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5137929
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) \(y=-x\)
5100759
A line passes through the points \(A(-1, -2)\) and \(B(3, 4)\). Which equation represents the line? a) \(y = x + 1\) b) \(y = 2x - 2\) c) \(y = 4x + 2\) d) \(y = 1.5x - 0.5\)

Hints

- How can you find the slope of a line from two points? - Once you know the slope, how can you use either point to find the y-intercept? - You can also substitute both points into each answer choice to test it.

Solution

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

Answer

d) \(y = 1.5x - 0.5\)
5128959
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.
5129169
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

\(f(x) = 2x + 4\)
5129179
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) \(g(x) = -0.75x + 3\)
5129529
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

\(f(x) = -\frac{3}{4}x - 3\)
5129679
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) \(m = -\frac{3}{4}\)
5129949
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

\(y = 2x - 1\)
5129989
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

\(y = -1.5x + 6\)
5130009
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

\(y = -2x + 10\)
5130679
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 \(x = 4\).
5131989
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) \(f(x) = \frac{3}{4}x\)
5288439
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) \(g(x)=-1.5x+4\)
5332609
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) \(P(0, -2.5)\), \(m = -1.5\) The graph for part b) is shown as a visual check.
Figure for problem 533260

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 \(y = -2x + b\): \(5 = -2 \cdot 2 + b\), 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 \(y = \frac{4}{3}x + b\): \(0 = \frac{4}{3} \cdot 3 + b\), 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) \(b = -2.5\); \(y = -1.5x - 2.5\)
5335269
Use the x- and y-intercepts shown on the graph to write equations for lines \(f\) and \(g\).
Figure for problem 533526

Hints

- Read both intercepts for each line. - Use \(m=\frac{y_2-y_1}{x_2-x_1}\), then substitute the y-intercept into \(y=mx+b\).

Solution

1. Line \(f\) has y-intercept \((0, 4)\) and x-intercept \((4, 0)\). Its slope is \(m=\frac{0-4}{4-0}=-1\), so its equation is \(y=-x+4\). 2. Line \(g\) has y-intercept \((0, -2)\) and x-intercept \((1, 0)\). Its slope is \(m=\frac{0-(-2)}{1-0}=2\), so its equation is \(y=2x-2\).

Answer

Line \(f\): \(y=-x+4\) Line \(g\): \(y=2x-2\)
5349389
The graph of a linear equation is shown. Which equation matches the graph? A: \(2x + y = 6\) B: \(2x - y = 6\) C: \(x + 2y = 6\) D: \(x - 2y = 6\)
Figure for problem 534938

Hints

- Look at where the line crosses the coordinate axes. - Substitute the coordinates of an intercept into each equation. - A point lies on the graph of an equation only if its coordinates make the equation true.

Solution

1. The graph crosses the y-axis at \((0, 6)\) and the x-axis at \((3, 0)\). 2. Test choice A: \(2 \cdot 0 + 6 = 6\) and \(2 \cdot 3 + 0 = 6\), so both intercepts satisfy \(2x + y = 6\). 3. The other choices fail for at least one of these intercepts. Therefore, choice A matches the graph.

Answer

A: \(2x + y = 6\)
5349439
Find the equations of the three lines \(f\), \(g\), and \(h\) shown in the coordinate plane.
Figure for problem 534943

Hints

- First check whether each line has a special position: horizontal, vertical, or through the origin. - For a slanted line, use a slope triangle. - Pay attention to where each line crosses the axes.

Solution

1. Line \(f\) passes through the origin and \((2, 3)\). Its slope is \(\frac{3}{2}\), so \(f\) has equation \(y = \frac{3}{2}x\). 2. Line \(g\) is horizontal and crosses the y-axis at \(-2\), so its equation is \(y = -2\). 3. Line \(h\) is vertical and crosses the x-axis at \(-4\), so its equation is \(x = -4\).

Answer

\(f: y = \frac{3}{2}x\), \(g: y = -2\), and \(h: x = -4\)
5349459
Write equations for lines \(p\), \(q\), and \(r\). Use clear grid points to determine slopes and intercepts where appropriate.
Figure for problem 534945

Hints

- For a nonvertical line, first identify the y-intercept. - Use two grid points to calculate rise over run. - A vertical line is written using its constant x-coordinate.

Solution

1. Line \(p\) has y-intercept \(-2\). It rises \(2\) units for every \(3\) units to the right, so its slope is \(\frac{2}{3}\). Thus \(p: y = \frac{2}{3}x - 2\). 2. Line \(q\) has y-intercept \(1\). It falls \(1\) unit for every \(1\) unit to the right, so its slope is \(-1\). Thus \(q: y = -x + 1\). 3. Line \(r\) is vertical and passes through all points with x-coordinate \(-1\). Thus \(r: x = -1\).

Answer

\(p: y = \frac{2}{3}x - 2\) \(q: y = -x + 1\) \(r: x = -1\)
5128919
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 \(q\).
5128939
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) \(g(x) = -1.2x + 6\)
5129279
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.
5129549
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 \(h(x) = -0.4x + 1.8\).
5129579
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) \(h(x) = 2x - 3\)
5129599
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.
5129609
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 \(\frac{1}{3}\).
5129819
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 \(p(-6) = 7\).
5129959
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) \((10, 0)\)
5129969
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.
5129979
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

\(y = \frac{1}{2}x - 3\)
5129999
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) \(y_E = -8\)
5130029
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) \((7, 0)\)
5130049
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. \((-1, -3)\)
5130069
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.
5130089
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) \(h(x) = x + 2\)
5130099
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) \(x = 5\)
5130109
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) \((0, 0)\)
5130119
A student 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 the student's error, then find the correct equation of the line.

Hints

- In the slope formula, which coordinate difference belongs in the numerator? - Make sure the coordinate differences use the same point order. - After finding the correct slope, use either point to find the y-intercept.

Solution

1. The student 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

The student divided the change in \(x\) by the change in \(y\). The correct slope is \(-2\), so the line is \(y = -2x + 9\).
5130159
The points \(A(1, 4)\), \(B(3, 10)\), and \(C(x, 19)\) lie on the same line. Find the x-coordinate of \(C\).

Hints

- Use \(A\) and \(B\) to find the slope of the line. - Points on the same line must satisfy the same linear equation. - After finding the equation from \(A\) and \(B\), substitute the y-coordinate of \(C\). - Solve the resulting equation for \(x\).

Solution

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

Answer

\(x = 6\)
5130179
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 \(y = 3x - 1\).
5130379
A line has slope \(m = -3\) and passes through \(P(2, -4)\). Another point \(Q(x, 8)\) also lies on the line. Find the x-coordinate of \(Q\).

Hints

- Use the known slope and point to write an equation for the line. - Substitute the known y-coordinate of \(Q\) into that equation. - Check your result by verifying that the slope between \(P\) and \(Q\) is \(-3\).

Solution

1. Use \(P(2, -4)\) to write the line in slope-intercept form: \(-4 = -3 \cdot 2 + b\), so \(b = 2\). 2. The line has equation \(y = -3x + 2\). 3. Since \(Q\) has y-coordinate \(8\), solve \(8 = -3x + 2\). Then \(6 = -3x\), so \(x = -2\).

Answer

\(x = -2\)
5130419
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 \(C\).
5130539
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 \((8, 0)\).
5130629
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 \(x = -3.5\).
5130819
A linear function \(f\) has slope \(-2\) and passes through \(P(4, 1)\). Find the value of \(x\) for which \(f(x) = 11\).

Hints

- Start with slope-intercept form using the given slope. - Use the given point to find the y-intercept. - Then set \(f(x)\) equal to the target value and solve for \(x\).

Solution

1. Write \(f(x) = -2x + b\). Substitute \((4, 1)\): \(1 = -2 \cdot 4 + b\), so \(b = 9\). 2. Thus \(f(x) = -2x + 9\). 3. Set the function equal to \(11\): \(11 = -2x + 9\). Solving gives \(x = -1\).

Answer

\(x = -1\)
5130829
A linear function \(f\) has y-intercept \(3\) and passes through \(Q(5, 13)\). Find its zero.

Hints

- What does the given y-intercept tell you about \(b\) in \(f(x) = mx + b\)? - How can you use the given point to find the slope? - What equation do you solve to find a zero?

Solution

1. The y-intercept is \(b = 3\), so write \(f(x) = mx + 3\). 2. Use \(Q(5, 13)\): \(13 = 5m + 3\). Then \(10 = 5m\), so \(m = 2\). 3. Thus \(f(x) = 2x + 3\). 4. Set \(f(x) = 0\): \(2x + 3 = 0\). Then \(2x = -3\), so \(x = -1.5\).

Answer

The zero is \(x = -1.5\).
5130839
The linear function \(g\) is given by \(g(x) = 0.5x - 2\). Another linear function \(f\) has slope \(3\) and the same zero as \(g\). Find \(f(10)\).

Hints

- First find where \(g\) crosses the x-axis. - If two functions have the same zero, what point do their graphs share? - Use that point and the given slope to write the equation of \(f\).

Solution

1. Find the zero of \(g\): \(0.5x - 2 = 0\), so \(0.5x = 2\) and \(x = 4\). 2. Since \(f\) has the same zero and slope \(3\), write \(f(x) = 3x + b\) and use \((4, 0)\): \(0 = 3 \cdot 4 + b\). 3. Then \(b = -12\), so \(f(x) = 3x - 12\). 4. Evaluate at \(x = 10\): \(f(10) = 3 \cdot 10 - 12 = 18\).

Answer

\(f(10) = 18\)
5130859
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. \(g_3: y = 0.5x - 1\)
5130869
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. \(h: y = -x + 5\)
5130889
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.
5131559
Line \(p\) passes through \(A(1, 4)\) and \(B(3, 0)\). a) Find the equation of \(p\). b) Find the x-intercept of \(p\). 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 from the two given points. - What does an x-intercept tell you about the y-coordinate? - For the new line, first change the slope as described. - Then use the shared x-intercept to find the new y-intercept.

Solution

1. The slope of \(p\) is \(m = \frac{0 - 4}{3 - 1} = -2\). Using \((3, 0)\), \(0 = -2 \cdot 3 + b\), so \(b = 6\). Thus \(p: y = -2x + 6\). 2. Since \(B(3, 0)\) is already on the x-axis, the x-intercept is \((3, 0)\). 3. The slope of \(q\) is half of \(-2\), so \(m_q = -1\). Since \(q\) also passes through \((3, 0)\), \(0 = -1 \cdot 3 + b_q\), giving \(b_q = 3\). Thus \(q: y = -x + 3\).

Answer

a) \(p: y = -2x + 6\) b) \((3, 0)\) c) \(q: y = -x + 3\)
5132009
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.
5132089
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.
5132099
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) \((-1, 0)\)
5132149
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) \((-4.5, 0)\)
5132209
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 \(l: y = 0.5x - 2\)
5132219
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 \(4\).
5136959
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) \(y_P = 9\)
5139439
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) \(r(x) = -0.5x\)
5141489
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) \(l(x) = -2x + 9\)
5154839
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) \((0, 1)\)
5199969
Line \(k\) passes through \(A(1, -1)\) and \(B(-1, 3)\). Find the missing coordinates so that \(M(0, \square)\) and \(N(\square, 5)\) also lie on \(k\).

Hints

- Find the slope from the two given points. - Use one point to write the equation of the line. - Substitute the known coordinate for each incomplete point.

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 \(x = 0\): \(y = -2(0) + 1 = 1\). Therefore, \(M = (0, 1)\). 4. For \(N\), substitute \(y = 5\): \(5 = -2x + 1\), so \(x = -2\). Therefore, \(N = (-2, 5)\).

Answer

\(M = (0, 1)\) and \(N = (-2, 5)\)
5228679
A linear function \(f\) passes through \(A(-3, 11)\) and \(B(2, -4)\). Write an equation for \(f\), and find the coordinates of its x-intercept.

Hints

- Use the slope formula with the two given points. - Write the line in the form \(y=mx+b\). - Substitute one point to find \(b\). - At the x-intercept, the function value is \(0\).

Solution

1. Find the slope: \(m=\frac{-4-11}{2-(-3)}=\frac{-15}{5}=-3\). 2. Substitute \(B(2, -4)\) into \(y=mx+b\): \(-4=-3\cdot 2+b\), so \(b=2\). 3. Therefore, \(f(x)=-3x+2\). 4. Set \(f(x)=0\) to find the x-intercept: \(0=-3x+2\), so \(x=\frac{2}{3}\).

Answer

Equation: \(f(x)=-3x+2\) x-intercept: \(\left(\frac{2}{3}, 0\right)\)
5228689
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: \((14, 0)\)
5229089
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

\(f(x)=x-3\)
5288139
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. \(\left(\frac{2}{3}, 0\right)\)
5288149
A line \(h\) passes through \(Q(3, -2)\). 1. Find its slope if the line also passes through the origin. Use point-slope 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

- The origin has coordinates \((0, 0)\). - Use the slope formula with the origin and the given point. - In part 2, express \(m\) in terms of \(x_1\) and \(y_1\) before substituting into the formula for \(b\).

Solution

1. Using \(Q(3, -2)\), point-slope form is \(y+2=m(x-3)\). Substitute the origin: \(0+2=m(0-3)\), so \(2=-3m\) and \(m=-\frac{2}{3}\). 2. The slope through \((0, 0)\) and \((x_1, y_1)\) is \(m=\frac{y_1}{x_1}\). Substitute into \(b=y_1-mx_1\): \(b=y_1-\frac{y_1}{x_1}x_1=y_1-y_1=0\).

Answer

1. \(m=-\frac{2}{3}\) 2. Since \(m=\frac{y_1}{x_1}\), \(b=y_1-\frac{y_1}{x_1}x_1=0\).
5288159
The graph of a linear function \(f\) passes through \(P_1(-0.5, 4)\) and \(P_2(1, 1)\). Determine algebraically whether \(Q(3, -3)\) lies on the graph.

Hints

- Use the two given points to find the slope. - Substitute one point to find the y-intercept. - Evaluate the resulting function at \(x=3\) and compare with the given y-coordinate.

Solution

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

Answer

Yes. Since \(f(x)=-2x+3\) and \(f(3)=-3\), the point \(Q(3, -3)\) lies on the graph.
5288169
Points \(A(-2, 7)\) and \(B(4, -2)\) lie on a line \(g\). A third point \(C(x, -5)\) also lies on the line. Find \(x\).

Hints

- Collinear points lie on a line with the same slope. - First write the equation of the line through \(A\) and \(B\). - Substitute \(y=-5\) and solve for the missing x-coordinate.

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 \(y=-1.5x+b\): \(7=-1.5\cdot(-2)+b\), so \(b=4\). The line is \(y=-1.5x+4\). 3. Use the y-coordinate of \(C\): \(-5=-1.5x+4\). Then \(-9=-1.5x\), so \(x=6\).

Answer

\(x=6\)
5288189
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) \(x=\frac{a}{3}\)
5288239
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 \(g\). Find \(y_1\), \(x_2\), \(y_3\), and \(x_4\).

Hints

- Find the slope from \(A\) and \(B\). - Substitute one point to find the y-intercept. - Use the completed line equation to solve for each missing coordinate.

Solution

1. The slope is \(m=\frac{1-4}{5-(-2.5)}=\frac{-3}{7.5}=-0.4\). 2. Substitute \(B(5, 1)\) into \(y=-0.4x+b\): \(1=-0.4\cdot 5+b\), so \(b=3\). Thus, \(g\) has equation \(y=-0.4x+3\). 3. For \(P_1\), \(y_1=-0.4\cdot 10+3=-1\). 4. For \(P_2\), \(6=-0.4x_2+3\), so \(x_2=-7.5\). 5. For \(P_3\), \(y_3=-0.4\cdot 0+3=3\). 6. For \(P_4\), \(0=-0.4x_4+3\), so \(x_4=7.5\).

Answer

\(y_1=-1\); \(x_2=-7.5\); \(y_3=3\); \(x_4=7.5\)
5288249
The graph of a linear function \(f\) passes through \(C(1, 1)\) and \(D(3, 6)\). Complete the missing entries in the table 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;">\(\dots\)</td> <td style="padding: 5px;">0</td> <td style="padding: 5px;">\(\dots\)</td> </tr> <tr> <th style="padding: 5px; background-color: #f2f2f2;">\(y\)</th> <td style="padding: 5px;">\(\dots\)</td> <td style="padding: 5px;">11</td> <td style="padding: 5px;">\(\dots\)</td> <td style="padding: 5px;">0</td> </tr> </table>

Hints

- Use \(C\) and \(D\) to write the equation \(f(x)=mx+b\). - Substitute a given x-value to find a missing y-value. - When a y-value is given, set the function equal to that value and solve for \(x\).

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: \(f(4)=2.5\cdot 4-1.5=8.5\), 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

In table order, the missing entries are \(8.5\), \(5\), \(-1.5\), and \(0.6\).
5322109
The coordinate plane shows the lines \(f\) and \(g\) and the points \(P\) and \(Q\). a) Find the equations of \(f\) and \(g\). b) A line \(h\) is parallel to \(f\) and passes through \(P\). Find the equation of \(h\). c) Find the equation of the line \(k\) that passes through \(P\) and \(Q\).
Figure for problem 532210

Hints

- Where does each displayed line cross the y-axis? - Use two grid points on a line to find its slope. - What must be true about the slopes of parallel lines? - For a line through two points, find the slope first and then use one point to find the y-intercept.

Solution

1. Line \(f\) crosses the y-axis at \((0, 2)\). Using \((0, 2)\) and \((3, 3)\), its slope is \(\frac{1}{3}\). Thus \(f(x) = \frac{1}{3}x + 2\). 2. Line \(g\) crosses the y-axis at \((0, -1)\). Using \((0, -1)\) and \((2, -2)\), its slope is \(-\frac{1}{2}\). Thus \(g(x) = -\frac{1}{2}x - 1\). 3. Because \(h\) is parallel to \(f\), its slope is \(\frac{1}{3}\). Using \(P(-3, 3)\), \(3 = \frac{1}{3} \cdot (-3) + b\), so \(b = 4\). Therefore, \(h(x) = \frac{1}{3}x + 4\). 4. The slope through \(P(-3, 3)\) and \(Q(1, -1)\) is \(\frac{-1 - 3}{1 - (-3)} = -1\). Using \(Q\), \(-1 = -1 + b\), so \(b = 0\). Therefore, \(k(x) = -x\).

Answer

a) \(f(x) = \frac{1}{3}x + 2\) and \(g(x) = -\frac{1}{2}x - 1\) b) \(h(x) = \frac{1}{3}x + 4\) c) \(k(x) = -x\)
5332779
A linear function \(f\) has slope \(m = -0.4\) and zero \(x = 5\). Find algebraically the x-value for which \(f(x) = -2\).

Hints

- A zero at \(x = 5\) means the point \((5, 0)\) lies on the graph. - Use \(y = mx + b\) to find the y-intercept first. - Once you know the equation, substitute \(-2\) for the function value and solve for \(x\).

Solution

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

Answer

\(f(x) = -2\) when \(x = 10\).
5332789
A linear function \(g\) has slope \(m=2.5\) and passes through \(P(2, 1)\). At what x-value does \(g(x)=11\)?
Figure for problem 533278

Hints

- Use slope-intercept form, \(y=mx+b\), and substitute the given point to find \(b\). - After finding the function rule, set \(g(x)=11\) and solve for \(x\).

Solution

1. Write the function in slope-intercept form: \(g(x)=2.5x+b\). 2. Substitute the point \((2, 1)\): \(1=2.5(2)+b\), 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

\(x=6\)
5333339
Find the equation of each line \(a\) and \(b\) shown in the coordinate plane. Write each equation first in slope-intercept form \(y = mx + b\), then in standard form \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers.
Figure for problem 533333

Hints

- Choose two grid points on each line. - Use rise over run to find the slope. - Read the y-intercept where the line crosses the vertical axis. - Clear fractions before rearranging into \(Ax + By = C\).

Solution

1. Line \(a\) passes through \((0, -2)\) and \((3, 0)\). Its slope is \(m = \frac{0 - (-2)}{3 - 0} = \frac{2}{3}\), and its y-intercept is \(-2\). Thus \(y = \frac{2}{3}x - 2\). Multiplying by \(3\) and rearranging gives \(2x - 3y = 6\). 2. Line \(b\) passes through \((0, 3)\) and \((3, 0)\). Its slope is \(-1\), and its y-intercept is \(3\). Thus \(y = -x + 3\), which is equivalent to \(x + y = 3\).

Answer

Line \(a\): \(y = \frac{2}{3}x - 2\) and \(2x - 3y = 6\) Line \(b\): \(y = -x + 3\) and \(x + y = 3\)
5136939
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 \((1, 3)\).

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