Zainab is pricing up a delivery of stock that arrived with the boxes this morning, and has two equations on the supplier's invoice that she needs to solve c...
Question 1 Report
Zainab is pricing up a delivery of stock that arrived with the boxes this morning, and has two equations on the supplier's invoice that she needs to solve carefully before entering the new prices into the till system ahead of the coming week's opening, so no customer is ever overcharged.
- Solve \(6x + 11 = 2x + 39\). (2)
- Solve \(\dfrac{2x - 3}{5} = 3\). (2)
Answer Details
This question tests solving one equation with the unknown on both sides and one equation with the unknown inside a fraction.
(a) For \(6x+11=2x+39\), subtract \(2x\) from both sides and subtract 11 from both sides:
\[6x-2x=39-11 \implies 4x=28 \text{ <b>[1]</b>} \implies x=7 \text{ <b>[1]</b>}\](b) For \(\dfrac{2x-3}{5}=3\), multiply both sides by 5 first:
\[2x-3=15 \text{ <b>[1]</b>} \implies 2x=18 \implies x=9 \text{ <b>[1]</b>}\]In part (b), multiplying by 5 clears the denominator because the entire expression \(2x-3\) is divided by 5, not just the \(2x\) term, so the whole numerator must be multiplied by 5 together.
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