Finding the Unknown | IT

Question 22

7.3 Mind the Mistake, Mend the Mistake

The following are some equations along with the steps used to solve them to find the value of the letter-number. Go through each solution and decide whether the steps are correct. If there is a mistake, describe the mistake, correct it and solve the equation.

Solution

Understand the Question

To solve linear equations with one variable, apply the following algebraic rules:
- Transposition: When shifting a term to the other side of an equation, reverse its operation ( $+$ becomes $-$ , $-$ becomes $+$ , $\times$ becomes $\div$ , and $\div$ becomes $\times$ ).
- Distributive Property: Multiply the factor outside the parentheses by every term inside: $a(b + c) = ab + ac$ .
- Combining Like Terms: Group and simplify variable terms together and constant terms together before isolating the variable.
For each problem, identify where the error occurred, explain the correction, and solve for the unknown variable.

(i) Examine the solution for $4x + 6 = 10$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 1

Mistake: When transposing $+6$ to the right side, it was added instead of subtracted ( $4x = 10 + 6$ ).
Correction: Subtract $6$ from both sides: $4x = 10 - 6$ .

Step 2 · Solve the Equation

\begin{aligned} 4x + 6 &= 10 \\[0.4em] 4x &= 10 - 6 \\[0.4em] 4x &= 4 \\[0.4em] x &= \dfrac{4}{4} \\[0.4em] x &= 1 \end{aligned}

Answer

(i) $x = 1$

(ii) Examine the solution for $7 - 8z = 5$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 2

Mistake: In the division step, $8$ was divided by $2$ ( $z = 4$ ) instead of dividing $2$ by $8$ .
Correction: Divide $-2$ by $-8$ to get $z = \dfrac{-2}{-8} = \dfrac{2}{8}$ .

Step 2 · Solve the Equation

\begin{aligned} 7 - 8z &= 5 \\[0.4em] -8z &= 5 - 7 \\[0.4em] -8z &= -2 \\[0.4em] z &= \dfrac{-2}{-8} \\[0.4em] z &= \dfrac{1}{4} \end{aligned}

Answer

(ii) $z = \dfrac{1}{4}$

(iii) Examine the solution for $2v - 4 = 6$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 3

Mistake: The coefficient $2$ was subtracted from the right side ( $v - 4 = 6 - 2$ ) instead of transposing the constant term first.
Correction: Add $4$ to the right side first ( $2v = 6 + 4$ ), then divide by $2$ .

Step 2 · Solve the Equation

\begin{aligned} 2v - 4 &= 6 \\[0.4em] 2v &= 6 + 4 \\[0.4em] 2v &= 10 \\[0.4em] v &= \dfrac{10}{2} \\[0.4em] v &= 5 \end{aligned}

Answer

(iii) $v = 5$

(iv) Examine the solution for $5z + 2 = 3z - 4$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 4

Mistake: Signs were not changed when transposing terms across the equals sign ( $5z + 3z = -4 + 2$ ).
Correction: Transpose $3z$ to the left as $-3z$ and $+2$ to the right as $-2$ , giving $5z - 3z = -4 - 2$ .

Step 2 · Solve the Equation

\begin{aligned} 5z + 2 &= 3z - 4 \\[0.4em] 5z - 3z &= -4 - 2 \\[0.4em] 2z &= -6 \\[0.4em] z &= \dfrac{-6}{2} \\[0.4em] z &= -3 \end{aligned}

Answer

(iv) $z = -3$

(v) Examine the solution for $15w - 4w = 26$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 5

Mistake: Like terms on the left side were separated unnecessarily ( $15w = 26 + 4w$ ) instead of combining them directly.
Correction: Combine like terms on the LHS directly: $15w - 4w = 11w$ .

Step 2 · Solve the Equation

\begin{aligned} 15w - 4w &= 26 \\[0.4em] 11w &= 26 \\[0.4em] w &= \dfrac{26}{11} \end{aligned}

Answer

(v) $w = \dfrac{26}{11}$

(vi) Examine the solution for $3x + 1 = -12$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 6

Mistake: Dividing only part of the equation by $3$ before isolating the variable term ( $x + 1 = \dfrac{-12}{3}$ ).
Correction: Subtract $1$ from both sides first to isolate $3x$ , then divide by $3$ .

Step 2 · Solve the Equation

\begin{aligned} 3x + 1 &= -12 \\[0.4em] 3x &= -12 - 1 \\[0.4em] 3x &= -13 \\[0.4em] x &= \dfrac{-13}{3} \\[0.4em] x &= -\dfrac{13}{3} \end{aligned}

Answer

(vi) $x = -\dfrac{13}{3}$

(vii) Examine the solution for $4(4q + 2) = 50$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Mistake: The factor $4$ was only multiplied with the first term $4q$ and not with $2$ ( $4(4q) = 50 - 2$ ).
Correction: Distribute $4$ to both terms inside the parentheses: $4(4q) + 4(2) = 16q + 8$ .

Step 2 · Solve the Equation

\begin{aligned} 4(4q + 2) &= 50 \\[0.4em] 16q + 8 &= 50 \\[0.4em] 16q &= 50 - 8 \\[0.4em] 16q &= 42 \\[0.4em] q &= \dfrac{42}{16} \\[0.4em] q &= \dfrac{21}{8} \end{aligned}

Answer

(vii) $q = \dfrac{21}{8}$

(viii) Examine the solution for $-2(3 - 4x) = 14$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 8

Mistake: The variable was incorrectly changed to $v$ , and multiplying negative numbers gave an incorrect sign ( $-6v - 8x = 14$ ).
Correction: Multiplying $-2$ by $-4x$ yields $+8x$ , giving $-6 + 8x = 14$ .

Step 2 · Solve the Equation

\begin{aligned} -2(3 - 4x) &= 14 \\[0.4em] -6 + 8x &= 14 \\[0.4em] 8x &= 14 + 6 \\[0.4em] 8x &= 20 \\[0.4em] x &= \dfrac{20}{8} \\[0.4em] x &= \dfrac{5}{2} \end{aligned}

Answer

(viii) $x = \dfrac{5}{2}$

(ix) Examine the solution for $3(7y + 4) = 9 + 5y$ , describe the mistake, correct it, and solve the equation.

Step 1 · Identify and Correct the Mistake

Diagram 9

Mistake: Only the first term on the RHS was divided by $3$ ( $7y + 4 = \dfrac{9}{3} + 5y$ ), and transposition sign errors occurred later.
Correction: Expand the LHS using the distributive property: $3(7y) + 3(4) = 21y + 12$ .

Step 2 · Solve the Equation

\begin{aligned} 3(7y + 4) &= 9 + 5y \\[0.4em] 21y + 12 &= 9 + 5y \\[0.4em] 21y - 5y &= 9 - 12 \\[0.4em] 16y &= -3 \\[0.4em] y &= \dfrac{-3}{16} \\[0.4em] y &= -\dfrac{3}{16} \end{aligned}

Answer

(ix) $y = -\dfrac{3}{16}$

Common Mistakes

Sign Error on Transposition: Forgetting that moving a term across the equals sign changes its sign (e.g., $+6$ becomes $-6$ , not $+6$ ).
Incomplete Distribution: Multiplying the coefficient outside the parentheses by only the first term inside (e.g., $4(4q + 2) \ne 16q + 2$ ).
Multiplying Negatives: Forgetting that multiplying two negative numbers gives a positive result (e.g., $-2 \times (-4x) = +8x$ , not $-8x$ ).
Fraction Division Inversion: Dividing the coefficient by the constant rather than dividing both sides by the coefficient of the variable (e.g., for $-8z = -2$ , writing $z = 8 \div 2 = 4$ instead of $z = \dfrac{-2}{-8} = \dfrac{1}{4}$ ).

Question 22

7.3 Mind the Mistake, Mend the Mistake

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7.3 Mind the Mistake, Mend the Mistake