Free Math Worksheets

- 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 a), distribute and combine: \(4x - 12 + 7 = 19\), so \(4x - 5 = 19\). Add \(5\), then divide by \(4\): \(x = 6\). 2. For b), distribute: \(12 - 3y - 6 = 0\), so \(6 - 3y = 0\). Subtract \(6\), then divide by \(-3\): \(y = 2\). 3. For c), distribute and combine: \(5z + 2z - 8 = 13\), so \(7z - 8 = 13\). Add \(8\), then divide by \(7\): \(z = 3\).

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 \(2\): \(x = 8\).

\(x = 8\)

- Use the distributive property to remove the parentheses. - Combine the constant terms before solving. - Isolate the variable term, then divide by its coefficient. - Substitute your answer into the expression to check that it equals \(0\).

1. Write an equation: \(3(x + 5) - 12 = 0\). 2. Distribute \(3\): \(3x + 15 - 12 = 0\). 3. Combine like terms: \(3x + 3 = 0\). 4. Subtract \(3\): \(3x = -3\). 5. Divide by \(3\): \(x = -1\).

\(x = -1\)

- Distribute the negative sign to every term in the parentheses. - Simplify each side before isolating \(x\). - Two equations have the same solution set when exactly the same values make them true.

1. Distribute the subtraction in equation 1): \(18 - x - 7 = 5\). Combine constants: \(11 - x = 5\). Subtract \(11\): \(-x = -6\), so \(x = 6\). 2. Distribute the subtraction in equation 2): \(24 = 35 - x - 5\). Combine constants: \(24 = 30 - x\). Subtract \(30\): \(-6 = -x\), so \(x = 6\). 3. Both equations have solution set \(S = \{6\}\).

1) \(S = \{6\}\) 2) \(S = \{6\}\) The equations have the same solution set.

- Apply the factor outside the parentheses to each term inside. - Combine the constant terms. - Isolate the variable term before dividing. - Divide by the coefficient of \(x\).

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

\(x = 6\)

- 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 \(0.9\): \(x = 1\).

\(x = 1\)

- 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. Alex’s method: Distribute to get \(6x + 12 = 30\). Subtract \(12\): \(6x = 18\). Divide by \(6\): \(x = 3\). 2. Priya’s method: Divide both sides by \(6\) to get \(x + 2 = 5\). Subtract \(2\): \(x = 3\). 3. Both methods give the same solution. Dividing first is especially efficient when the factor outside the parentheses divides every term on the other side without creating complicated values.

a) \(x = 3\) b) \(x = 3\) c) Dividing first is efficient when the outside factor divides all terms on the other side cleanly.

- Distribute every factor, including a negative sign before parentheses. - Combine like terms on each side. - Move variable terms to one side before isolating the variable. - Check the zero solution in part b carefully.

1. For a), distribute and combine: \(3x + 12 - 2x = 5x - 10\), so \(x + 12 = 5x - 10\). Then \(22 = 4x\), so \(x = 5.5\). 2. For b), remove the parentheses: \(8 - 2a - 5 = 3a + 3\), so \(3 - 2a = 3a + 3\). Then \(0 = 5a\), so \(a = 0\). 3. For c), distribute and combine: \(4k - 4 + 6 - 2k = 10\), so \(2k + 2 = 10\). Then \(2k = 8\), so \(k = 4\).

a) \(x = 5.5\) b) \(a = 0\) c) \(k = 4\)

- Use the perimeter formula for a rectangle. - Add the expressions for the length and width first. - Set the perimeter expression equal to \(52\). - Use the distributive property when removing parentheses.

1. Use the perimeter formula for a rectangle: \(2(\text{length} + \text{width}) = \text{perimeter}\). 2. Substitute the given expressions: \(2((2x + 5) + (x + 3)) = 52\). 3. Combine like terms: \(2(3x + 8) = 52\). 4. Distribute: \(6x + 16 = 52\). 5. Subtract \(16\): \(6x = 36\). 6. Divide by \(6\): \(x = 6\).

The value of \(x\) is \(6\).

- 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 \(3\): \(x = \frac{8}{3}\).

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

- Represent the width with a variable. - Express the length in terms of the width. - Use the perimeter formula for a rectangle. - Distribute and solve the equation.

1. Let \(x\) inches be the width. The length is \(x + 13\) inches. 2. Use the perimeter formula: \(126 = 2(x + x + 13)\). 3. Distribute and combine like terms: \(126 = 4x + 26\). 4. Subtract \(26\): \(100 = 4x\). 5. Divide by \(4\): \(x = 25\). 6. The length is \(25\,\text{in.} + 13\,\text{in.} = 38\,\text{in.}\).

The width is \(25\,\text{in.}\), and the length is \(38\,\text{in.}\).

- 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.

Method a: 1. Distribute \(4\): \(4x - 6 = 10\). 2. Add \(6\) to both sides: \(4x = 16\). 3. Divide by \(4\): \(x = 4\). Method b: 4. Divide both sides by \(4\): \(x - 1.5 = 10 \div 4 = 2.5\). 5. Add \(1.5\) to both sides: \(x = 4\). Both methods give the same solution.

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

- Use the fact that all sides of a square are equal. - Determine how much longer the original long side is than the short side. - Write a perimeter equation for the original rectangle. - Solve for the shorter side first.

1. Let \(x\) inches be the shorter side. 2. Because adding \(3\) to the shorter side and subtracting \(3\) from the longer side makes them equal, the original longer side is \(x + 6\) inches. 3. Use the perimeter formula: \(2(x + x + 6) = 40\). 4. Distribute and combine like terms: \(4x + 12 = 40\). 5. Subtract \(12\) and divide by \(4\): \(4x = 28\), so \(x = 7\). 6. The longer side is \(7\,\text{in.} + 6\,\text{in.} = 13\,\text{in.}\).

The original side lengths are \(7\,\text{in.}\) and \(13\,\text{in.}\).

- Simplify the expression by distributing first. - Pay close attention to the negative sign before the parentheses. - What happens to \(3 - 2x\) as \(x\) increases? - Compare \(10\) with your answer from part 1).

1. Set the expression equal to \(-1\): \(5 - 2(x + 1) = -1\). 2. Distribute \(-2\): \(5 - 2x - 2 = -1\). 3. Combine like terms: \(3 - 2x = -1\). 4. Subtract \(3\): \(-2x = -4\). Divide by \(-2\): \(x = 2\). 5. The simplified expression is \(3 - 2x\). Because the coefficient of \(x\) is negative, the value of the expression decreases as \(x\) increases. 6. Since \(10 > 2\), the value at \(x = 10\) must be less than the value at \(x = 2\), so it is less than \(-1\).

1) \(x = 2\) 2) Less than \(-1\), because \(3 - 2x\) decreases as \(x\) increases.

- 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.

- Use the distributive property when parentheses are present. - Track the negative factor carefully in equation 3). - Move variable terms to one side when they appear on both sides. - Substitute each result to check it.

1. Distribute \(3\): \(3a + 15 = 27\). Subtract \(15\): \(3a = 12\). Divide by \(3\): \(a = 4\). 2. Subtract \(5b\): \(3b - 12 = 9\). Add \(12\): \(3b = 21\). Divide by \(3\): \(b = 7\). 3. Distribute \(-2\): \(14 - 2x + 2 = 6\). Combine constants: \(16 - 2x = 6\). Subtract \(16\): \(-2x = -10\). Divide by \(-2\): \(x = 5\).

1) \(a = 4\) 2) \(b = 7\) 3) \(x = 5\)

- Represent the width with a variable. - Express the length using the given relationship. - Use the perimeter formula for a rectangle. - Distribute before solving the equation.

1. Let \(x\) feet be the width. The length is \(3x - 2\) feet. 2. Use the perimeter formula: \(2(x + 3x - 2) = 44\). 3. Combine like terms: \(2(4x - 2) = 44\). 4. Distribute: \(8x - 4 = 44\). 5. Add \(4\): \(8x = 48\). Divide by \(8\): \(x = 6\). 6. The length is \(3 \cdot 6\,\text{ft} - 2\,\text{ft} = 16\,\text{ft}\).

The garden bed is \(6\,\text{ft}\) wide and \(16\,\text{ft}\) long.

- Rewrite fractions and mixed numbers in a convenient form. - Apply a factor to every term inside parentheses. - Combine like terms after removing parentheses. - Check each answer by substitution.

1. For a), rewrite \(\frac{1}{5}\) as \(0.2\): \(0.6t + 0.2t - 1.2 = 2\). Combine like terms: \(0.8t - 1.2 = 2\). Add \(1.2\): \(0.8t = 3.2\). Divide by \(0.8\): \(t = 4\). 2. For b), rewrite \(1\frac{1}{2}\) as \(1.5\): \(1.5w - 0.5w - 4 = 6\). Combine like terms: \(w - 4 = 6\). Add \(4\): \(w = 10\). 3. For c), distribute \(\frac{2}{3}\): \(4x - 6 + 0.5x = 7.5\). Combine like terms: \(4.5x - 6 = 7.5\). Add \(6\): \(4.5x = 13.5\). Divide by \(4.5\): \(x = 3\).

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

- Express the number of children using the total number of people. - Multiply each admission price by the corresponding number of people. - Distribute before combining like terms. - Check that the two group sizes add to \(32\).

1. Let \(x\) be the number of adults. Then the number of children is \(32 - x\). 2. Write the cost equation: \(10x + 7(32 - x) = 254\). 3. Distribute: \(10x + 224 - 7x = 254\). 4. Combine like terms: \(3x + 224 = 254\). 5. Subtract \(224\): \(3x = 30\), so \(x = 10\). 6. The number of children is \(32 - 10 = 22\).

The group has \(10\) adults and \(22\) children.

- 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 \(0.7\): \(x = 2\).

\(x = 2\)

- Express the number of pencils using the total number of items. - Multiply each unit price by its quantity. - Add the two costs to form the total-cost equation. - Check both the item count and total price.

1. Let \(x\) be the number of markers. Then the number of pencils is \(45 - x\). 2. Write the cost equation: \(1.20x + 0.50(45 - x) = 38.60\). 3. Distribute and combine like terms: \(1.2x + 22.5 - 0.5x = 38.6\), so \(0.7x + 22.5 = 38.6\). 4. Subtract \(22.5\): \(0.7x = 16.1\). 5. Divide by \(0.7\): \(x = 23\). 6. The number of pencils is \(45 - 23 = 22\).

a) \(1.20x + 0.50(45 - x) = 38.60\) b) The teacher buys \(23\) markers and \(22\) pencils.

- Represent one ticket count with a variable. - Use the total number of tickets to express the other count. - Write an equation for total revenue. - Compare the two quantities after solving.

1. Let \(x\) be the number of adult tickets. Then the number of child tickets is \(150 - x\). 2. Write the revenue equation: \(5x + 2(150 - x) = 420\). 3. Distribute and combine like terms: \(5x + 300 - 2x = 420\), so \(3x + 300 = 420\). 4. Subtract \(300\): \(3x = 120\), so \(x = 40\). 5. The number of child tickets is \(150 - 40 = 110\). Since \(110 > 40\), more child tickets were sold.

More child tickets were sold: \(110\) child tickets and \(40\) adult tickets.

- Express one item count using the total number of snacks. - Multiply each price by its quantity. - Distribute before combining like terms. - Check both the total item count and revenue.

1. Let \(x\) be the number of sandwiches. Then the number of pretzels is \(60 - x\). 2. Write the revenue equation: \(1.80x + 1.20(60 - x) = 93.60\). 3. Distribute and combine like terms: \(1.8x + 72 - 1.2x = 93.6\), so \(0.6x + 72 = 93.6\). 4. Subtract \(72\): \(0.6x = 21.6\). 5. Divide by \(0.6\): \(x = 36\). 6. The number of pretzels is \(60 - 36 = 24\).

The stand sold \(36\) sandwiches and \(24\) pretzels.

- 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. Class B’s hourly rate is \(x\), and Class A’s is \(x + 3\). Write \(4(x + 3) + 5x = 93\). 2. Distribute and combine like terms: \(4x + 12 + 5x = 93\), so \(9x + 12 = 93\). 3. Subtract \(12\): \(9x = 81\). Divide by \(9\): \(x = 9\). 4. Class B plants \(9\) trees per hour, and Class A plants \(9 + 3 = 12\) trees per hour. 5. In \(4\) hours, Class B would plant \(9 \cdot 4 = 36\) 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.

- Start with the total number of questions. - If \(x\) questions are correct, how many are incorrect? - Write one expression that includes points earned and points lost. - Think about how the score changes when one more answer is correct instead of incorrect.

1. Let \(x\) be the number of correct answers. Then \(12 - x\) is the number of incorrect answers. 2. For a score \(P\), write \(4x - 2(12 - x) = P\). 3. Distribute and combine like terms: \(4x - 24 + 2x = P\), so \(6x - 24 = P\). 4. For part a), \(6x - 24 = 30\). Then \(6x = 54\), so \(x = 9\). 5. For part b), \(6x - 24 = 6\). Then \(6x = 30\), so \(x = 5\).

a) Team Math Masters answered \(9\) questions correctly. b) Team Puzzle Solvers answered \(5\) questions correctly.

- Write a formula for the score in terms of the number of hits. - Use the fact that there are \(15\) total attempts. - What values can the number of hits reasonably take? - Solve the score equation for \(50\) and interpret the result.

1. Let \(x\) be the number of hits. Then \(15 - x\) is the number of misses. 2. For a score \(P\), write \(7x - 3(15 - x) = P\). 3. Distribute and combine like terms: \(7x - 45 + 3x = P\), so \(10x - 45 = P\). 4. For part a), \(10x - 45 = 55\). Then \(10x = 100\), so \(x = 10\). 5. For part b), \(10x - 45 = 50\). Then \(10x = 95\), so \(x = 9.5\). 6. The number of hits must be a whole number from \(0\) through \(15\). Therefore, a score of \(50\) points is impossible.

a) The player made \(10\) hits. b) A score of \(50\) points would require \(9.5\) hits. Because the number of hits must be a whole number, that score is impossible.

- Represent the slower vehicle's speed with a variable. - Write the faster vehicle's speed in terms of that variable. - Use distance equals rate times time for each vehicle. - The two distances add to the full highway length.

1. Let \(v\,\text{mph}\) be the truck's average speed. The car's average speed is \(v + 20\). 2. In \(3\) hours, the truck travels \(3v\) miles and the car travels \(3(v + 20)\) miles. 3. Their distances add to \(360\), so \(3v + 3(v + 20) = 360\). 4. Distribute and combine like terms: \(3v + 3v + 60 = 360\), so \(6v + 60 = 360\). 5. Subtract \(60\) and divide by \(6\): \(v = 50\). 6. The car's speed is \(50 + 20 = 70\,\text{mph}\).

The truck averages \(50\,\text{mph}\), and the car averages \(70\,\text{mph}\).

- Convert all salt masses to the same unit. - Adding pure water does not change the amount of salt. - Write the new total volume in terms of \(x\). - Concentration times volume equals the salt mass.

1. Convert the salt mass: \(1.5\,\text{kg} = 1500\,\text{g}\). 2. Let \(x\) be the number of liters of pure water added. The new volume is \(50 + x\,\text{L}\). 3. The salt mass stays at \(1500\,\text{g}\), so write \(20(50 + x) = 1500\). 4. Distribute and solve: \(1000 + 20x = 1500\), so \(20x = 500\) and \(x = 25\).

A suitable equation is \(20(50 + x) = 1500\). Add \(25\,\text{L}\) of pure water.

- Use the same height in both area expressions. - Express the longer banner's length in terms of the shorter banner's length. - Add the two areas and set the sum equal to the combined area. - Distribute before solving.

1. Let \(l\) inches be the length of the shorter banner. The longer banner is \(l + 10\) inches. 2. Add the two areas: \(24l + 24(l + 10) = 1680\). 3. Distribute and combine like terms: \(48l + 240 = 1680\). 4. Subtract \(240\): \(48l = 1440\). 5. Divide by \(48\): \(l = 30\). 6. The longer banner is \(30\,\text{in.} + 10\,\text{in.} = 40\,\text{in.}\).

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

- 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.

- Account for the freight train's earlier departure. - Write a distance expression for each train. - The two distances add to the distance between the stations. - Convert the solved time from hours to minutes.

1. Let \(x\) be the passenger train's travel time in hours before the trains meet. 2. The freight train leaves \(0.5\) hour earlier, so it travels for \(x + 0.5\) hours. 3. Their distances add to \(150\) miles: \(50(x + 0.5) + 75x = 150\). 4. Distribute and combine like terms: \(50x + 25 + 75x = 150\), so \(125x + 25 = 150\). 5. Subtract \(25\): \(125x = 125\), so \(x = 1\). 6. One hour is \(60\) minutes.

The passenger train meets the freight train \(60\) minutes after leaving Station B.

- Add the distances the groups hike toward each other. - Convert the delayed start to hours. - In part b), Group A's hiking time is longer than Group B's. - For part c), focus on the remaining distance when Group B starts.

1. For part a), let \(x\) be the meeting time in hours. Write \(4x + 5x = 18\). 2. Solve: \(9x = 18\), so \(x = 2\) hours. 3. For part b), \(45\) minutes is \(0.75\) hour. Let \(y\) be Group B's hiking time in hours. Group A hikes for \(y + 0.75\) hours. 4. Write \(4(y + 0.75) + 5y = 18\). 5. Solve: \(4y + 3 + 5y = 18\), so \(9y = 15\) and \(y = \frac{5}{3}\) hours. 6. Convert to minutes: \(\frac{5}{3} \cdot 60 = 100\) minutes. 7. In part b), Group A reduces the distance between the groups before Group B begins, so the groups have less distance left to close while both are hiking.

a) The groups meet after \(2\) hours. b) Group B hikes for \(100\) minutes before the meeting. c) Group A has already reduced the distance between the groups before Group B starts, so Group B needs less hiking time.

- Use one time variable measured from the earlier start. - Subtract \(0.5\) hour for the later hiker's travel time. - Add the two distances and set the sum equal to the trail length. - Use the earlier hiker's speed to find the meeting location.

1. Let \(t\) be the number of hours after \(2{:}00\) p.m. when they meet. 2. Mr. Smith travels \(4t\) miles. Because Ms. Weber starts \(0.5\) hour later, she travels \(6(t - 0.5)\) miles. 3. Their distances add to \(17\): \(4t + 6(t - 0.5) = 17\). 4. Distribute and solve: \(4t + 6t - 3 = 17\), so \(10t = 20\) and \(t = 2\). 5. Two hours after \(2{:}00\) p.m. is \(4{:}00\) p.m. 6. Mr. Smith travels \(4 \cdot 2 = 8\) miles.

a) They meet at \(4{:}00\) p.m. b) They meet \(8\) miles from Mr. Smith's starting point.

- Convert all power values to watts. - Use one variable for the number of one panel type. - The two panel counts add to \(28\). - In part b, keep the Type A count unchanged and recalculate the required Type B output.

1. Convert the totals: \(10\,\text{kW} = 10{,}000\,\text{W}\) and \(12\,\text{kW} = 12{,}000\,\text{W}\). 2. Let \(x\) be the number of Type A panels. Then \(28 - x\) is the number of Type B panels. Write \(320x + 400(28 - x) = 10{,}000\). 3. Distribute and solve: \(320x + 11{,}200 - 400x = 10{,}000\), so \(-80x = -1200\) and \(x = 15\). 4. There are \(15\) Type A panels and \(28 - 15 = 13\) Type B panels. 5. The \(15\) Type A panels produce \(15 \cdot 320 = 4800\,\text{W}\). The Type B panels must produce \(12{,}000 - 4800 = 7200\,\text{W}\). 6. The new number of Type B panels is \(7200 \div 400 = 18\), so \(18 - 13 = 5\) additional panels are needed.

a) There are \(15\) Type A panels and \(13\) Type B panels. b) \(5\) additional Type B panels are needed.