Question 12 Marks

Solve the following simultaneous equations:
$x+y=3$; $3x-2y=4$

Answer

Multiply 1st eq by 2: $2x + 2y = 6$
Add to 2nd eq: $(2x + 2y) + (3x - 2y) = 6 + 4$
$5x = 10 \implies x = 2$
Substitute $x=2$ in 1st eq: $2 + y = 3 \implies y = 1$
Solution: $(2, 1)$

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