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zimovet [89]
3 years ago
11

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

Mathematics
1 answer:
inn [45]3 years ago
3 0
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)\}
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