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=\frac3x`
`(x\\eq0)` so


`x+y=x+\frac3x=\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)\}`