NECO 2024 · Paper 2 · Q1

Solve the pair of equations $4a + 12b = 16$ and $2a + 3b = 7$ . Hence, find the positive value of $K$ given that $a + b = K^2$ .

(a)
Enter $a$ , $b$ and $K$ (to 3 decimal places).
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Number the equations: $4a + 12b = 16$ (1) and $2a + 3b = 7$ (2).
Divide (1) by 4 to make it simpler: $a + 3b = 4$ (3).
Both (2) and (3) contain $3b$ , so take (3) from (2): $a = 3$ .
Put $a = 3$ into (3): $3 + 3b = 4$ , so $3b = 1$ and $b = \frac13$ .
Now $K^2 = a + b = 3 + \frac13 = \frac{10}{3}$ .
The positive value is $K = \sqrt{\frac{10}{3}} \approx 1.826$ .

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