Topics › KS3 › Algebra

Solving linear equations

eq-01 Present from here

Solve $4x = 20$

Option A: $x = 16$

Option B: $x = 5$

Option C: $x = 80$

Option D: $x = \frac{1}{5}$

Answer and misconceptions

Option A: $x = 16$ ✗
Subtracts the coefficient instead of dividingThinks 4x means 4 + x, so takes 4 away from 20.

Option B: $x = 5$ ✓ Correct

Option C: $x = 80$ ✗
Multiplies instead of using the inverseDoes the same operation (× 4) to 20 instead of the inverse.

Option D: $x = \frac{1}{5}$ ✗
Divides the wrong way roundKnows to divide but calculates 4 ÷ 20.
eq-02 Present from here

Solve $x + 9 = 4$

Option A: $x = 5$

Option B: $x = 13$

Option C: $x = -5$

Option D: $x = -13$

Answer and misconceptions

Option A: $x = 5$ ✗
Subtracts the smaller number from the largerCalculates 9 − 4 because they believe you cannot take 9 from 4.

Option B: $x = 13$ ✗
Does the same operation instead of the inverseAdds 9 to 4 instead of subtracting it.

Option C: $x = -5$ ✓ Correct

Option D: $x = -13$ ✗
Makes both numbers negative when moving themBelieves moving terms makes everything negative: −4 − 9 = −13.
eq-03 Present from here

Solve $3x + 5 = 20$

Option A: $x = 12$

Option B: $x = \frac{25}{3}$

Option C: $x = 45$

Option D: $x = 5$

Answer and misconceptions

Option A: $x = 12$ ✗
Subtracts the coefficient instead of dividingGets 3x = 15 correctly, then works out 15 − 3.

Option B: $x = \frac{25}{3}$ ✗
Adds the constant instead of subtracting itGets 3x = 25 by adding 5 to both sides, then divides by 3.

Option C: $x = 45$ ✗
Multiplies by the coefficient instead of dividingGets 3x = 15 correctly, then works out 15 × 3.

Option D: $x = 5$ ✓ Correct
eq-04 Present from here

Solve $5x + 2 = 3x + 10$

Option A: $x = 4$

Option B: $x = 8$

Option C: $x = 1.5$

Option D: $x = 6$

Answer and misconceptions

Option A: $x = 4$ ✓ Correct

Option B: $x = 8$ ✗
Forgets to divide by the coefficientReaches 2x = 8 and gives 8 as the value of x.

Option C: $x = 1.5$ ✗
Moves terms across without changing the signWrites 5x + 3x = 10 + 2, so 8x = 12.

Option D: $x = 6$ ✗
Adds the constant across instead of subtractingGets 2x = 10 + 2 = 12, so x = 6.
eq-05 Present from here

Solve $2(x - 3) = 10$

Option A: $x = 6.5$

Option B: $x = 11$

Option C: $x = 8$

Option D: $x = 2$

Answer and misconceptions

Option A: $x = 6.5$ ✗
Multiplies only the first term in the bracketExpands to 2x − 3 = 10, so 2x = 13.

Option B: $x = 11$ ✗
Subtracts the number outside the bracketTakes 2 from both sides instead of dividing: x − 3 = 8.

Option C: $x = 8$ ✓ Correct

Option D: $x = 2$ ✗
Subtracts instead of adding when moving a termExpands to 2x − 6 = 10 correctly, then writes 2x = 10 − 6.