JAMB 1997 · UME · Q12

Solve the simultaneous equations $\frac2x - \frac3y = 2$ and $\frac4x + \frac3y = 10$ .

Worked solution (try it first)

Add the two equations so the $\frac3y$ terms cancel: $\frac6x = 12$ .
Multiply both sides by $x$ and divide by 12: $x = \frac{6}{12} = \frac12$ .
Put $x = \frac12$ into the first equation: $\frac2x = 4$ , so $4 - \frac3y = 2$ and $\frac3y = 2$ .
So $y = \frac32$ , and the answer is $x = \frac12$ , $y = \frac32$ , option B.