- Focus only on the first step needed to remove the parentheses. - Check whether the outside factor is applied to every term inside the parentheses.

1. The factor \(2\) multiplies both terms inside the parentheses. 2. Therefore, \(2(x + 3)\) rewrites as \(2x + 6\). 3. The correct choice is B.

B) \(2x + 6 = 10\)

- Use the two side labels in the diagram to form the rectangle's area expression. - Check which choice applies the outside factor to every term inside the parentheses.

1. The side labels in the diagram give the area equation \(3(x + 2) = 15\). 2. Distributing \(3\) gives \(3x + 6 = 15\). 3. The correct choice is B.

B) \(3x + 6 = 15\)

- What does an equals sign tell you about both sides? - Which choice changes both sides in the same way?

1. Dividing both sides by the same nonzero number keeps the two sides equal. 2. Dividing by \(3\) gives \(x+2=5\). Subtracting \(3\) from only one side would change the equation's solutions.

Divide both sides by \(3\), giving \(x+2=5\). An operation must preserve equality.

- How does the subtraction affect both terms in the parentheses? - Check your rewrite by trying one value of \(x\).

1. Multiply both \(x\) and \(3\) by \(-2\). 2. This gives the equivalent equation \(5-2x-6=7\).

\(5-2x-6=7\)

- Use the distributive property to remove both sets of parentheses. - Pay close attention to the subtraction before the second set of parentheses. - Combine like terms before isolating the variable.

1. Distribute: \(6b + 9 - 8 - 4b = 5\). 2. Combine like terms: \(2b + 1 = 5\). 3. Subtract \(1\): \(2b = 4\). 4. Divide by \(2\): \(b = 2\).

\(b = 2\)

- Distribute the factor to every term inside the parentheses. - Pay special attention to the negative factor in part b. - Combine like terms before isolating the variable. - Substitute your result to check it.

1. For the first equation, distribute and combine: \(4x-12+7=19\), so \(4x-5=19\). Add \(5\) and divide by \(4\) to get \(x=6\). 2. For the second equation, distribute: \(12-3y-6=0\), so \(6-3y=0\). Subtract \(6\) and divide by \(-3\) to get \(y=2\). 3. For the third equation, distribute and combine: \(5z+2z-8=13\), so \(7z-8=13\). Add \(8\) and divide by \(7\) to get \(z=3\).

a) \(x = 6\) b) \(y = 2\) c) \(z = 3\)

- A minus sign before parentheses changes the signs inside. - Simplify the left side before isolating the variable. - Use inverse operations one step at a time.

1. Remove the parentheses, including the negative sign: \(7x - 2x - 5 = 15\). 2. Combine like terms: \(5x - 5 = 15\). 3. Add \(5\): \(5x = 20\). 4. Divide by \(5\): \(x = 4\).

\(x = 4\)

- Distribute \(-3\) to both terms inside the parentheses. - Combine like terms before isolating the variable. - Use inverse operations on both sides.

1. Distribute the negative factor: \(5x - 3x - 12 = 4\). 2. Combine like terms: \(2x - 12 = 4\). 3. Add \(12\): \(2x = 16\). 4. Divide by \(2\): \(x = 8\).

\(x = 8\)

- Replace \(A\) with its target value to make an equation. - Take care with the negative sign outside the parentheses. - How can you isolate \(x\) after simplifying?

1. Set the expression equal to the target: \(5-2(x+1)=-1\). 2. Distribute the negative factor: \(5-2x-2=-1\), so \(3-2x=-1\). 3. Subtract \(3\): \(-2x=-4\). Divide by \(-2\): \(x=2\).

\(A=3-2x\), so \(x=2\).

- Set the expression equal to zero. - Look for an easy way to remove the parentheses. - Substitute your answer back to check that the expression equals zero.

1. Write the equation: \(3(x+5)-12=0\). 2. Distribute \(3\): \(3x+15-12=0\). 3. Combine constants: \(3x+3=0\). 4. Subtract \(3\) and divide by \(3\): \(x=-1\).

\(3x+3=0\), so \(x=-1\).

- Compare the value of a parenthesized expression with the value after subtraction. - Check your rewritten equations using a trial input. - Verify the final value in both original equations.

1. Distribute the negative sign in the first equation: \(18-x-7=5\). Combine constants to get \(11-x=5\), so \(x=6\). 2. Distribute the negative sign in the second equation: \(24=35-x-5\). Simplify to \(24=30-x\), so \(x=6\). 3. Both equations have the same solution, \(x=6\).

1) \(11-x=5\), so \(x=6\). 2) \(24=30-x\), so \(x=6\). Yes, both have the same solution.

- How can you remove the parentheses? - Combine the constant terms on the left. - Check your result in the original equation.

1. Distribute \(6\): \(6x-24+8=20\). 2. Combine constants: \(6x-16=20\). 3. Add \(16\) to obtain \(6x=36\). 4. Divide by \(6\): \(x=6\).

\(6x-16=20\), so \(x=6\).

- Use the distributive property to remove each set of parentheses. - Combine the terms containing \(x\) and the constants separately. - Isolate the variable term before dividing. - Check by substitution.

1. Distribute: \(0.5x - 0.6 + 0.4x + 0.2 = 0.5\). 2. Combine like terms: \(0.9x - 0.4 = 0.5\). 3. Add \(0.4\): \(0.9x = 0.9\). 4. Divide by \(0.9\): \(x = 1\).

\(x = 1\)

- Both first steps are allowed because they preserve equality. - Compare the numbers you would need to work with after each first step. - Check the solution in the original equation, not just your rewritten one.

1. Divide both sides by \(5\) to get \(2x-1=7\). 2. Add \(1\): \(2x=8\). Divide by \(2\): \(x=4\). 3. Expanding first gives \(10x-5=35\), then \(10x=40\), then \(x=4\). Both methods use three algebra steps. Dividing first keeps the coefficients smaller.

\(x=4\). Dividing by \(5\) first keeps the numbers smaller; both methods take three steps.

- Try the same value of \(x\) in Lee's first two expressions. Do they match? - Find the correct expression before solving for \(x\). - Check your answer in the original equation.

1. Lee multiplied only \(x\), not \(-3\), by \(4\). Distribution should give \(4x-12=8\). 2. Add \(12\) to get \(4x=20\), so \(x=5\). 3. Checking: \(4(5-3)=8\).

Lee forgot to multiply \(-3\) by \(4\). The correct solution is \(x=5\).

- Carry out both methods one step at a time. - Compare the number and complexity of the steps. - Look closely at the relationship between \(6\) and \(30\).

1. Using Alex’s method, distribute: \(6x+12=30\). 2. Subtract \(12\): \(6x=18\). Divide by \(6\): \(x=3\). 3. Using Priya’s method, divide both sides of the original equation by \(6\): \(x+2=5\). 4. Subtract \(2\): \(x=3\). 5. Both methods give the same solution. Dividing first is especially efficient when removing the outside factor immediately simplifies both sides without creating awkward values.

a) \(x = 3\) b) \(x = 3\) c) Dividing first is efficient when removing the outside factor immediately makes the equation simpler.

- How do you find the perimeter of a rectangle? - Add the length and width expressions before expanding. - Once you have a single equation, how can you solve it for \(x\)?

1. The perimeter is twice the sum of length and width: \(2[(2x+5)+(x+3)]=52\). 2. Combine inside parentheses to get \(2(3x+8)=52\). 3. Expand: \(6x+16=52\). 4. Subtract \(16\) and divide by \(6\): \(x=6\).

\(2(3x+8)=52\) expands to \(6x+16=52\), so \(x=6\).

- Recall what an isosceles triangle tells you about its side lengths. - Add all three side lengths to write the perimeter equation. - Remember that the expression for a congruent side is used twice. - Distribute before combining like terms.

1. Add the base and the two congruent sides: \((y - 2) + 2(2y + 1) = 35\). 2. Distribute: \(y - 2 + 4y + 2 = 35\). 3. Combine like terms: \(5y = 35\). 4. Divide by \(5\): \(y = 7\).

The value of \(y\) is \(7\).

- Distribute the factor outside the parentheses. - Combine decimal coefficients carefully. - Substitute the result into the original equation to check it.

1. Distribute: \(6x - 10 - 0.4x = 18\). 2. Combine like terms: \(5.6x - 10 = 18\). 3. Add \(10\): \(5.6x = 28\). 4. Divide by \(5.6\): \(x = 5\). 5. Check: \(5(1.2 \cdot 5 - 2) - 0.4 \cdot 5 = 5(4) - 2 = 18\).

\(x = 5\)

- Multiply by a common denominator to clear all fractions. - Apply the subtraction sign to the entire second numerator. - Combine like terms before isolating \(x\).

1. Multiply every term by \(6\): \(2(2x - 1) - (x + 3) = 3\). 2. Distribute and remove parentheses: \(4x - 2 - x - 3 = 3\). 3. Combine like terms: \(3x - 5 = 3\). 4. Add \(5\): \(3x = 8\). 5. Divide by \(3\): \(x = \frac{8}{3}\).

\(x = \frac{8}{3}\)

- Use a variable for the width and express the length using it. - What is the perimeter of a rectangle in terms of its two dimensions? - Check that twice the sum of your dimensions is the given perimeter.

1. Let \(x\) inches be the width, so the length is \(x+13\) inches. 2. Write the perimeter equation: \(126=2(x+x+13)\). 3. Expand and combine terms: \(126=4x+26\). 4. Subtract \(26\) to get \(100=4x\), so \(x=25\). 5. The length is \(25+13=38\) inches.

Width: \(25\,\text{in.}\); length: \(38\,\text{in.}\). One expanded perimeter equation is \(126=4x+26\), where \(x\) is the width.

- Apply the factor outside the parentheses to each term inside. - Dividing both sides by the same nonzero number preserves equality. - Keep both sides balanced at every step. - Compare the final value from each method.

1. Using the distributive property first gives \(4x-6=10\). Add \(6\): \(4x=16\). Divide by \(4\): \(x=4\). 2. Dividing both sides of the original equation by \(4\) first gives \(x-1.5=10\div4=2.5\). Add \(1.5\): \(x=4\). 3. Both valid methods give the same solution, \(x=4\).

a) \(x = 4\) b) \(x = 4\)

- What must be true about the new side lengths of a square? - How much longer than the short side was the original long side? - Write the original rectangle's perimeter using just one variable.

1. Let \(x\) inches be the shorter side. The longer side must be \(x+6\) inches, since the changes of \(3\) inches in opposite directions make them equal. 2. Write the perimeter equation: \(2[x+(x+6)]=40\). 3. Expand: \(4x+12=40\). Subtract \(12\) and divide by \(4\) to obtain \(x=7\). 4. The longer side is \(7+6=13\) inches. Changing them gives sides of \(10\) inches each, as required.

The original sides are \(7\,\text{in.}\) and \(13\,\text{in.}\). An expanded equation is \(4x+12=40\).

- Let the smallest certificate count be \(x\). - Use parentheses because the factor of \(4\) applies to the entire merit-certificate count. - The three expressions must add to \(220\). - Check that the three results total \(220\).

1. Let \(x\) be the number of winner certificates. 2. The number of merit certificates is \(x + 20\), and the number of participation certificates is \(4(x + 20)\). 3. Write the total equation: \(x + (x + 20) + 4(x + 20) = 220\). 4. Distribute and combine like terms: \(x + x + 20 + 4x + 80 = 220\), so \(6x + 100 = 220\). 5. Subtract \(100\): \(6x = 120\). Divide by \(6\): \(x = 20\). 6. There are \(20\) winner certificates, \(20 + 20 = 40\) merit certificates, and \(4 \cdot 40 = 160\) participation certificates.

There are \(20\) winner certificates, \(40\) merit certificates, and \(160\) participation certificates.

- Check that your first rewrite gives the same value for a test input. - The minus sign in the third equation affects the entire parentheses. - Substitute each solution in the original equation.

1. Distribute \(3\): \(3a+15=27\). Subtract \(15\): \(3a=12\), so \(a=4\). 2. Distribute \(4\): \(8b-12=12\). Add \(12\): \(8b=24\), so \(b=3\). 3. Distribute \(-2\): \(14-2x+2=6\), so \(16-2x=6\). Subtract \(16\) and divide by \(-2\): \(x=5\).

1) \(3a+15=27\), so \(a=4\). 2) \(8b-12=12\), so \(b=3\). 3) \(16-2x=6\), so \(x=5\).

- Express the length in terms of the width. - Which perimeter equation uses both dimensions? - Check your final pair of lengths against the perimeter.

1. Let \(x\) feet be the width. Then the length is \(3x-2\) feet. 2. Use the perimeter equation \(2[x+(3x-2)]=44\). 3. Combine terms and distribute: \(2(4x-2)=44\), so \(8x-4=44\). 4. Add \(4\) and divide by \(8\): \(x=6\). 5. The length is \(3\cdot6-2=16\) feet.

Width: \(6\,\text{ft}\); length: \(16\,\text{ft}\). The expanded equation is \(8x-4=44\), where \(x\) is the width.

- Apply each outside factor or subtraction to every term inside its parentheses. - Rewrite fractions or mixed numbers in a convenient equivalent form when useful. - Combine like terms after removing parentheses. - Check each answer by substitution.

1. For a), distribute \(2\): \(0.6t + 0.2t - 1.2 = 2\). Combine like terms: \(0.8t - 1.2 = 2\). Then \(0.8t = 3.2\), so \(t = 4\). 2. For b), rewrite \(1\frac{1}{2}\) as \(1.5\) and distribute the subtraction: \(1.5w - 0.5w - 4 = 6\). Combine like terms: \(w - 4 = 6\), so \(w = 10\). 3. For c), distribute \(\frac{2}{3}\): \(4x - 6 + 0.5x = 7.5\). Combine like terms: \(4.5x - 6 = 7.5\). Then \(4.5x = 13.5\), so \(x = 3\).

a) \(t = 4\) b) \(w = 10\) c) \(x = 3\)

- Use the total to express the number of children using the number of adults. - Multiply each ticket price by the corresponding number of tickets. - How can you check both the total group size and total cost?

1. Let \(x\) be the number of adults, so there are \(32-x\) children. 2. Write the cost equation: \(10x+7(32-x)=254\). 3. Distribute: \(10x+224-7x=254\), hence \(3x+224=254\). 4. Subtract \(224\): \(3x=30\), so \(x=10\). 5. There are \(32-10=22\) children. The cost is \(10\cdot10+7\cdot22=254\) dollars.

\(10\) adults and \(22\) children. \(10x+7(32-x)=254\) expands to \(3x+224=254\).

- Use the small-package weight as the variable. - Express the other package weights in terms of the small-package weight. - Multiply each package weight by the number of packages of that size. - Add the three weight expressions to equal \(86\,\text{lb}\).

1. Let \(x\) be the weight of one small package in pounds. 2. A medium package weighs \(x + 2\), and a large package weighs \(2x\). 3. Write the total-weight equation \(10x + 5(x + 2) + 2(2x) = 86\). 4. Distribute and combine like terms: \(10x + 5x + 10 + 4x = 86\), so \(19x + 10 = 86\). 5. Subtract \(10\): \(19x = 76\). Divide by \(19\): \(x = 4\). 6. The medium package weighs \(4 + 2 = 6\,\text{lb}\), and the large package weighs \(2 \cdot 4 = 8\,\text{lb}\).

A small package weighs \(4\,\text{lb}\), a medium package weighs \(6\,\text{lb}\), and a large package weighs \(8\,\text{lb}\).

- Recall the sum of the interior angles of a triangle. - Express the other two angles in terms of \(\alpha\). - Pay attention to which angle is tripled. - Distribute before combining like terms.

1. Let \(x\) degrees be the measure of angle \(\alpha\). 2. Then \(\beta = x + 20\), and \(\gamma = 3(x + 20)\). 3. The interior angles of a triangle add to \(180^\circ\), so \(x + (x + 20) + 3(x + 20) = 180\). 4. Distribute and combine like terms: \(5x + 80 = 180\). 5. Subtract \(80\) and divide by \(5\): \(5x = 100\), so \(x = 20\). 6. Therefore, \(\alpha = 20^\circ\), \(\beta = 40^\circ\), and \(\gamma = 120^\circ\).

The angle measures are \(\alpha = 20^\circ\), \(\beta = 40^\circ\), and \(\gamma = 120^\circ\).

- Use one variable for the equal numbers of muffins and pretzels. - Express the number of juice boxes in terms of that variable. - Multiply each price by its quantity and add. - Distribute the juice-box price across the parentheses.

1. Let \(x\) be the number of muffins sold. Then \(x\) pretzels and \(x + 10\) juice boxes are sold. 2. Write the revenue equation: \(1.50x + 1.10x + 0.80(x + 10) = 144\). 3. Distribute and combine like terms: \(1.5x + 1.1x + 0.8x + 8 = 144\), so \(3.4x + 8 = 144\). 4. Subtract \(8\): \(3.4x = 136\). 5. Divide by \(3.4\): \(x = 40\). 6. Therefore, \(40\) muffins, \(40\) pretzels, and \(50\) juice boxes are sold.

The class sells \(40\) muffins, \(40\) pretzels, and \(50\) juice boxes.

- Pay close attention to the negative factor before the second set of parentheses. - Combine all terms containing \(x\). - Combine all constant terms separately.

1. Distribute each factor: \(12x - 18 - 12x - 15 + 8x - 4 = 9\). 2. Combine like terms: \(8x - 37 = 9\). 3. Add \(37\): \(8x = 46\). 4. Divide by \(8\): \(x = \frac{46}{8} = \frac{23}{4} = 5.75\).

\(x = \frac{23}{4} = 5.75\)

- Distribute the negative decimal factor carefully. - Combine variable terms and constants separately. - Use subtraction to isolate the variable term. - Divide by the coefficient of \(x\).

1. Distribute, paying attention to the subtraction: \(0.8x + 1.6 - 0.6x + 1.2 + 0.5x = 4.2\). 2. Combine the variable terms and constants: \(0.7x + 2.8 = 4.2\). 3. Subtract \(2.8\): \(0.7x = 1.4\). 4. Divide by \(0.7\): \(x = 2\).

\(x = 2\)

- A single variable can describe one count; the other count depends on it. - What stays fixed when you change how many markers were bought? - Test your counts against both totals.

1. Let \(x\) be the number of markers, so \(45-x\) is the number of pencils. The cost equation is \(1.20x+0.50(45-x)=38.60\). 2. Distribute: \(1.20x+22.50-0.50x=38.60\), giving \(0.70x+22.50=38.60\). 3. Subtract \(22.50\): \(0.70x=16.10\), so \(x=23\). 4. The number of pencils is \(45-23=22\). The costs total \(27.60+11.00=38.60\) dollars.

a) \(1.20x+0.50(45-x)=38.60\), which expands to \(0.70x+22.50=38.60\). b) \(23\) markers and \(22\) pencils.

- Use one variable for the number of adult tickets. - Express the number of child tickets using the total sold. - The revenue from the two ticket types must add to the total. - Check which count is greater after solving.

1. Let \(x\) be the number of adult tickets, so there are \(150-x\) child tickets. 2. The revenue equation is \(5x+2(150-x)=420\). 3. Distribute and simplify: \(5x+300-2x=420\), hence \(3x+300=420\). 4. Subtract \(300\) to obtain \(3x=120\), so \(x=40\). 5. There are \(150-40=110\) child tickets, which is more than the adult tickets.

More child tickets: \(110\) child tickets and \(40\) adult tickets. With \(x\) adult tickets, \(5x+2(150-x)=420\) expands to \(3x+300=420\).

- Use the total snack count to express one quantity in terms of the other. - Multiply each price by its number of sales. - Check that your two counts give the total revenue.

1. Let \(x\) be the number of sandwiches, so \(60-x\) is the number of pretzels. 2. The revenue equation is \(1.80x+1.20(60-x)=93.60\). 3. Distribute and simplify: \(1.80x+72-1.20x=93.60\), hence \(0.60x=21.60\). 4. Divide by \(0.60\): \(x=36\). The number of pretzels is \(60-36=24\).

\(36\) sandwiches and \(24\) pretzels. \(1.80x+1.20(60-x)=93.60\) expands to \(0.60x+72=93.60\).

- Use a variable for the lower pump rate. - Amount moved equals rate times time. - Add the amounts moved by both pumps to equal the total. - Keep Pump A’s rate \(10\) greater than Pump B’s.

1. Let \(x\) be Pump B’s rate in gallons per minute. Pump A’s rate is \(x + 10\). 2. Write the total-volume equation \(45(x + 10) + 60x = 9900\). 3. Distribute and combine like terms: \(45x + 450 + 60x = 9900\), so \(105x + 450 = 9900\). 4. Subtract \(450\): \(105x = 9450\). 5. Divide by \(105\): \(x = 90\). 6. Pump A’s rate is \(90 + 10 = 100\,\text{gal/min}\).

Pump A moves \(100\,\text{gal/min}\), and Pump B moves \(90\,\text{gal/min}\).

- Let \(x\) represent Class B’s trees per hour. - Multiply each hourly rate by the number of hours worked. - Add the two totals to equal \(93\). - For part c, use Class B’s rate with a new time.

1. Let Class B’s hourly rate be \(x\). Then Class A’s hourly rate is \(x+3\), so the total-tree equation is \(4(x+3)+5x=93\). 2. Distribute and combine: \(4x+12+5x=93\), so \(9x=81\) and \(x=9\). 3. Class B plants \(9\) trees per hour, and Class A plants \(9+3=12\) trees per hour. 4. In \(4\) hours, Class B would plant \(9\cdot4=36\) trees.

a) \(4(x + 3) + 5x = 93\) b) Class A plants \(12\) trees per hour, and Class B plants \(9\) trees per hour. c) Class B would plant \(36\) trees in \(4\) hours.

- If \(x\) are correct, how many are incorrect? - How do correct and incorrect answers affect the total score? - Use the same expression for both teams, but a different final score.

1. Let \(x\) be the number correct; there are \(12-x\) incorrect answers. The score is \(4x-2(12-x)\). 2. For Math Masters, solve \(4x-2(12-x)=30\). Expanding gives \(6x-24=30\), so \(x=9\). 3. For Puzzle Solvers, solve \(4x-2(12-x)=6\). Expanding gives \(6x-24=6\), so \(x=5\).

a) \(4x-2(12-x)=30\) becomes \(6x-24=30\), so \(9\) correct answers. b) \(4x-2(12-x)=6\) becomes \(6x-24=6\), so \(5\) correct answers.

- Express the number of misses using the total attempts. - Write the total score as points earned minus points lost. - What values are possible for a count of hits? - Compare the algebraic result with that restriction.

1. Let \(x\) be the number of hits, so \(15-x\) are misses. The score is \(7x-3(15-x)\). 2. For a score of \(55\), \(7x-3(15-x)=55\), which simplifies to \(10x-45=55\). Thus \(x=10\). 3. A score of \(50\) gives \(10x-45=50\), so \(x=9.5\). 4. Hits must be a whole number from \(0\) to \(15\); therefore, \(50\) points is impossible.

a) \(7x-3(15-x)=55\) expands to \(10x-45=55\), so \(10\) hits. b) \(7x-3(15-x)=50\) expands to \(10x-45=50\), which would require \(9.5\) hits; impossible.

- Use one variable for the slower speed. - Distance equals speed multiplied by time. - How are the distances traveled related to the initial separation?

1. Let \(v\) be the truck's speed in miles per hour; the car's speed is \(v+20\). 2. In \(3\) hours their combined distance is \(3v+3(v+20)=360\). 3. Distribute and combine: \(3v+3v+60=360\), so \(6v=300\) and \(v=50\). 4. The car averages \(50+20=70\) miles per hour. They cover \(150+210=360\) miles.

Truck: \(50\,\text{mph}\); car: \(70\,\text{mph}\). \(3v+3(v+20)=360\) expands to \(6v+60=360\).

- Keep the units of salt mass consistent with the stated concentration. - Does adding pure water change the amount of salt? - How can you express the new volume using one variable?

1. Convert the salt mass: \(1.5\,\text{kg}=1500\,\text{g}\). 2. Let \(x\) be the liters of pure water added. The total volume is \(50+x\) liters, while the salt mass remains \(1500\) grams. 3. Write \(20(50+x)=1500\). Distribute: \(1000+20x=1500\). 4. Solve \(20x=500\), so \(x=25\) liters. The new concentration is \(1500/75=20\) grams per liter.

\(20(50+x)=1500\) expands to \(1000+20x=1500\). Add \(25\,\text{L}\) of water.

- Express the longer length using the shorter length. - Which area formula applies to each banner? - Add the two area expressions before solving.

1. Let \(l\) inches be the shorter length; the longer length is \(l+10\) inches. 2. Their areas add to \(24l+24(l+10)=1680\). 3. Distribute and combine: \(48l+240=1680\). 4. Subtract \(240\) and divide by \(48\): \(l=30\). 5. The longer length is \(40\) inches. The areas add to \(24\cdot30+24\cdot40=1680\) square inches.

\(24l+24(l+10)=1680\) expands to \(48l+240=1680\). The lengths are \(30\,\text{in.}\) and \(40\,\text{in.}\).

- Distribute the negative factor carefully. - Combine variable terms and constants separately. - Isolate the variable term before dividing. - Align decimal place values when calculating.

1. Distribute: \(6x - 3 - 4x - 8 + 0.5x = 6.5\). 2. Combine like terms: \(2.5x - 11 = 6.5\). 3. Add \(11\): \(2.5x = 17.5\). 4. Divide by \(2.5\): \(x = 7\).

\(x = 7\)

- Use the first piece as the variable. - The third piece is twice the entire length of the second piece. - Add expressions for all three pieces and set the sum equal to \(150\). - Check that your three lengths add to the original beam length.

1. Let \(x\) be the length of the first piece in inches. 2. The second piece is \(x + 10\), and the third piece is \(2(x + 10)\). 3. The pieces total \(150\,\text{in.}\), so write \(x + (x + 10) + 2(x + 10) = 150\). 4. Distribute and combine like terms: \(x + x + 10 + 2x + 20 = 150\), so \(4x + 30 = 150\). 5. Subtract \(30\): \(4x = 120\). Divide by \(4\): \(x = 30\). 6. The piece lengths are \(30\,\text{in.}\), \(40\,\text{in.}\), and \(80\,\text{in.}\).

The pieces are \(30\,\text{in.}\), \(40\,\text{in.}\), and \(80\,\text{in.}\) long.

- Which train starts earlier, and by how much? - Express both travel times using the passenger train's time. - What must the two traveled distances add to? - Convert the final number of hours to minutes.

1. Let \(x\) be the passenger train's travel time in hours before the meeting. The freight train travels for \(x+0.5\) hours. 2. Their distances sum to \(150\): \(50(x+0.5)+75x=150\). 3. Distribute: \(50x+25+75x=150\), giving \(125x=125\). 4. Thus \(x=1\) hour, or \(60\) minutes after the passenger train departs.

\(60\) minutes after the passenger train leaves. With \(x\) in hours after the passenger train leaves, \(50(x+0.5)+75x=150\) expands to \(125x+25=150\).

- When both start together, how fast does the distance between them shrink? - Convert the delay to hours and use one variable for Eli's group's walking time. - Maya's group has been walking longer in part b). - For part c), how far apart are the groups when Eli's group starts?

1. With the same start time, \(4x+5x=18\), so \(x=2\) hours. 2. For the delayed start, \(45\) minutes is \(0.75\) hour. Let \(y\) be Eli's group's walking time in hours, so Maya's group walks for \(y+0.75\) hours. 3. Write \(4(y+0.75)+5y=18\). Distribute: \(4y+3+5y=18\), yielding \(9y=15\) and \(y=\frac{5}{3}\) hour, or \(100\) minutes. 4. When Eli's group begins later, Maya's group has already shortened the distance between them, so Eli's group needs less walking time.

a) \(2\) hours. b) \(4(y+0.75)+5y=18\) expands to \(9y+3=18\), so \(100\) minutes. c) Maya's group has already shortened the distance before Eli's group begins walking.

- Use one time variable measured from the earlier start. - How long has the second hiker been walking compared with the first? - The distances they walk together cover the whole trail. - Which speed gives the distance from Mr. Smith's starting point?

1. Let \(t\) be the number of hours after \(2{:}00\) p.m. Mr. Smith walks \(4t\) miles and Ms. Weber walks \(6(t-0.5)\) miles. 2. Their distances sum to \(17\): \(4t+6(t-0.5)=17\). 3. Distribute: \(4t+6t-3=17\). Then \(10t=20\) and \(t=2\), so they meet at \(4{:}00\) p.m. 4. Mr. Smith travels \(4\cdot2=8\) miles. They meet \(8\) miles from his starting point.

a) \(4{:}00\) p.m. b) \(8\) miles from Mr. Smith's starting point. With \(t\) in hours after \(2{:}00\) p.m., \(4t+6(t-0.5)=17\) expands to \(10t-3=17\).

- Convert both total power values to watts before writing equations. - Express the two panel counts using one variable and their total. - In part b, keep the Type A count the same and find the needed Type B output. - Compare the required Type B count with the installed count.

1. Convert \(10\,\text{kW}\) to \(10{,}000\,\text{W}\). Let \(x\) be the number of Type A panels, so there are \(28-x\) Type B panels. 2. Solve \(320x+400(28-x)=10{,}000\). Distributing gives \(320x+11{,}200-400x=10{,}000\), hence \(-80x=-1200\) and \(x=15\). 3. There are \(15\) Type A and \(13\) Type B panels. The Type A panels produce \(15\cdot320=4800\,\text{W}\). 4. At \(12{,}000\,\text{W}\), the Type B panels must produce \(7200\,\text{W}\), requiring \(18\) Type B panels. 5. The owner needs \(18-13=5\) more Type B panels.

a) \(320x+400(28-x)=10{,}000\) expands to \(-80x+11{,}200=10{,}000\), giving \(15\) Type A and \(13\) Type B panels. b) \(5\) additional Type B panels.