Question 13

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Jambmaths question: 

Solve for x and yin the equation below$\begin{align}  & {{x}^{2}}-{{y}^{2}}=4 \\ & x+y=2 \\\end{align}$

Option A: 

$x=0,y=-2$

Option B: 

$x=0,y=2$

Option C: 

$x=2,y=0$

Option D: 

$x=-2,y=0$

Jamb Maths Solution: 

$\begin{align}  & {{x}^{2}}-{{y}^{2}}=(x-y)(x+y)=4----(i) \\ & x+y=2-------(ii) \\ & \text{Substitute }2\text{ for }x+y\text{ in equation }(i) \\ & 2(x-y)=4 \\ & x-y=2----(iii) \\ & \text{Adding }(ii)\text{ and }(iii)\,\text{together} \\ & 2x=4 \\ & x=2 \\ & \text{Substitute 2 for }x\text{ in equation }(ii) \\ & 2+y=2 \\ & y=0 \\ & (x,y)=(2,0) \\\end{align}$

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