- Substitute the proposed value for every occurrence of \(x\). - Evaluate the left and right sides separately. - The value is a solution only when both sides are equal.

1. Substitute \(x = 3\): \(7 \cdot 3 - 5 = 16\). The equation is true, so \(3\) is a solution. 2. Substitute \(x = 4\): \(18 - 3 \cdot 4 = 6\). The equation is true, so \(4\) is a solution. 3. Substitute \(x = 1\): \(2(1 + 4) = 10\). The equation is true, so \(1\) is a solution. 4. Substitute \(x = 5\): the left side is \(4 \cdot 5 + 1 = 21\), while the right side is \(2 \cdot 5 + 9 = 19\). The equation is false, so \(5\) is not a solution.

1) Yes 2) Yes 3) Yes 4) No

- Which operation was done after the multiplication by \(3\)? - How can you undo that operation while keeping the two sides equal?

1. Subtract \(4\) from both sides to undo the added \(4\). 2. This gives \(3x=15\).

Subtract \(4\) from both sides: \(3x=15\).

- What operation is keeping \(x\) from being alone? - Do the inverse operation to both sides.

1. Divide both sides of \(2x=12\) by \(2\). 2. The solution is \(x=6\); in the original equation, \(2\cdot6+5=17\).

\(x=6\)

- Imagine the two sides of the equation must stay balanced. - What happened to the right side when Omar changed the left side?

1. Subtracting only on the left changes the equality: \(4x=15\) is not equivalent to \(4x+3=15\). 2. Subtract \(3\) from both sides to get \(4x=12\).

No. Subtract \(3\) from both sides; the correct next equation is \(4x=12\).

- Which operation takes place after dividing \(x\) by \(4\)? - Undo that operation on both sides.

1. Adding \(2\) to both sides removes the \(-2\) after the division. 2. The equivalent equation is \(\frac{x}{4}=5\), with the division ready to undo next.

\(\frac{x}{4}=5\); adding \(2\) undoes the subtraction without changing the solution.

- Which two bricks add to make the brick above them? - Write the sum with a letter for the missing number.

1. If the unknown brick is \(x\), the wall gives \(x+4=9\). 2. Subtract \(4\) from both sides to get \(x=5\).

\(5\), since \(x+4=9\) gives \(x=5\).

- Isolate the term containing \(x\) first. - How does the sign of the right-hand side affect the solution? - Which of your answers could be counted on an ordinary whole-number tally?

1. For a), subtract \(5\): \(2x=-14\), so \(x=-7\), a negative integer. 2. For b), add \(3\): \(4x=-12\), so \(x=-3\), also a negative integer. 3. For c), add \(4\): \(3x=33\), so \(x=11\), a positive whole number. 4. Only \(11\) is a whole number, because whole numbers do not include negatives.

a) \(x=-7\), negative; not a whole number. b) \(x=-3\), negative; not a whole number. c) \(x=11\), positive; a whole number.

- Think about the value the entire expression in parentheses must have. - Work backward through the operations while preserving equality. - Substitute your result into the original equation to verify it.

1. Multiply both sides by \(0.5\): \(-12.4+\square=-5\). 2. Add \(12.4\): \(\square=7.4\).

\(7.4\)

- Treat the expression inside the grouping symbols as one quantity at first. - Undo the outer operation before the inner operation. - Multiplication and division are inverse operations. - Check each answer in the original equation.

1. For a), divide by \(3\): \(x + 1.2 = 5\). Subtract \(1.2\): \(x = 3.8\). 2. For b), multiply by \(4\): \(18.6 - x = 14\). Subtract \(18.6\): \(-x = -4.6\). Multiply by \(-1\): \(x = 4.6\).

a) \(x = 3.8\) b) \(x = 4.6\)

- Identify the operation performed last on the variable. - Undo addition or subtraction before undoing multiplication or division. - Perform the same inverse operation on both sides. - Follow the sign rules when dividing by a negative number.

1. For a), subtract \(14\): \(5x = 25\). Divide by \(5\): \(x = 5\). 2. For b), subtract \(18\): \(-2x = -8\). Divide by \(-2\): \(x = 4\). 3. For c), subtract \(5\): \(\frac{x}{3} = 7\). Multiply by \(3\): \(x = 21\).

a) \(x = 5\) b) \(x = 4\) c) \(x = 21\)

- In part a, you can divide before working inside the parentheses. - Perform each operation on both sides of the equation. - Pay close attention to a negative coefficient on the variable. - Substitute each result to check it.

1. For a), divide by \(4\): \(x + 2 = 7\). Subtract \(2\): \(x = 5\). 2. For b), subtract \(21\): \(-9 = 3x\). Divide by \(3\): \(x = -3\). 3. For c), subtract \(15\): \(-5x = 20\). Divide by \(-5\): \(x = -4\).

a) \(x = 5\) b) \(x = -3\) c) \(x = -4\)

- Isolate the variable term before dividing. - Follow the sign rules when dividing by a negative number. - Multiplying by a reciprocal can undo multiplication by a fraction. - Apply each operation to both sides.

1. For a), subtract \(14\): \(6x = 36\). Divide by \(6\): \(x = 6\). 2. For b), subtract \(18\): \(-4y = -16\). Divide by \(-4\): \(y = 4\). 3. For c), add \(3\): \(0.5z = 7.5\). Divide by \(0.5\): \(z = 15\). 4. For d), subtract \(4\): \(\frac{2}{3}b = 6\). Multiply by \(\frac{3}{2}\): \(b = 9\).

a) \(x = 6\) b) \(y = 4\) c) \(z = 15\) d) \(b = 9\)

- Use a variable for the starting number before doing any arithmetic. - Translate the operations in the order they occur. - Isolate the variable by undoing the outside addition and then the division. - Check in the original equation.

1. Let \(x\) be the number and write \(\frac{x}{4}+15=30\). 2. Subtract \(15\) from both sides: \(\frac{x}{4}=15\). 3. Multiply both sides by \(4\): \(x=60\). 4. Check: \(60\div4+15=30\).

Equation: \(\frac{x}{4}+15=30\) Solution: \(x=60\)

- Use one letter for the starting number so you can represent the whole calculation. - What must both sides equal after your first change to the equation? - Does the value you find produce the stated result when checked from the start?

1. Let \(x\) be the number. The equation is \(6x+4=40\). 2. Subtract \(4\) from both sides: \(6x=36\). 3. Divide by \(6\): \(x=6\). 4. Check: \(6\cdot6+4=36+4=40\), as required.

Equation: \(6x+4=40\). Solution: \(x=6\).

- Use a variable for the starting number. - Preserve the order of multiplication and subtraction when writing the equation. - Undo the subtraction before the multiplication. - Check by substituting into the equation.

1. Let \(x\) be the starting number and write \(4x-24=100\). 2. Add \(24\) to both sides: \(4x=124\). 3. Divide both sides by \(4\): \(x=31\). 4. Check: \(4\cdot31-24=100\).

Equation: \(4x-24=100\) Starting number: \(31\)

- The subtraction happens before the division, so the numerator must represent the result after subtracting. - Use inverse operations on the equation in reverse order. - Keep the grouping intact until the division has been undone. - Check the complete operation sequence.

1. Let \(x\) be the starting number and write \(\frac{x-50}{5}=40\). 2. Multiply both sides by \(5\): \(x-50=200\). 3. Add \(50\) to both sides: \(x=250\). 4. Check: \((250-50)\div5=40\).

Equation: \(\frac{x-50}{5}=40\) Starting number: \(250\)

- Treat the entire expression in parentheses as one unknown quantity first. - What must be subtracted from \(1200\) to get \(700\)? - Once you know the value of the parentheses, solve for \(x\).

1. Treat the expression in parentheses as one quantity. Since \(1200 - 700 = 500\), the equation becomes \(x + 150 = 500\). 2. Subtract \(150\): \(x = 500 - 150 = 350\). 3. Check: \(1200 - (350 + 150) = 1200 - 500 = 700\).

\(x = 350\)

- Undo addition or subtraction first. - Then undo multiplication to isolate \(x\). - Check each solution in the original equation.

1. For a), add \(6\): \(7x=35\). Divide by \(7\): \(x=5\). 2. For b), subtract \(4\): \(9x=54\). Divide by \(9\): \(x=6\). 3. For c), add \(14\): \(8x=64\). Divide by \(8\): \(x=8\). 4. For d), subtract \(27\): \(5x=25\). Divide by \(5\): \(x=5\).

a) \(x=5\) b) \(x=6\) c) \(x=8\) d) \(x=5\)

- Solve each equation by working backward. - Undo addition or subtraction first. - Then undo multiplication by dividing. - Check each solution by substitution.

1. In a), subtract \(6\): \(4x = 24\). Then divide by \(4\): \(x = 6\). 2. In b), add \(5\): \(9x = 45\). Then divide by \(9\): \(x = 5\). 3. In c), subtract \(3\): \(7x = 49\). Then divide by \(7\): \(x = 7\).

a) \(x = 6\) b) \(x = 5\) c) \(x = 7\)

- Undo the operation outside the parentheses first. - Use inverse operations to isolate \(x\). - Check each solution in the original equation.

1. a) Divide by \(4\): \(x + 8 = 10\). Then subtract \(8\): \(x = 2\). 2. b) Divide by \(6\): \(x - 2 = 6\). Then add \(2\): \(x = 8\). 3. c) Subtract \(15\): \(3x = 30\). Then divide by \(3\): \(x = 10\). 4. d) Subtract \(7\): \(4x = 24\). Then divide by \(4\): \(x = 6\).

a) \(x = 2\) b) \(x = 8\) c) \(x = 10\) d) \(x = 6\)

- Undo the outside operation first when a factor multiplies parentheses. - In part c), isolate the variable term before dividing by its coefficient. - Check each solution in the original equation.

1. For a), divide both sides by \(6\): \(x+5=12\). Subtract \(5\): \(x=7\). 2. For b), divide both sides by \(4\): \(x-12=12\). Add \(12\): \(x=24\). 3. For c), subtract \(60\): \(-3x=-39\). Divide by \(-3\): \(x=13\).

a) \(x=7\) b) \(x=24\) c) \(x=13\)

- Undo the multiplication outside the parentheses first. - Then use addition or subtraction to isolate \(x\). - Check each solution by substitution.

1. a) Divide by \(2\): \(x + 18 = 25\). Subtract \(18\): \(x = 7\). 2. b) Divide by \(2\): \(x - 3 = 17\). Add \(3\): \(x = 20\). 3. c) Divide by \(5\): \(x - 4 = 5\). Add \(4\): \(x = 9\).

a) \(x = 7\) b) \(x = 20\) c) \(x = 9\)

- The increase happens before the division, so group the increased quantity together. - Undo the outer division before undoing the addition. - Keep the equation balanced at each step. - Check the original operation order.

1. Let \(x\) be the mystery number and write \(\frac{x+120}{3}=100\). 2. Multiply both sides by \(3\): \(x+120=300\). 3. Subtract \(120\): \(x=180\). 4. Check: \((180+120)\div3=100\).

Equation: \(\frac{x+120}{3}=100\) Mystery number: \(180\)

- Translate the division and subtraction in the order stated. - Undo the subtraction before undoing the division. - Apply the same inverse operation to both sides. - Check the value in the original equation.

1. Let \(x\) be the number and write \(\frac{x}{4}-80=20\). 2. Add \(80\) to both sides: \(\frac{x}{4}=100\). 3. Multiply both sides by \(4\): \(x=400\). 4. Check: \(400\div4-80=20\).

Equation: \(\frac{x}{4}-80=20\) Solution: \(x=400\)

- Isolate the variable term before dividing. - Substitute the proposed value into the original equation. - A value is a solution only when both sides are equal.

1. Add \(3\): \(4y = 20\). 2. Divide by \(4\): \(y = 5\). 3. For \(y = 4\), the left side is \(4 \cdot 4 - 3 = 13\). 4. Since \(13 \ne 17\), \(y = 4\) is not a solution.

a) \(y = 5\) b) \(y = 4\) is not a solution because the left side equals \(13\), not \(17\).

- Move the constant term away from the variable term first. - Divide by the coefficient of the variable. - Follow the sign rules when dividing by a negative number. - Apply each operation to both sides.

1. For a), add \(12\): \(5x = 30\). Divide by \(5\): \(x = 6\). 2. For b), subtract \(7\): \(-2y = 6\). Divide by \(-2\): \(y = -3\). 3. For c), subtract \(4\): \(\frac{1}{2}z = -3\). Multiply by \(2\): \(z = -6\). 4. For d), add \(0.4\): \(1.2w = 2.4\). Divide by \(1.2\): \(w = 2\).

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

- Solve each equation independently. - Isolate the variable term before dividing. - Compare the three final values.

1. For equation 1), add \(2.8\): \(4x = 8\). Divide by \(4\): \(x = 2\). 2. For equation 2), subtract \(15\): \(-3x = -9\). Divide by \(-3\): \(x = 3\). 3. For equation 3), subtract \(1.5\): \(0.5x = 1\). Divide by \(0.5\): \(x = 2\). 4. Equations 1) and 3) both have solution \(x = 2\).

1) \(x = 2\) 2) \(x = 3\) 3) \(x = 2\) Equations 1) and 3) have the same solution.

- Use a variable for the points per correct answer. - Write an expression that combines the question points and the bonus. - Remove the bonus before dividing by the number of correct answers.

1. Let \(x\) be the points earned for each correct answer. 2. Write the equation \(6x + 25 = 145\). 3. Subtract the bonus: \(6x = 120\). 4. Divide by \(6\): \(x = 20\).

Each correct answer is worth \(20\) points.

- Move the constant term away from the variable term first. - Divide by the coefficient of \(x\). - Apply each operation to both sides. - Remember that \(-x\) means \(-1 \cdot x\).

1. Subtract \(9\): \(3x = -9\). Divide by \(3\): \(x = -3\). 2. Add \(4\): \(-2x = 10\). Divide by \(-2\): \(x = -5\). 3. Add \(5\): \(20 = 4x\). Divide by \(4\): \(x = 5\). 4. Subtract \(10\): \(-x = 3\). Multiply by \(-1\): \(x = -3\).

1) \(x = -3\) 2) \(x = -5\) 3) \(x = 5\) 4) \(x = -3\)

- Combine like terms before solving. - Use an inverse operation to remove the constant term. - Divide by the variable coefficient at the end.

1. For a), combine like terms: \(9z - 14 = 4\). 2. Add \(14\): \(9z = 18\). Divide by \(9\): \(z = 2\). 3. For b), combine like terms: \(11w - 5 = 50\). 4. Add \(5\): \(11w = 55\). Divide by \(11\): \(w = 5\).

a) \(z = 2\) b) \(w = 5\)

- Combine terms that contain the same variable. - Combine constant terms separately. - After simplifying, use two inverse operations to isolate the variable.

1. Combine like terms: \(9x+9=81\). Subtract \(9\): \(9x=72\). Divide by \(9\): \(x=8\). 2. Combine the constants: \(3y+9=24\). Subtract \(9\): \(3y=15\). Divide by \(3\): \(y=5\). 3. Combine like terms: \(7z-4=31\). Add \(4\): \(7z=35\). Divide by \(7\): \(z=5\).

1) \(x=8\) 2) \(y=5\) 3) \(z=5\)

- Combine variable terms and constant terms separately. - Isolate the variable term before dividing. - Track negative signs carefully. - Check each solution in the original equation.

1. Combine like terms: \(10x - 5 = 25\). 2. Add \(5\): \(10x = 30\). Divide by \(10\): \(x = 3\). 3. Combine like terms: \(-8y + 12 = 36\). 4. Subtract \(12\): \(-8y = 24\). Divide by \(-8\): \(y = -3\).

1) \(x = 3\) 2) \(y = -3\)

- Combine the terms containing \(x\) first. - Combine the constant terms. - Isolate the variable term before dividing. - Use division to undo multiplication by \(11\).

1. Combine like terms: \(11x - 6 = 27\). 2. Add \(6\) to both sides: \(11x = 33\). 3. Divide both sides by \(11\): \(x = 3\).

\(x = 3\)

- Divide by the factor outside the parentheses first. - Use the inverse operation to remove the constant inside the parentheses. - Keep the equation balanced at every step. - Check each answer by substitution.

1. Divide by \(6\): \(x + 4 = 8.5\). Subtract \(4\): \(x = 4.5\). 2. Divide by \(9\): \(x - 12 = 2\). Add \(12\): \(x = 14\). 3. Divide by \(4\): \(3x + 2 = 14\). Subtract \(2\): \(3x = 12\). Divide by \(3\): \(x = 4\). 4. Divide by \(0.4\): \(x - 5 = 5\). Add \(5\): \(x = 10\).

1) \(x = 4.5\) 2) \(x = 14\) 3) \(x = 4\) 4) \(x = 10\)

- Identify which symbol has no assigned value. - Substitute the known values before solving. - Undo multiplication before undoing the subtraction inside the parentheses.

1. The remaining unknown variable is \(x\). 2. Substitute the known values: \(3=0.25(x-5)\). 3. Divide by \(0.25\): \(12=x-5\). 4. Add \(5\): \(x=17\).

The unknown variable is \(x\), and \(x=17\).

- Identify the operation applied to the variable term last. - Undo that operation on both sides first. - Substitute your solution into the original equation to check it.

1. Add \(7\) to both sides: \(7x=49\). 2. Divide both sides by \(7\): \(x=7\). 3. The expression \(7x-7\) means \(7\) is subtracted after the product \(7x\) is formed. Reversing the operations in reverse order isolates the variable term first.

\(x=7\). Add \(7\) first to isolate \(7x\), then divide by \(7\).

- Which part can you remove from the left side while keeping the coefficient of \(x\)? - How could you check the sign of your proposed answer? - Does substituting the answer make both sides equal?

1. Subtract \(5\) from both sides: \(-2x=8\). 2. Divide by \(-2\): \(x=-4\). 3. Check: \(-2(-4)+5=8+5=13\).

Subtract \(5\): \(-2x=8\). Divide by \(-2\): \(x=-4\). Check: \(-2(-4)+5=13\).

- What part of the left side can you isolate first? - What must the numerator be for that fraction to have the required value? - Can you substitute your answer to check both operations?

1. Subtract \(4\): \(\frac{x}{3}=5\). 2. Multiply by \(3\): \(x=15\). 3. Check: \(\frac{15}{3}+4=5+4=9\).

\(x=15\). Check: \(\frac{15}{3}+4=9\).

- Read the operation at each branch of the tree. - Begin with the final operation and undo it. - Does the input you find reproduce the displayed result?

1. The tree represents \(4x+6=22\). 2. Subtract \(6\) to get \(4x=16\). 3. Divide by \(4\): \(x=4\). 4. Check: \(4\cdot4+6=22\).

\(4x+6=22\); \(x=4\).

- Undo addition or subtraction before undoing multiplication in forms such as \(px+q=r\). - In a form such as \(p(x+q)=r\), undo the outside multiplication first. - Keep track of the negative coefficient in part b. - Check each result by substitution.

1. For a), subtract \(12.45\): \(2.5x=17.75\). Divide by \(2.5\): \(x=7.1\). 2. For b), subtract \(45.6\): \(-1.2x=-33.6\). Divide by \(-1.2\): \(x=28\). 3. For c), divide by \(1.2\): \(x+2.5=7\). Subtract \(2.5\): \(x=4.5\).

a) \(x=7.1\) b) \(x=28\) c) \(x=4.5\)

- Rewrite all fractions with denominator \(12\). - What expression appears in the numerator of the second equivalent fraction? - Solve the resulting equation for \(x\).

1. Rewrite the fractions with denominator \(12\): \(\frac{1}{6}=\frac{2}{12}\) and \(\frac{x}{4}=\frac{3x}{12}\). 2. The numerator equation is \(2+3x=11\). 3. Subtract \(2\): \(3x=9\). Divide by \(3\): \(x=3\). 4. Check: \(\frac{1}{6}+\frac{3}{4}=\frac{2}{12}+\frac{9}{12}=\frac{11}{12}\).

\(x=3\)

- Identify the operations that have been applied to the missing number in each equation. - Undo the equation in a way that preserves equality on both sides. - Substitute each result to check the original equation.

1. Add \(1.5\): \(2\square=-5\). Divide by \(2\): \(\square=-2.5\). 2. Subtract \(0.4\): \((-0.2)\cdot\square=1\). Divide by \(-0.2\): \(\square=-5\). 3. Subtract \(0.5\): \(\square\div(-10)=0.25\). Multiply by \(-10\): \(\square=-2.5\). 4. Subtract \(0.8\): \(-2\square=0.8\). Divide by \(-2\): \(\square=-0.4\).

a) \(-2.5\) b) \(-5\) c) \(-2.5\) d) \(-0.4\)

- Identify the operations applied to the missing number in each equation. - Reverse those operations while keeping both sides equivalent. - Substitute each result to verify the original equation.

1. Subtract \(1.5\): \(2\square=-4\). Divide by \(2\): \(\square=-2\). 2. Add \(2.4\): \(6\square=-2.4\). Divide by \(6\): \(\square=-0.4\). 3. Subtract \(1\): \(\square\cdot\left(-\frac{2}{3}\right)=4\). Divide by \(-\frac{2}{3}\): \(\square=4\cdot\left(-\frac{3}{2}\right)=-6\).

a) \(-2\) b) \(-0.4\) c) \(-6\)

- How could you isolate the term containing \(x\)? - Check each result by substituting it into the original equation. - Which answers could describe a count of objects without using a fraction?

1. For a), subtract \(2\): \(-1.5x=0.75\), so \(x=-0.5\), not a whole number. 2. For b), add \(1\): \(0.5x=3\), so \(x=6\), a whole number. 3. For c), add \(1.2\): \(-x=3.6\), so \(x=-3.6\), not a whole number.

a) \(x=-0.5\); not a whole number. b) \(x=6\); a whole number. c) \(x=-3.6\); not a whole number.

- First solve each equation without worrying about the restriction. - Which solutions could represent a positive count of items? - Does an algebraically correct solution always satisfy an extra restriction?

1. For a), \(4x=20\), so \(x=5\), which is allowed. 2. For b), \(3x=-7\), so \(x=-\frac{7}{3}\), which is not a positive whole number. 3. For c), \(2x=7\), so \(x=3.5\), which is not a whole number. 4. For d), \(-5x=-5\), so \(x=1\), which is allowed.

a) \(x=5\) b) No allowed solution. c) No allowed solution. d) \(x=1\)

- Solve the equations before deciding which answers are allowed. - Does an integer have to be positive? - If the solution lies between two consecutive integers, could either integer solve the original equation exactly?

1. For a), \(3x=3\), so \(x=1\), an allowed integer. 2. For b), \(5x=15\), so \(x=3\), an allowed integer. 3. For c), \(2x=-7\), so \(x=-3.5\), which is not an integer. 4. For d), \(4x=10\), so \(x=2.5\), which is not an integer.

a) \(x=1\) b) \(x=3\) c) No allowed solution. d) No allowed solution.

- Solve the equation before thinking about restrictions. - Is your answer an integer in the first case? - A restriction can exclude a numerical solution without changing the equation's arithmetic.

1. Subtract \(6\): \(2x=5\). Divide by \(2\): \(x=2.5\). 2. If only integers are allowed, \(2.5\) is excluded. Youssef is right about there being no allowed solution. 3. If fractions and decimals are allowed, \(2.5\) solves the equation. Youssef is wrong.

1. Youssef is right: no integer solves the equation. 2. Youssef is wrong: \(x=2.5\) is a solution.

- Isolate the variable term first. - Use decimal points and track negative signs carefully. - For a grouped expression, undo the outer operation first. - Check each solution in the original equation.

1. For a), add \(4\): \(0.5x = 5\). Divide by \(0.5\): \(x = 10\). 2. For b), subtract \(7\): \(-1.5x = 6\). Divide by \(-1.5\): \(x = -4\). 3. For c), divide by \(2\): \(3x + 4 = -2\). Subtract \(4\): \(3x = -6\). Divide by \(3\): \(x = -2\).

a) \(x = 10\) b) \(x = -4\) c) \(x = -2\)

- Simplify each side before isolating the variable. - A minus sign before parentheses changes the signs inside. - In some equations, dividing first removes the grouping symbols efficiently.

1. For a), divide by \(2\): \(x - 5 = 7\). Add \(5\): \(x = 12\). 2. For b), combine like terms: \(4x + 8 = 20\). Subtract \(8\), then divide by \(4\): \(x = 3\). 3. For c), divide by \(3\): \(5 = y + 2\). Subtract \(2\): \(y = 3\). 4. For d), remove the parentheses: \(5 - x + 2 = 10\). Then \(7 - x = 10\), so \(-x = 3\) and \(x = -3\).

a) \(x = 12\) b) \(x = 3\) c) \(y = 3\) d) \(x = -3\)

- Try transforming the first equation into the second using the same operation on both sides. - When distributing, multiply every term inside the parentheses. - You may solve both equations and compare their solution sets.

1. For a), divide every term in \(7x - 14 = 21\) by \(7\). This gives \(x - 2 = 3\), exactly the second equation. Both equations have solution \(x = 5\), so they are equivalent. 2. For b), distributing in the first equation gives \(3a + 12 = 15\), not \(3a + 4 = 15\). The first equation has solution \(a = 1\), while the second has solution \(a = \frac{11}{3}\). Therefore, they are not equivalent.

a) Yes. The equations are equivalent. b) No. The equations are not equivalent.

- Apply an operation to every term on both sides of an equation. - Multiplying by \(2\) must affect both terms on the left in part b. - Solving both equations can confirm whether their solution sets match.

1. For a), subtract \(12\) from both sides of the first equation to get \(-4y = -4\). Multiplying both sides by \(-1\) gives \(4y = 4\). The transformation is valid, and both equations have solution \(y = 1\). 2. For b), multiplying the first equation by \(2\) must multiply every term on both sides. The correct result is \(z + 10 = 22\), not \(z + 5 = 22\). The transformation is not valid.

a) Yes, the equations are equivalent. b) No, the equations are not equivalent.

- An equivalent transformation must be applied to both sides and to every affected term. - Solve the original equation first. - You can test whether each new equation has the same solution.

1. The original equation gives \(5x = 25\), so \(x = 5\). 2. For a), adding \(10\) to both sides gives \(5x = 25\). This is equivalent. 3. For b), only the term \(5x\) was divided by \(5\). Dividing the entire left side would give \(x - 2\), not \(x - 10\). This equation has solution \(x = 13\), so it is not equivalent. 4. For c), dividing every term of the original equation by \(5\) gives \(x - 2 = 3\). This is equivalent. 5. For d), subtracting \(15\) from both sides gives \(5x - 25 = 0\). This is equivalent.

a) Yes; add \(10\) to both sides. b) No; the entire left side was not divided by \(5\). c) Yes; divide every term by \(5\). d) Yes; subtract \(15\) from both sides.

- Solve Equations A and B separately. - A known solution makes an equation true when substituted. - Substitute \(x = 4\) into the equation with the missing number.

1. For Equation A, add \(8\): \(5x = 20\). Divide by \(5\): \(x = 4\). 2. For Equation B, divide by \(2\): \(x + 1 = 5\). Subtract \(1\): \(x = 4\). 3. Substitute \(x = 4\) into \(4x + \square = 20\): \(16 + \square = 20\). Subtract \(16\): \(\square = 4\).

a) Both equations have solution \(x = 4\). b) The missing number is \(4\).

- Combine known constants before isolating the variable term. - Undo addition or subtraction before undoing multiplication or division. - Use a common denominator when combining fractions.

1. For a), combine the known terms: \(12.5+2.5=15\). Then \(15-0.5x=10\). Subtract \(15\): \(-0.5x=-5\). Divide by \(-0.5\): \(x=10\). 2. For b), combine the known fractions: \(\frac{3}{8}-\frac{1}{8}=\frac{1}{4}\). Then \(\frac{1}{4}x+\frac{1}{4}=\frac{7}{8}\). Subtract \(\frac{1}{4}\): \(\frac{1}{4}x=\frac{5}{8}\). Multiply by \(4\): \(x=\frac{5}{2}=2.5\). 3. For c), subtract \(0.8\): \(0.4x=2\). Divide by \(0.4\): \(x=5\). 4. For d), add \(0.5\): \(\frac{x}{4}=2\). Multiply by \(4\): \(x=8\).

a) \(x=10\) b) \(x=\frac{5}{2}\), or \(x=2.5\) c) \(x=5\) d) \(x=8\)

- Rewrite fractions as decimals when that makes the constants easier to combine. - Undo the outer operation before the inner operation when parentheses are present. - Keep track of the negative coefficient in part c).

1. For a), \(\frac{3}{4}=0.75\), and \(0.75-0.25=0.5\). The equation becomes \(0.5+0.5x=1\). Subtract \(0.5\): \(0.5x=0.5\). Divide by \(0.5\): \(x=1\). 2. For b), divide both sides by \(2\): \(x+1.5=3.5\). Subtract \(1.5\): \(x=2\). 3. For c), subtract \(10\): \(-\frac{x}{2}=-1.5\). Multiply by \(-2\): \(x=3\).

a) \(x=1\) b) \(x=2\) c) \(x=3\)

- Work backward through the operations. - Which operation undoes division by \(4\)? - What operation undoes subtracting \(24\)? - Check your answer by running it through the machine from the beginning.

1. Work backward from \(10\). Undo division by \(4\): \(10 \cdot 4 = 40\). 2. Undo subtracting \(24\): \(40 + 24 = 64\). 3. Undo multiplication by \(8\): \(64 \div 8 = 8\). 4. Check: \((8 \cdot 8 - 24) \div 4 = 40 \div 4 = 10\).

\(8\)

- Reverse the chain of operations one step at a time. - What must you do first to undo division by \(12\)? - At each step, use the inverse operation.

1. Work backward from \(11\). Undo division by \(12\): \(11 \cdot 12 = 132\). 2. Undo subtracting \(28\): \(132 + 28 = 160\). 3. Undo multiplication by \(4\): \(160 \div 4 = 40\). 4. Undo adding \(17\): \(40 - 17 = 23\). 5. Check: \(((23 + 17) \cdot 4 - 28) \div 12 = 132 \div 12 = 11\).

\(23\)

- Work backward through the operations. - Undo the last operation first. - Record each intermediate value. - Check by applying the original steps.

1. Undo the subtraction: \(160 + 30 = 190\). 2. Undo the addition: \(190 - 150 = 40\). 3. Undo the division: \(40 \cdot 5 = 200\). 4. Check: \(200 \div 5 + 150 - 30 = 160\).

The original number is \(200\).

- Translate each statement into its own equation before solving. - In each equation, undo the added or subtracted constant before the multiplication. - Keep the variables separate because the two people may have different numbers. - Compare only after both equations are solved.

1. Luke’s statement gives \(6x+4=40\). Subtract \(4\) and divide by \(6\): \(x=6\). 2. Sara’s statement gives \(5y-3=32\). Add \(3\) and divide by \(5\): \(y=7\). 3. Since \(7>6\), Sara chose the greater number.

Luke: \(6x+4=40\), so \(x=6\). Sara: \(5y-3=32\), so \(y=7\). Sara chose the greater number.

- Write one equation for each operation sequence before calculating. - Luke’s variable is multiplied before a subtraction; Sarah’s is divided before an addition. - Use inverse operations in the reverse order from the equation. - Compare the solutions after both are found.

1. Luke’s equation is \(2x-20=80\). Add \(20\) and divide by \(2\): \(x=50\). 2. Sarah’s equation is \(\frac{y}{2}+20=80\). Subtract \(20\) and multiply by \(2\): \(y=120\). 3. Since \(120>50\), Sarah chose the greater number.

Luke: \(2x-20=80\), so \(x=50\). Sarah: \(\frac{y}{2}+20=80\), so \(y=120\). Sarah chose the greater number.

- Follow the order of operations described in each statement. - A sum that is multiplied needs parentheses. - Simplify fixed numerical expressions, then use two inverse operations to isolate the variable.

1. Write \(2n+17=45\). Subtract \(17\): \(2n=28\). Divide by \(2\): \(n=14\). 2. Write \(100-2z=4\cdot18\). Simplify to \(100-2z=72\). Subtract \(100\): \(-2z=-28\). Divide by \(-2\): \(z=14\). 3. Write \(5(k+2.2)=25\). Divide by \(5\): \(k+2.2=5\). Subtract \(2.2\): \(k=2.8\).

1) \(n=14\) 2) \(z=14\) 3) \(k=2.8\), or \(2\frac{4}{5}\)

- How can you rewrite the mixed number as an improper fraction? - Which term should you remove first to isolate the variable term? - How do you divide by a fraction? - How can substitution verify your solution?

1. Rewrite the mixed number: \(2\frac{1}{6} = \frac{13}{6}\). 2. Subtract \(\frac{1}{2}\) from both sides: \(\frac{2}{3}x = \frac{13}{6} - \frac{3}{6} = \frac{10}{6} = \frac{5}{3}\). 3. Divide by \(\frac{2}{3}\): \(x = \frac{5}{3} \div \frac{2}{3} = \frac{5}{3} \cdot \frac{3}{2} = \frac{5}{2} = 2.5\). 4. Check by substitution: \(\frac{2}{3} \cdot \frac{5}{2} + \frac{1}{2} = \frac{5}{3} + \frac{1}{2} = \frac{10}{6} + \frac{3}{6} = \frac{13}{6} = 2\frac{1}{6}\). The equation is true.

\(x = \frac{5}{2} = 2.5\)

- Use a variable for the original amount. - Express double the amount and half the amount algebraically. - Combine the variable terms before solving.

1. Let \(x\) be the original amount of money. 2. Write the equation \(2x + 0.5x + 12 = 107\). 3. Combine like terms: \(2.5x + 12 = 107\). 4. Subtract \(12\): \(2.5x = 95\). 5. Divide by \(2.5\): \(x = 38\).

The piggy bank originally contained \(\$38\).

- Convert both sides of the scale to the same unit. - Represent the four identical boxes with a variable term. - A level scale means the expressions on both sides are equal. - Remove the known weight before dividing among the four boxes.

1. Convert the right-side weight to ounces: \(3\,\text{lb} = 48\,\text{oz}\). 2. Let \(x\) be the weight of one candy box in ounces. Write \(4x + 8 = 48\). 3. Subtract \(8\): \(4x = 40\). 4. Divide by \(4\): \(x = 10\).

Each box of candy weighs \(10\,\text{oz}\).

- First isolate the term containing the variable. - Track the sign when dividing by a negative number. - You may rewrite a mixed number as a fraction or decimal. - Use the inverse of the operation applied to the variable.

1. Subtract \(12\): \(-5x = -30\). Divide by \(-5\): \(x = 6\). 2. Add \(3.2\): \(\frac{a}{8} = -1.5\). Multiply by \(8\): \(a = -12\). 3. Rewrite \(2\frac{1}{4}\) as \(2.25\). Subtract \(7\): \(2.25y = -4.5\). Divide by \(2.25\): \(y = -2\).

1) \(x = 6\) 2) \(a = -12\) 3) \(y = -2\)

- Combine the variable terms first. - Track the signs carefully in part b). - Isolate the variable term before dividing.

1. For a), combine like terms: \(5k - 4.2 = 15.8\). 2. Add \(4.2\): \(5k = 20\). Divide by \(5\): \(k = 4\). 3. For b), combine like terms: \(20 - 4m = 8\). 4. Subtract \(20\): \(-4m = -12\). Divide by \(-4\): \(m = 3\).

a) \(k = 4\) b) \(m = 3\)

- Group terms containing the same variable. - Combine constant terms separately. - Treat decimal coefficients the same way as whole-number coefficients. - Substitute your answers to check them.

1. Combine like terms: \(2z - 8 = -12\). 2. Add \(8\): \(2z = -4\). Divide by \(2\): \(z = -2\). 3. Combine like terms: \(2k - 8 = 14\). 4. Add \(8\): \(2k = 22\). Divide by \(2\): \(k = 11\).

1) \(z = -2\) 2) \(k = 11\)

- Start with the known final result. - Use the inverse of each operation as you move upward. - Check by evaluating the tree from top to bottom.

1. Undo the final addition: \(68 - 18 = 50\). 2. Undo multiplication by \(5\): \(50 \div 5 = 10\). 3. Undo subtracting \(12\): \(10 + 12 = 22\). 4. Check: \((22 - 12) \cdot 5 + 18 = 68\).

\(? = 22\)

- Evaluate the branch with only known numbers first. - Find the value the other branch must have to make the final sum. - Then use an inverse operation to find \(x\).

1. Evaluate the known branch: \(15 \cdot 8 = 120\). 2. The other branch must equal \(150 - 120 = 30\). 3. Solve \(x - 25 = 30\): \(x = 30 + 25 = 55\). 4. Check: \(15 \cdot 8 + (55 - 25) = 150\).

\(x = 55\)

- In the arithmetic approach, ask what one of the five equal groups must equal. - In the algebraic approach, identify the outside operation on \(x-2\). - Compare the order in which the two methods reverse the original operations.

1. Arithmetic reasoning: if \(5\) equal groups total \(35\), each group is \(35\div5=7\). Since each group is \(x-2\), \(x\) must be \(2\) more than \(7\), so \(x=9\). 2. Algebraically, divide both sides by \(5\): \(x-2=7\). Add \(2\): \(x=9\). 3. Both methods first undo multiplication by \(5\) and then undo subtraction of \(2\).

a) \(x=9\) b) \(x=9\) c) Both methods undo multiplication by \(5\) first and subtraction of \(2\) second.

- Express the top brick in terms of the three bottom bricks. - The two outside bottom bricks each contribute once to the top, while the middle bottom brick contributes twice. - Subtract the outside bricks from the top. The remainder must equal twice the middle brick. - Check whether the remainder is divisible by \(2\) and gives a nonnegative result. - Also check whether the sum of the two outside bricks already exceeds the top brick.

1. Let the bottom row be \(a\), \(b\), and \(c\), and let the top be \(T\). Then \(T = a + 2b + c\), so \(2b = T - a - c\). 2. Wall a): \(2b = 500 - 150 - 250 = 100\), so \(b = 50\). The middle row is \(200\) and \(300\). 3. Wall b): \(2b = 500 - 150 - 249 = 101\). This would give \(b = 50.5\), which is not a nonnegative integer, so the wall cannot be completed under the stated condition. 4. Wall c): \(2b = 800 - 300 - 450 = 50\), so \(b = 25\). The middle row is \(325\) and \(475\). 5. Wall d): \(2b = 800 - 300 - 520 = -20\), so \(b = -10\). This is not nonnegative, so the wall cannot be completed under the stated condition.

1. Walls a) and c) can be completed. a) Bottom middle: \(50\); middle row: \(200\), \(300\) c) Bottom middle: \(25\); middle row: \(325\), \(475\) 2. Walls b) and d) cannot be completed using nonnegative integers. b) The equation gives \(2b = 101\), so \(b = 50.5\), which is not an integer. d) The equation gives \(2b = -20\), so \(b = -10\), which is not nonnegative.