Additional Mathematics – Simultaneous equations | e-Consult

Clear fractions by multiplying the first equation by 6 and the second by 20:

• \(3x - 2y = 24\)  (1)
• \(8x + 15y = -20\)  (2)

From (1), \(3x = 24 + 2y \Rightarrow x = \dfrac{24 + 2y}{3}\).

Substitute this expression for \(x\) into (2):

\(8\left(\dfrac{24 + 2y}{3}\right) + 15y = -20\)

\(\dfrac{192 + 16y}{3} + 15y = -20\)

Multiply by 3:

\(192 + 16y + 45y = -60\)

\(61y = -252 \Rightarrow y = -\dfrac{252}{61}\).

Then \(x = \dfrac{24 + 2(-252/61)}{3} = \dfrac{320}{61}\).

Thus the solution is \(\displaystyle x = \frac{320}{61},\; y = -\frac{252}{61}\).