Question

Q8 Q3.) Solve the system by the substitution method.

Answer 1

Recall that $(x+y)^2=x^2+2xy+y^2$. So we can add twice the first equation to the second one to get

$x^2+y^2+2xy=10+2\cdot3\iff(x+y)^2=16\implies x+y=\pm4$

Since $xy=3$, we have $y=\dfrac3x$ $(x\neq0)$ so

$x+y=x+\dfrac3x=\pm4\implies x^2\mp4x+3=0$

If $x+y=4$, then

$x^2-4x+3=(x-3)(x-1)=0\implies x=3,x=1\implies y=1,y=3$

If $x+y=-4$, then

$x^2+4x+3=(x+1)(x+3)=0\implies x=-1,x=-3\implies y=-3,y=-1$

So the solution set is

$(x,y)\in\{(-3,1),(1,3),(-1,-3),(-3,-1)\}$