Question 22

We will check each solution for mistakes and then correct them.

Problem 1: $4x + 6 = 10$

The student made a mistake when moving a term.

Step 1 — Identify the mistake

The student moved the constant term +6 to the right side. They added 6 instead of subtracting 6. The incorrect step was: $4x = 10 + 6$

Step 2 — Correct the mistake

When we move a term to the other side, we change its sign. We should subtract 6 from both sides. The correct step is: $4x = 10 - 6$

Step 3 — Solve the equation

Let us now solve the equation correctly. $4x + 6 = 10$

Subtract 6 from both sides. $4x = 10 - 6$

Simplify the right side. $4x = 4$

Divide both sides by 4. $x = \frac{4}{4}$

$x = \mathbf{1}$

$\boxed{x = 1}$

Problem 2: $7 - 8z = 5$

The student made a mistake in the final division.

Step 1 — Identify the mistake

The student correctly isolated 8z. They then divided 2 by 8 incorrectly. The incorrect step was: $z = 4$

Step 2 — Correct the mistake

We need to divide 2 by 8. The correct division is $2 \div 8$. The correct step is: $z = \frac{2}{8}$

Step 3 — Solve the equation

Let us now solve the equation correctly. $7 - 8z = 5$

Subtract 7 from both sides. $-8z = 5 - 7$

Simplify the right side. $-8z = -2$

Divide both sides by -8. $z = \frac{-2}{-8}$

$z = \frac{\mathbf{1}}{\mathbf{4}}$

$\boxed{z = \frac{1}{4}}$

Problem 3: $2v - 4 = 6$

The student made a mistake in handling the coefficient.

Step 1 — Identify the mistake

The student incorrectly divided by 2. They subtracted 2 from 6 instead of dividing the sum. The incorrect step was: $v - 4 = 6 - 2$

Step 2 — Correct the mistake

First, we should move the constant term -4. Then, we divide the entire right side by 2. The correct first step is: $2v = 6 + 4$

Step 3 — Solve the equation

Let us now solve the equation correctly. $2v - 4 = 6$

Add 4 to both sides. $2v = 6 + 4$

Simplify the right side. $2v = 10$

Divide both sides by 2. $v = \frac{10}{2}$

$v = \mathbf{5}$

$\boxed{v = 5}$

Problem 4: $5z + 2 = 3z - 4$

The student made mistakes when moving terms.

Step 1 — Identify the mistake

The student moved terms across the equals sign. They did not change the signs correctly. The incorrect step was: $5z + 3z = -4 + 2$

Step 2 — Correct the mistake

When moving $3z$ to the left, it becomes $-3z$. When moving $+2$ to the right, it becomes $-2$. The correct step is: $5z - 3z = -4 - 2$

Step 3 — Solve the equation

Let us now solve the equation correctly. $5z + 2 = 3z - 4$

Move $3z$ to the left and $+2$ to the right. $5z - 3z = -4 - 2$

Combine like terms on both sides. $2z = -6$

Divide both sides by 2. $z = \frac{-6}{2}$

$z = \mathbf{-3}$

$\boxed{z = -3}$

Problem 5: $15w - 4w = 26$

The student made a mistake in combining like terms.

Step 1 — Identify the mistake

The student had like terms on the left side. They moved $-4w$ to the right side. The incorrect step was: $15w = 26 + 4w$

Step 2 — Correct the mistake

We should combine the like terms on the left side first. $15w$ and $-4w$ are like terms. The correct step is: $11w = 26$

Step 3 — Solve the equation

Let us now solve the equation correctly. $15w - 4w = 26$

Combine the terms on the left side. $11w = 26$

Divide both sides by 11. $w = \frac{26}{11}$

$w = \frac{\mathbf{26}}{\mathbf{11}}$

$\boxed{w = \frac{26}{11}}$

Problem 6: $3x + 1 = -12$

The student made a mistake in the order of operations.

Step 1 — Identify the mistake

The student divided by 3 too early. They divided only the right side by 3. The incorrect step was: $x + 1 = \frac{-12}{3}$

Step 2 — Correct the mistake

First, we should move the constant term +1. Then, we divide the entire right side by 3. The correct first step is: $3x = -12 - 1$

Step 3 — Solve the equation

Let us now solve the equation correctly. $3x + 1 = -12$

Subtract 1 from both sides. $3x = -12 - 1$

Simplify the right side. $3x = -13$

Divide both sides by 3. $x = \frac{-13}{3}$

$x = \mathbf{-\frac{13}{3}}$

$\boxed{x = -\frac{13}{3}}$

Problem 7: $4(4q + 2) = 50$

The student made a mistake in applying the distributive property.

Step 1 — Identify the mistake

The student did not multiply 4 by 2. They only multiplied 4 by $4q$. The incorrect step was: $4(4q) = 50 - 2$

Step 2 — Correct the mistake

We must distribute the 4 to both terms inside the bracket. $4 \times 4q$ is $16q$. $4 \times 2$ is $8$. The correct step is: $16q + 8 = 50$

Step 3 — Solve the equation

Let us now solve the equation correctly. $4(4q + 2) = 50$

Apply the distributive property. $16q + 8 = 50$

Subtract 8 from both sides. $16q = 50 - 8$

Simplify the right side. $16q = 42$

Divide both sides by 16. $q = \frac{42}{16}$

Simplify the fraction. $q = \frac{\mathbf{21}}{\mathbf{8}}$

$\boxed{q = \frac{21}{8}}$

Problem 8: $-2(3 - 4x) = 14$

The student made two mistakes in distributing the number.

Step 1 — Identify the mistake

The student changed the variable from $x$ to $v$. They also made a sign error when multiplying $-2$ by $-4x$. The incorrect step was: $-6v - 8x = 14$

Step 2 — Correct the mistake

We must keep the variable as $x$. When multiplying $-2$ by $-4x$, the result is $+8x$. The correct step is: $-6 + 8x = 14$

Step 3 — Solve the equation

Let us now solve the equation correctly. $-2(3 - 4x) = 14$

Apply the distributive property. $-6 + 8x = 14$

Add 6 to both sides. $8x = 14 + 6$

Simplify the right side. $8x = 20$

Divide both sides by 8. $x = \frac{20}{8}$

Simplify the fraction. $x = \frac{\mathbf{5}}{\mathbf{2}}$

$\boxed{x = \frac{5}{2}}$

Problem 9: $3(7y + 4) = 9 + 5y$

The student made mistakes in distribution and moving terms.

Step 1 — Identify the mistake

The student divided only 9 by 3. They should have distributed 3 to both terms. The incorrect step was: $7y + 4 = \frac{9}{3} + 5y$

Also, later, when moving +4, they made a sign error. The incorrect step was: $2y = 4 - 3$

Step 2 — Correct the mistake

We must distribute the 3 to both terms inside the bracket. $3 \times 7y$ is $21y$. $3 \times 4$ is $12$. The correct first step is: $21y + 12 = 9 + 5y$

When moving +4 to the right side, it should become -4. The correct step should be $2y = 3 - 4$.

Step 3 — Solve the equation

Let us now solve the equation correctly. $3(7y + 4) = 9 + 5y$

Apply the distributive property. $21y + 12 = 9 + 5y$

Move $5y$ to the left and $+12$ to the right. $21y - 5y = 9 - 12$

Combine like terms on both sides. $16y = -3$

Divide both sides by 16. $y = \frac{-3}{16}$

$y = \mathbf{-\frac{3}{16}}$

$\boxed{y = -\frac{3}{16}}$

Answer

(i) $x = 1$ (ii) $z = \frac{1}{4}$ (iii) $v = 5$ (iv) $z = -3$ (v) $w = \frac{26}{11}$ (vi) $x = -\frac{13}{3}$ (vii) $q = \frac{21}{8}$ (viii) $x = \frac{5}{2}$ (ix) $y = -\frac{3}{16}$