Since \(g\) has slope \(3\) and passes through the origin, \(g(x) = 3x\). Shifting it down 4 units gives \(h(x) = 3x - 4\). 2. The new slope is \(3 \cdot \left(-\frac{1}{3}\right) = -1\), and the y-intercept remains \(-4\). Thus, \(k(x) = -x - 4\). 3. At the intersection, \(3x = -x - 4\). Solving gives \(4x = -4\), so \(x = -1\). Then \(y = 3 \cdot (-1) = -3\). The intersection is \((-1, -3)\).

Answer

1. \(g(x) = 3x\) and \(h(x) = 3x - 4\) 2. \(k(x) = -x - 4\) 3. \((-1, -3)\)

Write linear equations 61

The table represents a linear function \(f\). <table> <tbody> <tr><td>\(x\)</td><td>\(-2\)</td><td>\(0\)</td><td>\(3\)</td></tr> <tr><td>\(f(x)\)</td><td>\(-7\)</td><td>\(-3\)</td><td>?</td></tr> </tbody> </table> a) Find the equation of \(f\) and the missing value when \(x = 3\). b) Use a calculation to determine whether \(P(10, 17)\) lies on the graph of \(f\).

Hints

- What does the table tell you about the y-intercept when \(x = 0\)? - How can you find the slope from two table entries? - To test a point, substitute its x-coordinate into the function and compare the result with its y-coordinate.

Solution

1. From the table, \(f(0) = -3\), so the y-intercept is \(b = -3\). 2. Using \((-2, -7)\) and \((0, -3)\), the slope is \(m = \frac{-3 - (-7)}{0 - (-2)} = 2\). Thus \(f(x) = 2x - 3\). 3. The missing value is \(f(3) = 2 \cdot 3 - 3 = 3\). 4. Check \(P\): \(f(10) = 2 \cdot 10 - 3 = 17\), so \(P(10, 17)\) lies on the graph.

Answer

a) \(f(x) = 2x - 3\), and \(f(3) = 3\). b) Yes. Since \(f(10) = 17\), \(P(10, 17)\) lies on the graph.

Write linear equations 62

The table is intended to represent a linear function. <table> <tbody> <tr><td>\(x\)</td><td>\(2\)</td><td>\(5\)</td><td>\(11\)</td></tr> <tr><td>\(y\)</td><td>\(4\)</td><td>\(5.5\)</td><td>\(y_3\)</td></tr> </tbody> </table> a) What value must \(y_3\) have so that the relationship is linear? b) A new linear function \(h\) has twice the slope of the original function and still passes through \((2, 4)\). Find the equation of \(h\).

Hints

- A linear function has a constant rate of change. Use the first two points to find it. - After doubling the slope, use the point \((2, 4)\) to find the new y-intercept.

Solution

1. From \((2, 4)\) and \((5, 5.5)\), the slope of the original function is \(m = \frac{5.5 - 4}{5 - 2} = 0.5\). 2. From \(x = 5\) to \(x = 11\), \(x\) increases by \(6\), so \(y\) increases by \(0.5 \cdot 6 = 3\). Thus \(y_3 = 8.5\). 3. The new slope is \(2 \cdot 0.5 = 1\). 4. Since \(h\) passes through \((2, 4)\), \(4 = 1 \cdot 2 + b\), so \(b = 2\). Therefore, \(h(x) = x + 2\).

Answer

a) \(y_3 = 8.5\) b) \(h(x) = x + 2\)

Write linear equations 63

A line passes through \(A(-1, 7)\) and \(B(2, -2)\). a) Find the equation of the line in slope-intercept form. b) A point \(C(x, -11)\) also lies on the line. Find the missing x-coordinate.

Hints

- How do you find slope from two points? - After finding the slope, substitute one point to find the y-intercept. - A point on the line must satisfy the line's equation.

Solution

1. Find the slope: \(m = \frac{-2 - 7}{2 - (-1)} = \frac{-9}{3} = -3\). 2. Substitute \((2, -2)\) into \(y = -3x + b\): \(-2 = -3 \cdot 2 + b\), so \(b = 4\). Thus the equation is \(y = -3x + 4\). 3. For \(C\), substitute \(y = -11\): \(-11 = -3x + 4\). Solving gives \(x = 5\).

Answer

a) \(y = -3x + 4\) b) \(x = 5\)

Write linear equations 64

Line \(g\) passes through \(P(0, 3)\) and \(Q(4, 5)\). A second line \(h\) is parallel to \(g\) and passes through \(R(2, 1)\). a) Find the equations of \(g\) and \(h\). b) Find the x-intercept of \(h\).

Hints

- What must be true about the slopes of parallel lines? - Which given point tells you the y-intercept of \(g\) immediately? - What is the y-coordinate at an x-intercept?

Solution

1. For \(g\), the y-intercept is \(3\) and the slope is \(m = \frac{5 - 3}{4 - 0} = \frac{1}{2}\). Thus \(g(x) = \frac{1}{2}x + 3\). 2. Since \(h\) is parallel to \(g\), it also has slope \(\frac{1}{2}\). Substitute \((2, 1)\): \(1 = \frac{1}{2} \cdot 2 + b\), so \(b = 0\). Therefore, \(h(x) = \frac{1}{2}x\). 3. Set \(h(x) = 0\): \(0 = \frac{1}{2}x\), so \(x = 0\).

Answer

a) \(g(x) = \frac{1}{2}x + 3\) and \(h(x) = \frac{1}{2}x\) b) \((0, 0)\)

Write linear equations 65

Eleni tries to find the equation of the line through \(C(2, 5)\) and \(D(4, 1)\) and calculates the slope as \(m = \frac{4 - 2}{1 - 5} = \frac{2}{-4} = -0.5\). Explain Eleni's error, then find the correct equation of the line.

Hints

- Check which coordinate difference belongs in the numerator of the slope ratio. - Keep the point order consistent in both coordinate differences. - After finding the correct slope, use one given point to determine the remaining constant.

Solution

1. Eleni reversed the slope ratio. Slope is change in \(y\) divided by change in \(x\), not change in \(x\) divided by change in \(y\). 2. The correct slope is \(m = \frac{1 - 5}{4 - 2} = \frac{-4}{2} = -2\). 3. Substitute \((2, 5)\) into \(y = -2x + b\): \(5 = -2 \cdot 2 + b\), so \(b = 9\). 4. Therefore, the correct equation is \(y = -2x + 9\).

Answer

Eleni divided the change in \(x\) by the change in \(y\). The correct slope is \(-2\), so the line is \(y = -2x + 9\).

Write linear equations 66

The points \(A(1,4)\), \(B(3,10)\), and \(C(x,19)\) lie on the same line. a) Write the equation of the line through \(A\) and \(B\) in slope-intercept form. b) Use your equation to find the x-coordinate of \(C\).

Hints

- Use \(A\) and \(B\) to determine the slope. - After finding the slope, use either point to determine the y-intercept. - Write the full line equation before substituting the y-coordinate of \(C\).

Solution

1. The slope through \(A\) and \(B\) is \(m=\frac{10-4}{3-1}=3\). 2. Substitute \(A(1,4)\) into \(y=3x+b\): \(4=3+b\), so \(b=1\). The line is \(y=3x+1\). 3. Since \(C\) has y-coordinate \(19\), solve \(19=3x+1\). This gives \(x=6\).

Answer

a) \(y=3x+1\) b) \(x=6\)

Write linear equations 67

The points \(S(-2, -7)\), \(T(0, -1)\), \(U(1, 2)\), and \(V(3, 7)\) are given. Three of the points lie on the same line, and one does not. Identify the point that is not on the line, and find the equation of the line through the other three points.

Hints

- Compare slopes between pairs of points. - A change in slope indicates that one of the points is not collinear with the others. - Which point gives the y-intercept immediately? - Use two points that are on the same line to write its equation.

Solution

1. Compare slopes: \(m_{ST} = \frac{-1 - (-7)}{0 - (-2)} = 3\), and \(m_{TU} = \frac{2 - (-1)}{1 - 0} = 3\). Thus \(S\), \(T\), and \(U\) are collinear. 2. The slope from \(U\) to \(V\) is \(m_{UV} = \frac{7 - 2}{3 - 1} = \frac{5}{2}\), which is not \(3\). Therefore, \(V\) is not on the line. 3. Since \(T(0, -1)\) is on the y-axis, the y-intercept is \(-1\). With slope \(3\), the equation is \(y = 3x - 1\).

Answer

\(V(3, 7)\) is not on the line. The other three points lie on \(y = 3x - 1\).

Write linear equations 68

A line has slope \(m=-3\) and passes through \(P(2,-4)\). a) Write its equation in slope-intercept form. b) Another point \(Q(x,8)\) lies on the line. Use your equation to find the x-coordinate of \(Q\).

Hints

- Start with slope-intercept form using the given slope. - Substitute the known point to determine the y-intercept. - Only after writing the equation should you substitute the y-coordinate of \(Q\).

Solution

1. Write \(y=-3x+b\) and substitute \(P(2,-4)\): \(-4=-6+b\), so \(b=2\). 2. The line is \(y=-3x+2\). 3. For \(Q(x,8)\), solve \(8=-3x+2\). Then \(6=-3x\), so \(x=-2\).

Answer

a) \(y=-3x+2\) b) \(x=-2\)

Write linear equations 69

A technician records three data points: \(A(10, 25)\), \(B(20, 45)\), and \(C(30, 62)\). a) Find the equation of the line \(g\) that passes exactly through \(A\) and \(B\). b) Use a calculation to determine whether \(C\) also lies on \(g\). c) The technician wants to adjust the slope while keeping the line through \(A\) so that the line is closer to \(C\). Should the slope be increased or decreased? Explain without writing a new equation.

Hints

- Find the slope from \(A\) and \(B\), then use one point to find the y-intercept. - A point lies on the line only if its coordinates satisfy the equation. - Imagine rotating the line around \(A\). What happens to its height at \(x = 30\) when the slope becomes smaller?

Solution

1. The slope through \(A\) and \(B\) is \(m = \frac{45 - 25}{20 - 10} = 2\). 2. Substitute \((10, 25)\) into \(y = 2x + b\): \(25 = 2 \cdot 10 + b\), so \(b = 5\). Thus \(g(x) = 2x + 5\). 3. Evaluate at \(x = 30\): \(g(30) = 65\). Since \(65 \ne 62\), point \(C\) is not on the line. 4. At \(x = 30\), the line through \(A\) and \(B\) is above \(C\). Keeping \(A\) fixed, decreasing the slope lowers the line to the right of \(A\), so the slope should be decreased.

Answer

a) \(g(x) = 2x + 5\) b) No. \(g(30) = 65\), not \(62\). c) Decrease the slope. With \(A\) fixed, a smaller slope gives a lower y-value at \(x = 30\), moving the line toward \(C\).

Write linear equations 70

Two linear functions, \(f\) and \(g\), have the same x-intercept. The function \(f\) is given by \(f(x) = -\frac{1}{2}x + 4\). The graph of \(g\) passes through \(P(0, -4)\). Find an equation for \(g\), and explain without calculation why the slope of \(g\) must be positive.

Hints

- What does it mean geometrically for two functions to have the same zero? - What does a point of the form \((0, y)\) tell you about \(y = mx + b\)? - Compare the y-coordinates of the two known points as x increases. Does the line rise or fall?

Solution

1. Find the zero of \(f\): \(-\frac{1}{2}x + 4 = 0\). Then \(\frac{1}{2}x = 4\), so \(x = 8\). 2. Since \(g\) has the same x-intercept, its graph passes through \((8, 0)\). 3. The point \(P(0, -4)\) gives the y-intercept, so \(b = -4\). 4. Use \((8, 0)\) in \(g(x) = mx - 4\): \(0 = 8m - 4\). Thus \(8m = 4\), so \(m = 0.5\). 5. Therefore, \(g(x) = 0.5x - 4\). 6. The slope must be positive because the line goes from the negative y-intercept \((0, -4)\) up to the x-axis at the positive x-value \(8\).

Answer

\(g(x) = 0.5x - 4\). The slope must be positive because the line rises from \((0, -4)\) to its x-intercept at \((8, 0)\).

Write linear equations 71

A linear function \(f\) passes through \(A(-2, 7)\) and \(B(2, -1)\). a) Find the equation of \(f\). b) Find the missing values in the table. <table> <tr><td>\(x\)</td><td>\(-2\)</td><td>\(2\)</td><td>\(5\)</td><td>?</td></tr> <tr><td>\(f(x)\)</td><td>\(7\)</td><td>\(-1\)</td><td>?</td><td>\(10\)</td></tr> </table>

Hints

- Use the slope formula with the two given points. - Pay close attention to signs when subtracting negative coordinates. - After finding the slope, substitute one point to find the y-intercept.

Solution

1. Find the slope: \(m = \frac{-1 - 7}{2 - (-2)} = \frac{-8}{4} = -2\). 2. Substitute \((2, -1)\) into \(y = -2x + b\): \(-1 = -2 \cdot 2 + b\), so \(b = 3\). Thus \(f(x) = -2x + 3\). 3. For \(x = 5\), \(f(5) = -2 \cdot 5 + 3 = -7\). 4. For \(f(x) = 10\), solve \(10 = -2x + 3\). This gives \(x = -3.5\).

Answer

a) \(f(x) = -2x + 3\) b) \(f(5) = -7\), and the x-value paired with \(10\) is \(x = -3.5\).

Write linear equations 72

A linear function \(f\) has slope \(-2\) and passes through \(P(4,1)\). a) Write \(f(x)\) in slope-intercept form. b) Use your function to find the value of \(x\) for which \(f(x)=11\).

Hints

- Use the given point to determine the missing intercept in \(f(x)=mx+b\). - Write the complete function before using the target output. - For part b), set your function equal to \(11\) and solve for the input.

Solution

1. Write \(f(x)=-2x+b\). Since \(f(4)=1\), \(1=-8+b\), so \(b=9\). 2. Therefore, \(f(x)=-2x+9\). 3. Solve \(11=-2x+9\). Then \(2=-2x\), so \(x=-1\).

Answer

a) \(f(x)=-2x+9\) b) \(x=-1\)

Write linear equations 73

A linear function \(f\) has y-intercept \(3\) and passes through \(Q(5,13)\). a) Write \(f(x)\) in slope-intercept form. b) Use your function to find its zero.

Hints

- Put the known y-intercept into slope-intercept form first. - Use the second point to determine the slope. - Write the complete function before setting its output equal to \(0\).

Solution

1. The y-intercept gives \(b=3\), so \(f(x)=mx+3\). 2. Use \(Q(5,13)\): \(13=5m+3\), so \(m=2\). 3. Thus, \(f(x)=2x+3\). 4. A zero satisfies \(0=2x+3\), so \(x=-\frac{3}{2}=-1.5\).

Answer

a) \(f(x)=2x+3\) b) The zero is \(x=-1.5\).

Write linear equations 74

The linear function \(g\) is \(g(x)=0.5x-2\). Another linear function \(f\) has slope \(3\) and the same zero as \(g\). a) Find the shared zero. b) Write \(f(x)\) in slope-intercept form. c) Use your function to find \(f(10)\).

Hints

- Find the x-intercept of \(g\) first. - The shared zero gives a point that must lie on \(f\). - Use that point and the given slope to write the complete equation of \(f\) before evaluating it.

Solution

1. Solve \(0=0.5x-2\), giving the shared zero \(x=4\). Thus, \(f\) passes through \((4,0)\). 2. Write \(f(x)=3x+b\). Substituting \((4,0)\) gives \(0=12+b\), so \(b=-12\). 3. Therefore, \(f(x)=3x-12\). 4. Evaluate \(f(10)=30-12=18\).

Answer

a) \(x=4\) b) \(f(x)=3x-12\) c) \(f(10)=18\)

Write linear equations 75

Three lines \(g_1\), \(g_2\), and \(g_3\) are intended to intersect at \(P(4, 1)\). Line \(g_1\) has equation \(y = -x + 5\). 1. Verify algebraically that \(P\) lies on \(g_1\). 2. Line \(g_2\) passes through the origin and through \(P\). Find its equation. 3. Line \(g_3\) has slope \(m = 0.5\) and also passes through \(P\). Find its equation.

Hints

- To test a point, substitute its coordinates into the equation. - What is the y-intercept of every line through the origin? - When the slope and one point are known, substitute them into slope-intercept form to find the y-intercept.

Solution

1. Substitute \((4, 1)\) into \(g_1\): \(1 = -4 + 5\), which is true. So \(P\) lies on \(g_1\). 2. A line through the origin has equation \(y = mx\). Using \(P\), \(1 = 4m\), so \(m = 0.25\). Thus \(g_2: y = 0.25x\). 3. Write \(g_3\) as \(y = 0.5x + b\). Substituting \((4, 1)\) gives \(1 = 0.5 \cdot 4 + b\), so \(b = -1\). Thus \(g_3: y = 0.5x - 1\).

Answer

1. Yes. \(1 = -4 + 5\), so \(P\) lies on \(g_1\). 2. \(g_2: y = 0.25x\) 3. \(g_3: y = 0.5x - 1\)

Write linear equations 76

Lines \(g\) and \(h\) intersect at \(S(3, 2)\). Line \(g\) passes through the origin. 1. Find the equation of \(g\). 2. Briefly explain why infinitely many different lines can intersect \(g\) at \(S\). 3. Line \(h\) has y-intercept \((0, 5)\). Find the equation of \(h\).

Hints

- For a line through the origin and one other point, how can you find its slope? - Think about how many different slopes a line through one fixed point can have. - Use the two known points on \(h\) to find its slope.

Solution

1. Since \(g\) passes through the origin, write \(g\) as \(y = mx\). Using \(S(3, 2)\), \(2 = 3m\), so \(m = \frac{2}{3}\). Thus \(g: y = \frac{2}{3}x\). 2. Infinitely many lines with different slopes can pass through a single point. Every such line other than \(g\) itself intersects \(g\) at \(S\). 3. Line \(h\) passes through \((0, 5)\) and \((3, 2)\), so its slope is \(m = \frac{2 - 5}{3 - 0} = -1\). Therefore, \(h: y = -x + 5\).

Answer

1. \(g: y = \frac{2}{3}x\) 2. Infinitely many lines with different slopes can pass through the same point \(S\). 3. \(h: y = -x + 5\)

Write linear equations 77

Line \(p\) passes through \(A(1,4)\) and \(B(3,0)\). a) Find the equation of \(p\). b) Find \(p(5)\). c) A second line \(q\) has the same x-intercept as \(p\), but its slope is half the slope of \(p\). Find the equation of \(q\).

Hints

- Find the slope of \(p\) from the two given points, then use either point to determine its y-intercept. - Once you have the equation of \(p\), substitute the requested input for part b). - For \(q\), use the changed slope together with the point where both lines cross the x-axis.

Solution

1. The slope of \(p\) is \(m=\frac{0-4}{3-1}=-2\). Using \((3,0)\), \(0=-2\cdot3+b\), so \(b=6\). Thus, \(p(x)=-2x+6\). 2. Evaluate the function: \(p(5)=-2\cdot5+6=-4\). 3. Point \(B(3,0)\) shows that the shared x-intercept is \((3,0)\). The slope of \(q\) is half of \(-2\), so \(m_q=-1\). 4. Using \((3,0)\), \(0=-1\cdot3+b_q\), so \(b_q=3\). Therefore, \(q(x)=-x+3\).

Answer

a) \(p(x)=-2x+6\) b) \(p(5)=-4\) c) \(q(x)=-x+3\)

Write linear equations 78

A line \(k\) passes through \(P(2, 1)\). Along the line, a run of \(\Delta x = 4\) corresponds to a rise of \(\Delta y = 2\). a) Find the slope \(m\). b) Find the y-intercept \(b\). c) Use a calculation to determine whether \(Q(10, 5)\) also lies on \(k\).

Hints

- How do rise and run determine the slope? - Once you know the slope and one point, substitute them into slope-intercept form. - To test a point, substitute its x-coordinate and compare the output with its y-coordinate.

Solution

1. The slope is \(m = \frac{\Delta y}{\Delta x} = \frac{2}{4} = 0.5\). 2. Substitute \((2, 1)\) into \(y = 0.5x + b\): \(1 = 0.5 \cdot 2 + b\), so \(b = 0\). Thus \(k(x) = 0.5x\). 3. Check \(Q\): \(k(10) = 0.5 \cdot 10 = 5\). This matches the y-coordinate of \(Q\), so \(Q\) lies on the line.

Answer

a) \(m = 0.5\) b) \(b = 0\), so \(k(x) = 0.5x\). c) Yes. \(k(10) = 5\), so \(Q(10, 5)\) lies on the line.

Write linear equations 79

A linear function \(h\) has x-intercept \((4, 0)\) and passes through \(P(-2, 3)\). a) Find the equation of \(h\). b) Give the y-intercept of the graph. c) Compare the slope of \(h\) with the slope of \(g(x) = -0.5x + 1\). What does the comparison tell you about the two graphs?

Hints

- What is the y-coordinate of a point on the x-axis? - Use the two known points to find the slope and then the y-intercept. - What relationship do two distinct lines have when their slopes are equal?

Solution

1. Use the points \((4, 0)\) and \((-2, 3)\): \(m = \frac{3 - 0}{-2 - 4} = -0.5\). 2. Substitute \((4, 0)\) into \(y = -0.5x + b\): \(0 = -0.5 \cdot 4 + b\), so \(b = 2\). Thus \(h(x) = -0.5x + 2\). 3. The y-intercept is \((0, 2)\). 4. Both \(h\) and \(g\) have slope \(-0.5\), but their y-intercepts are different, so their graphs are parallel.

Answer

a) \(h(x) = -0.5x + 2\) b) \((0, 2)\) c) Both slopes are \(-0.5\), so the graphs are parallel.

Write linear equations 80

Line \(g\) passes through \(A(-4, -1)\) and \(B(2, 2)\). A second line \(h\) has the same y-intercept as \(g\), but its slope is twice the slope of \(g\). a) Find the equations of \(g\) and \(h\). b) Find the x-intercept of \(h\).

Hints

- First find the complete equation of \(g\). - Which parts of the equation of \(h\) can you get directly from the equation of \(g\)? - What y-value do you use to find an x-intercept?

Solution

1. The slope of \(g\) is \(m_g = \frac{2 - (-1)}{2 - (-4)} = \frac{1}{2}\). 2. Substitute \((2, 2)\) into \(y = \frac{1}{2}x + b\): \(2 = \frac{1}{2} \cdot 2 + b\), so \(b = 1\). Thus \(g(x) = \frac{1}{2}x + 1\). 3. Line \(h\) has the same y-intercept, \(1\), and twice the slope, so its slope is \(1\). Therefore, \(h(x) = x + 1\). 4. Set \(h(x) = 0\): \(0 = x + 1\), so \(x = -1\).

Answer

a) \(g(x) = \frac{1}{2}x + 1\) and \(h(x) = x + 1\) b) \((-1, 0)\)

Write linear equations 81

Consider two lines \(g\) and \(h\). Line \(g\) passes through \(P(-2, 5)\) and has slope \(m = 2\). Line \(h\) passes through the origin and through \(Q(4, 10)\). a) Find the equation of each line. b) Determine algebraically whether the two lines are parallel. c) Find the x-intercept of \(g\).

Hints

- Use the given slope and point to write the equation of \(g\). - What is the y-intercept of a line through the origin? - What must be true about the slopes of parallel lines?

Solution

1. For \(g\), substitute \((-2, 5)\) into \(y = 2x + b\): \(5 = 2 \cdot (-2) + b\), so \(b = 9\). Thus \(g: y = 2x + 9\). 2. Since \(h\) passes through the origin, its y-intercept is \(0\). Its slope is \(m = \frac{10}{4} = 2.5\), so \(h: y = 2.5x\). 3. The slopes are \(2\) and \(2.5\), so the lines are not parallel. 4. For the x-intercept of \(g\), set \(y = 0\): \(0 = 2x + 9\), so \(x = -4.5\).

Answer

a) \(g: y = 2x + 9\); \(h: y = 2.5x\) b) No. Their slopes are different. c) \((-4.5, 0)\)

Write linear equations 82

Line \(k\) passes through \(A(-2, 4)\) and \(B(2, 6)\). A second line \(l\) is parallel to \(k\) and has x-intercept \((4, 0)\). Find the equations of both lines.

Hints

- Find the slope of \(k\) from its two points. - What point corresponds to the given x-intercept of \(l\)? - What do parallel lines have in common about their slopes?

Solution

1. The slope of \(k\) is \(m_k = \frac{6 - 4}{2 - (-2)} = 0.5\). 2. Substitute \((2, 6)\) into \(y = 0.5x + b_k\): \(6 = 0.5 \cdot 2 + b_k\), so \(b_k = 5\). Thus \(k: y = 0.5x + 5\). 3. Since \(l\) is parallel to \(k\), its slope is also \(0.5\). 4. Substitute its x-intercept \((4, 0)\): \(0 = 0.5 \cdot 4 + b_l\), so \(b_l = -2\). Thus \(l: y = 0.5x - 2\).

Answer

\(k: y = 0.5x + 5\) and \(l: y = 0.5x - 2\)

Write linear equations 83

Line \(f\) is determined by \(R(-1, 5)\) and \(S(2, -4)\). Line \(g\) is parallel to \(f\) and passes through the origin. Find the equations of both lines, then find the x-coordinate of the point on \(g\) whose y-coordinate is \(-12\).

Hints

- Find the slope of \(f\) from its two points. - What is the y-intercept of a line through the origin? - When an output value is given, substitute it into the equation and solve for the input.

Solution

1. The slope of \(f\) is \(m_f = \frac{-4 - 5}{2 - (-1)} = -3\). 2. Substitute \((-1, 5)\) into \(y = -3x + b_f\): \(5 = -3 \cdot (-1) + b_f\), so \(b_f = 2\). Thus \(f: y = -3x + 2\). 3. Since \(g\) is parallel to \(f\), its slope is \(-3\). Because it passes through the origin, \(g: y = -3x\). 4. For y-coordinate \(-12\), solve \(-12 = -3x\). This gives \(x = 4\).

Answer

\(f: y = -3x + 2\), \(g: y = -3x\), and the requested x-coordinate is \(4\).

Write linear equations 84

The ordered pairs \((2, 5)\) and \((-2, 3)\) are solutions of the same linear equation. a) Find the equation in the form \(y = mx + b\). b) Find \(y_P\) so that \((10, y_P)\) is also a solution.

Hints

- Use the two ordered pairs to find the slope. - After finding the slope, substitute either pair to find the y-intercept. - Then substitute \(x = 10\) to find the corresponding y-value.

Solution

1. Find the slope from the two solutions: \(m = \frac{3 - 5}{-2 - 2} = \frac{1}{2}\). 2. Substitute \((2, 5)\) into \(y = \frac{1}{2}x + b\): \(5 = \frac{1}{2} \cdot 2 + b\), so \(b = 4\). 3. The equation is \(y = \frac{1}{2}x + 4\). 4. For \(x = 10\), \(y_P = \frac{1}{2} \cdot 10 + 4 = 9\).

Answer

a) \(y = \frac{1}{2}x + 4\) b) \(y_P = 9\)

Write linear equations 85

The graphs of two linear functions \(p\) and \(q\) intersect at \(S(4, 2)\). The function \(p\) is \(p(x) = -0.5x + 4\), and line \(q\) has slope \(m = 2\). a) Find the equation of \(q\). b) Write an equation for a line \(r\) that is parallel to \(p\) and passes through the origin.

Hints

- Use the known slope and intersection point to find the missing y-intercept of \(q\). - What do parallel lines have in common about their slopes? - What is the y-intercept of a line through the origin?

Solution

1. Write \(q(x) = 2x + b\). Since \(q\) passes through \((4, 2)\), \(2 = 2 \cdot 4 + b\), so \(b = -6\). Thus \(q(x) = 2x - 6\). 2. A line parallel to \(p\) has the same slope, \(-0.5\). Passing through the origin gives y-intercept \(0\), so \(r(x) = -0.5x\).

Answer

a) \(q(x) = 2x - 6\) b) \(r(x) = -0.5x\)

Write linear equations 86

A line \(k\) is parallel to the line \(h\) with equation \(h(x) = \frac{2}{3}x - 5\) and passes through the point \(P(6, 1)\). a) Find the equation of line \(k\). b) Find the x-intercept of \(k\). c) Another line \(l\) has the same x-intercept as \(k\) but has slope \(m = -2\). Find the equation of \(l\).

Hints

- What do you know about the slopes of parallel lines? - What is true about the y-coordinate of a point on the x-axis? - How can you use a point and a slope to write the equation of a line?

Solution

1. Because \(k\) is parallel to \(h\), its slope is \(\frac{2}{3}\). Write \(k(x) = \frac{2}{3}x + b\). Using \(P(6, 1)\), \(1 = \frac{2}{3} \cdot 6 + b\), so \(b = -3\). Therefore, \(k(x) = \frac{2}{3}x - 3\). 2. For the x-intercept, set \(k(x) = 0\): \(0 = \frac{2}{3}x - 3\). Thus \(x = 4.5\), so the x-intercept is \((4.5, 0)\). 3. Line \(l\) has slope \(-2\) and passes through \((4.5, 0)\). Write \(l(x) = -2x + b\). Then \(0 = -2 \cdot 4.5 + b\), so \(b = 9\). Therefore, \(l(x) = -2x + 9\).

Answer

a) \(k(x) = \frac{2}{3}x - 3\) b) \((4.5, 0)\) c) \(l(x) = -2x + 9\)

Write linear equations 87

A line \(k\) passes through the points \(A(2, 5)\) and \(B(-2, -3)\). a) Find the equation of line \(k\). b) Use substitution to determine whether \(C(5, 11)\) lies on line \(k\). c) A second line \(l\) is parallel to \(k\) and passes through \(P(1, 1)\). Find the equation of \(l\). d) Find the y-intercept of line \(k\).

Hints

- How can you find the slope from the two given points? - What does it mean for a point to satisfy a line's equation? - What must be true about the slopes of parallel lines?

Solution

1. The slope of \(k\) is \(m = \frac{-3 - 5}{-2 - 2} = 2\). 2. Using \(A(2, 5)\), \(5 = 2 \cdot 2 + b\), so \(b = 1\). Therefore, \(k(x) = 2x + 1\). 3. Substitute \(x = 5\): \(k(5) = 2 \cdot 5 + 1 = 11\). This matches the y-coordinate of \(C\), so \(C\) lies on \(k\). 4. Since \(l\) is parallel to \(k\), its slope is \(2\). Using \(P(1, 1)\), \(1 = 2 \cdot 1 + b\), so \(b = -1\). Thus \(l(x) = 2x - 1\). 5. The y-intercept of \(k(x) = 2x + 1\) is \((0, 1)\).

Answer

a) \(k(x) = 2x + 1\) b) Yes, because \(k(5) = 11\). c) \(l(x) = 2x - 1\) d) \((0, 1)\)

Write linear equations 88

Line \(k\) passes through \(A(1,-1)\) and \(B(-1,3)\). a) Write the equation of \(k\) in slope-intercept form. b) Use your equation to find the missing coordinates so that \(M(0,y_M)\) and \(N(x_N,5)\) also lie on \(k\).

Hints

- Use \(A\) and \(B\) to determine the slope. - Use either point to determine the y-intercept and write the full equation. - Then substitute the known coordinate of \(M\) or \(N\) into that equation.

Solution

1. The slope through \(A\) and \(B\) is \(m=\frac{3-(-1)}{-1-1}=\frac{4}{-2}=-2\). 2. Using \(A(1,-1)\) in \(y=-2x+b\) gives \(-1=-2(1)+b\), so \(b=1\). Thus, \(k\) has equation \(y=-2x+1\). 3. For \(M\), substitute \(x=0\): \(y_M=-2(0)+1=1\). Therefore, \(M=(0,1)\). 4. For \(N\), substitute \(y=5\): \(5=-2x_N+1\), so \(x_N=-2\). Therefore, \(N=(-2,5)\).

Answer

a) \(k(x)=-2x+1\) b) \(M=(0,1)\) and \(N=(-2,5)\)

Write linear equations 89

The graph of a linear function \(g\) passes through \(P(8, 3)\). It has the same y-intercept as \(h(x)=-2x+7\). Write an equation for \(g\), and find its x-intercept.

Hints

- Identify the y-intercept from the equation for \(h\). - Treat the y-intercept as a point on \(g\). - Use that point and \(P\) to find the slope. - At the x-intercept, the function value is \(0\).

Solution

1. The y-intercept of \(h(x)=-2x+7\) is \((0, 7)\), so \(g\) also passes through \((0, 7)\). 2. Use \((0, 7)\) and \((8, 3)\) to find the slope: \(m=\frac{3-7}{8-0}=-\frac{4}{8}=-\frac{1}{2}\). 3. Therefore, \(g(x)=-\frac{1}{2}x+7\). 4. Set \(g(x)=0\): \(0=-\frac{1}{2}x+7\), so \(x=14\).

Answer

Equation: \(g(x)=-\frac{1}{2}x+7\) x-intercept: \((14, 0)\)

Write linear equations 90

A linear function \(f\) passes through \(A(5, 2)\) and has the same x-intercept as \(g(x)=\frac{1}{3}x-1\). Write an equation for \(f\).

Hints

- At an x-intercept, the output is \(0\). - First find where \(g\) crosses the x-axis. - Use the two points on \(f\) to find its slope. - Substitute a point and the slope into \(y=mx+b\).

Solution

1. Find the x-intercept of \(g\): \(0=\frac{1}{3}x-1\), so \(x=3\). Thus, \(f\) passes through \((3, 0)\). 2. Use \((3, 0)\) and \((5, 2)\) to find the slope: \(m=\frac{2-0}{5-3}=1\). 3. Substitute \((3, 0)\) into \(y=x+b\): \(0=3+b\), so \(b=-3\). 4. Therefore, \(f(x)=x-3\).

Answer

\(f(x)=x-3\)

Write linear equations 91

A line \(g\) passes through \(P(-4, 7)\) and has slope \(m=-1.5\). 1. Write the equation of \(g\) in point-slope form. 2. Rewrite the equation in slope-intercept form. 3. Find the coordinates of the x-intercept.

Hints

- Identify \(x_1\), \(y_1\), and \(m\) in the point-slope formula. - Distribute and isolate \(y\) to obtain slope-intercept form. - The y-coordinate of an x-intercept is \(0\).

Solution

1. Substitute \((x_1, y_1)=(-4, 7)\) and \(m=-1.5\) into \(y-y_1=m(x-x_1)\): \(y-7=-1.5(x+4)\). 2. Distribute and solve for \(y\): \(y-7=-1.5x-6\), so \(y=-1.5x+1\). 3. At the x-intercept, \(y=0\). Solve \(0=-1.5x+1\), giving \(x=\frac{2}{3}\). The x-intercept is \(\left(\frac{2}{3}, 0\right)\).

Answer

1. \(y-7=-1.5(x+4)\) 2. \(y=-1.5x+1\) 3. \(\left(\frac{2}{3}, 0\right)\)

Write linear equations 92

A line \(h\) passes through the origin and \(Q(3,-2)\). 1. Find its slope and write its equation first in point-slope form using \(Q\), then in slope-intercept form. 2. Show algebraically that every line through the origin and a point \(P_1(x_1,y_1)\), where \(x_1\ne0\), has y-intercept \(b=0\). Use \(b=y_1-mx_1\).

Hints

- Use the origin and \(Q\) to determine the slope. - Put the slope and \(Q\) into point-slope form before simplifying. - In the general proof, express \(m\) using the origin and \(P_1\), then substitute that expression into \(b=y_1-mx_1\).

Solution

1. The slope through \((0,0)\) and \(Q(3,-2)\) is \(m=\frac{-2-0}{3-0}=-\frac{2}{3}\). Using \(Q\), point-slope form is \(y+2=-\frac{2}{3}(x-3)\). Simplifying gives \(h(x)=-\frac{2}{3}x\). 2. In general, a line through \((0,0)\) and \(P_1(x_1,y_1)\) has \(m=\frac{y_1}{x_1}\). Therefore \(b=y_1-\frac{y_1}{x_1}x_1=0\).

Answer

1. \(m=-\frac{2}{3}\); \(y+2=-\frac{2}{3}(x-3)\); \(h(x)=-\frac{2}{3}x\) 2. Since \(m=\frac{y_1}{x_1}\), \(b=y_1-\frac{y_1}{x_1}x_1=0\).

Write linear equations 93

The graph of a linear function \(f\) passes through \(P_1(-0.5,4)\) and \(P_2(1,1)\). a) Write the equation of \(f\) in slope-intercept form. b) Use your equation to determine algebraically whether \(Q(3,-3)\) lies on the graph.

Hints

- Use the two given points to find the slope. - Substitute one point into \(y=mx+b\) to determine the y-intercept. - After writing the equation, evaluate it at the x-coordinate of \(Q\).

Solution

1. Find the slope: \(m=\frac{1-4}{1-(-0.5)}=\frac{-3}{1.5}=-2\). 2. Substitute \((1,1)\) into \(y=-2x+b\): \(1=-2+b\), so \(b=3\). Thus, \(f(x)=-2x+3\). 3. Evaluate \(f(3)=-2\cdot3+3=-3\). This equals the y-coordinate of \(Q\), so \(Q\) lies on the graph.

Answer

a) \(f(x)=-2x+3\) b) Yes. Since \(f(3)=-3\), \(Q(3,-3)\) lies on the graph.

Write linear equations 94

Points \(A(-2,7)\) and \(B(4,-2)\) lie on a line \(g\). A third point \(C(x,-5)\) also lies on the line. a) Write the equation of \(g\) in slope-intercept form. b) Use your equation to find the x-coordinate of \(C\).

Hints

- Use \(A\) and \(B\) to determine the slope first. - Substitute one point into \(y=mx+b\) to complete the equation. - Then set the function output equal to the y-coordinate of \(C\).

Solution

1. Find the slope through \(A\) and \(B\): \(m=\frac{-2-7}{4-(-2)}=\frac{-9}{6}=-1.5\). 2. Substitute \(A(-2,7)\) into \(y=-1.5x+b\): \(7=-1.5(-2)+b\), so \(b=4\). Thus, \(g(x)=-1.5x+4\). 3. Use the y-coordinate of \(C\): \(-5=-1.5x+4\). Then \(-9=-1.5x\), so \(x=6\).

Answer

a) \(g(x)=-1.5x+4\) b) \(x=6\)

Write linear equations 95

In a geothermal model, underground temperature increases approximately linearly with depth. The temperature is \(25\,\text{°C}\) at a depth of \(500\,\text{m}\) and \(70\,\text{°C}\) at a depth of \(2000\,\text{m}\). a) Write a linear function \(T(d)\) for temperature as a function of depth \(d\), measured in meters. b) What surface temperature does the model predict? c) At what depth does the model predict a temperature of \(100\,\text{°C}\)?

Hints

- Treat depth as the input and temperature as the output. - Use the two data points to find the rate of change. - The y-intercept represents the model's temperature at depth \(0\). - To find a depth from a temperature, set the function equal to that temperature and solve for \(d\).

Solution

1. Use the points \((500, 25)\) and \((2000, 70)\). The slope is \(m=\frac{70-25}{2000-500}=\frac{45}{1500}=0.03\) degrees Celsius per meter. 2. Substitute \((500, 25)\) into \(T(d)=0.03d+b\): \(25=0.03\cdot 500+b\), so \(b=10\). Therefore, \(T(d)=0.03d+10\). 3. At the surface, \(d=0\), so \(T(0)=10\). The predicted surface temperature is \(10\,\text{°C}\). 4. Set \(T(d)=100\): \(100=0.03d+10\). Thus, \(90=0.03d\), so \(d=3000\,\text{m}\).

Answer

a) \(T(d)=0.03d+10\) b) \(10\,\text{°C}\) c) \(3000\,\text{m}\)

Write linear equations 96

The line \(g\) passes through \(A(-2.5,4)\) and \(B(5,1)\). The points \(P_1(10,y_1)\), \(P_2(x_2,6)\), \(P_3(0,y_3)\), and \(P_4(x_4,0)\) also lie on \(g\). a) Write the equation of \(g\) in slope-intercept form. b) Find \(y_1\), \(x_2\), \(y_3\), and \(x_4\).

Hints

- Find the slope from \(A\) and \(B\), then determine the y-intercept. - Write the completed function rule before filling any missing coordinate. - Substitute a known input for a missing output, and set the rule equal to a known output for a missing input.

Solution

1. The slope is \(m=\frac{1-4}{5-(-2.5)}=\frac{-3}{7.5}=-0.4\). 2. Substitute \(B(5,1)\) into \(y=-0.4x+b\): \(1=-0.4\cdot5+b\), so \(b=3\). Thus, \(g(x)=-0.4x+3\). 3. For \(P_1\), \(y_1=-0.4\cdot10+3=-1\). 4. For \(P_2\), \(6=-0.4x_2+3\), so \(x_2=-7.5\). 5. For \(P_3\), \(y_3=-0.4\cdot0+3=3\). 6. For \(P_4\), \(0=-0.4x_4+3\), so \(x_4=7.5\).

Answer

a) \(g(x)=-0.4x+3\) b) \(y_1=-1\), \(x_2=-7.5\), \(y_3=3\), and \(x_4=7.5\)

Write linear equations 97

The graph of a linear function \(f\) passes through \(C(1,1)\) and \(D(3,6)\). a) Write the equation of \(f\) in slope-intercept form. b) Complete the missing entries so that every ordered pair lies on the graph of \(f\). <table border="1" style="border-collapse: collapse; text-align: center; width: 100%;"><tr><th style="padding: 5px; background-color: #f2f2f2;">\(x\)</th><td style="padding: 5px;">\(4\)</td><td style="padding: 5px;">?</td><td style="padding: 5px;">\(0\)</td><td style="padding: 5px;">?</td></tr><tr><th style="padding: 5px; background-color: #f2f2f2;">\(y\)</th><td style="padding: 5px;">?</td><td style="padding: 5px;">\(11\)</td><td style="padding: 5px;">?</td><td style="padding: 5px;">\(0\)</td></tr></table>

Hints

- Use \(C\) and \(D\) to determine the slope and y-intercept first. - Keep the completed equation visible while working through the table. - For a missing output, substitute x; for a missing input, set the function equal to the given y-value.

Solution

1. The slope is \(m=\frac{6-1}{3-1}=\frac{5}{2}=2.5\). 2. Substitute \(C(1,1)\) into \(y=2.5x+b\): \(1=2.5+b\), so \(b=-1.5\). Thus, \(f(x)=2.5x-1.5\). 3. Evaluate: \(f(4)=2.5\cdot4-1.5=8.5\), and \(f(0)=-1.5\). 4. Solve for the missing inputs: \(11=2.5x-1.5\) gives \(x=5\), and \(0=2.5x-1.5\) gives \(x=0.6\).

Answer

a) \(f(x)=2.5x-1.5\) b) In table order, the missing entries are \(8.5\), \(5\), \(-1.5\), and \(0.6\).

Write linear equations 98

The coordinate plane uses different scales on the x-axis and y-axis. a) Find the equation of the linear function \(g\) shown in the graph. b) Calculate \(g(6)\), then check your result by reading the corresponding point from the graph.

Figure for problem 532213

Hints

- Describe what the graph shows before calculating. - Pay close attention to the scale on each axis. - What are the coordinates of the axis intercepts? - How can two points be used to calculate the slope? - Which value can you read directly as the y-intercept?

Solution

1. The line crosses the y-axis at \((0, 10)\) and the x-axis at \((-2, 0)\). 2. The slope is \(m = \frac{10 - 0}{0 - (-2)} = 5\), and the y-intercept is \(10\). Therefore, \(g(x) = 5x + 10\). 3. \(g(6) = 5 \cdot 6 + 10 = 40\). The graph also shows the point \((6, 40)\), confirming the calculation.

Answer

a) \(g(x) = 5x + 10\) b) \(g(6) = 40\)

Write linear equations 99

A linear function \(f\) has slope \(m=-0.4\) and zero \(x=5\). a) Write the equation of \(f\) in slope-intercept form. b) Find algebraically the x-value for which \(f(x)=-2\).

Hints

- A zero at \(x=5\) gives a point on the line with output \(0\). - Use the known slope and that point to determine the y-intercept. - After writing the full equation, set the output equal to \(-2\) and solve.

Solution

1. Write \(f(x)=-0.4x+b\). Since \(x=5\) is a zero, \(f(5)=0\). 2. Substitute \(x=5\): \(0=-0.4\cdot5+b\), so \(b=2\). Therefore, \(f(x)=-0.4x+2\). 3. Set the function value equal to \(-2\): \(-2=-0.4x+2\). 4. Then \(-4=-0.4x\), so \(x=10\).

Answer

a) \(f(x)=-0.4x+2\) b) \(x=10\)

Write linear equations 100

A linear function \(g\) has slope \(m=2.5\) and passes through \(P(2,1)\). a) Write the equation of \(g\) in slope-intercept form. b) At what x-value does \(g(x)=11\)?

Hints

- The slope is known, but the y-intercept must still be determined from the given point. - Write the full function rule before working backward from the requested output. - Check the final input by substituting it into your equation.

Solution

1. Write \(g(x)=2.5x+b\). 2. Substitute \((2,1)\): \(1=2.5(2)+b\), so \(b=-4\). Therefore, \(g(x)=2.5x-4\). 3. Set the output equal to \(11\): \(11=2.5x-4\). Then \(15=2.5x\), so \(x=6\).

Answer

a) \(g(x)=2.5x-4\) b) \(x=6\)

Write linear equations 101

A line \(g\) passes through \(P(-1, 4)\) and \(Q(2, -2)\), as shown in the graph. a) Find the equation of line \(g\). b) Use substitution to determine whether \(R(4, -6)\) also lies on the line.

Figure for problem 533291

Hints

- Use the coordinates of \(P\) and \(Q\) to find the slope. - Once you know the slope, substitute one point into \(y = mx + b\) to find \(b\). - To test \(R\), substitute its x-coordinate into your equation and compare the result with its y-coordinate.

Solution

1. The slope is \(m = \frac{-2 - 4}{2 - (-1)} = -2\). Using \(Q(2, -2)\), \(-2 = -2 \cdot 2 + b\), so \(b = 2\). Therefore, \(g(x) = -2x + 2\). 2. Substitute \(x = 4\): \(g(4) = -2 \cdot 4 + 2 = -6\). This matches the y-coordinate of \(R\), so \(R\) lies on the line.

Answer

a) \(g(x) = -2x + 2\) b) Yes, \(R(4, -6)\) lies on \(g\).

Write linear equations 102

A linear relationship is recorded in the table. <table><tr><td>\(x\)</td><td>\(2\)</td><td>\(5\)</td><td>\(9\)</td></tr><tr><td>\(y\)</td><td>\(11\)</td><td>\(23\)</td><td>\(39\)</td></tr></table> Jordan proposes the model \(y=4x+11\). Determine whether Jordan's model is correct. If it is not, write the correct linear model and explain the error.

Hints

- First verify whether the table supports one constant slope. - Test the proposed equation against at least one displayed point. - Remember that the constant term of \(y=mx+b\) is the output when the input is zero, not just any listed output.

Solution

1. The slope from \((2,11)\) to \((5,23)\) is \(\frac{23-11}{5-2}=4\). The slope from \((5,23)\) to \((9,39)\) is also \(4\), so a linear model with slope \(4\) is justified. 2. Jordan's model gives \(y=4(2)+11=19\) when \(x=2\), so it does not match the table. 3. Write \(y=4x+b\) and use \((2,11)\): \(11=8+b\), so \(b=3\). 4. The correct model is \(y=4x+3\). Jordan used the observed y-value \(11\) as though it were the y-intercept, even though that value occurs at \(x=2\), not at \(x=0\).

Answer

Jordan's model is incorrect. The correct model is \(y=4x+3\). The value \(11\) is not the y-intercept because it corresponds to \(x=2\).

Write linear equations 103

The ordered pair \((3, -1)\) is a solution of a linear equation in \(x\) and \(y\). a) Write one possible equation of the form \(ax + by = 10\), where \(a\) and \(b\) are nonzero integers. b) Give two other ordered pairs that are solutions of the equation you wrote in part a).

Hints

- Substitute the given ordered pair for \(x\) and \(y\). - Can you choose one coefficient and then solve for the other? - Once you have an equation, choose an x-value and solve for the corresponding y-value.

Solution

1. Substitute \((3, -1)\) into \(ax + by = 10\): \(3a - b = 10\). 2. Choose a nonzero integer value for one coefficient. For example, let \(a = 4\). Then \(3 \cdot 4 - b = 10\), so \(b = 2\). One possible equation is \(4x + 2y = 10\). 3. To find more solutions, choose x-values and solve for \(y\). If \(x = 0\), then \(2y = 10\), so \(y = 5\). If \(x = 1\), then \(4 + 2y = 10\), so \(y = 3\).

Answer

a) Answers will vary. One possible equation is \(4x + 2y = 10\). b) For that equation, two possible solutions are \((0, 5)\) and \((1, 3)\).

Write linear equations 104

A family of linear functions \(f_a\) passes through \(P(a, 2a)\) and \(Q(3a, 8a)\), where \(a\ne0\). a) Write \(f_a(x)\) in terms of \(a\). b) Determine whether the slope depends on \(a\). c) Find the zero of \(f_a\) in terms of \(a\).

Hints

- Treat \(a\) as a nonzero number when using the slope formula. - Check whether the parameter cancels from the slope. - Substitute one point to find the y-intercept. - Set the function equal to \(0\) to find its zero.

Solution

1. The slope is \(m=\frac{8a-2a}{3a-a}=\frac{6a}{2a}=3\). Because \(a\ne0\), it cancels, so the slope does not depend on \(a\). 2. Substitute \(P(a, 2a)\) into \(y=3x+b\): \(2a=3a+b\), so \(b=-a\). Therefore, \(f_a(x)=3x-a\). 3. Set \(f_a(x)=0\): \(0=3x-a\), so \(x=\frac{a}{3}\).

Answer

a) \(f_a(x)=3x-a\) b) No. The slope is always \(3\). c) \(x=\frac{a}{3}\)

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