Question 8

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waecmaths question: 

Solve the following equations            $\begin{align}  & 2x+3y=7 \\ & x+5y=0 \\\end{align}$ 

Option A: 

x = 5, y = –1

Option B: 

$x=\tfrac{1}{5},y=-1$

Option C: 

$x=\tfrac{1}{5},y=-1$

Option D: 

 $x=-5,y=1$

waecmaths solution: 

$\begin{align}  & 2x+3y=7---(i) \\ & x+5y=0---(ii) \\ & x=-5y \\ & \text{Substitute }-5y\text{ for }x\text{ into }(i) \\ & 2(-5y)+3y=7 \\ & -10y+3y=7 \\ & -7y=7 \\ & y=-1 \\ & \therefore x=-5(-1)=5 \\ & x=5,\text{ }y=-1 \\\end{align}$

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