Question #30c88

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Answer 1 · 2016-10-24T22:47:18

$(x, y) in {(0, 0), (-3, 2)}$

${(x+2y = 1),(y^2=1-x):}$

In the second equation, add $x-y^2$ to both sides to isolate $x$ :

$x = 1-y^2$

Substitute this value for $x$ into the first equation.

$(1-y^2)+2y = 1$

Gather the terms on the left hand side.

$-y^2+2y+1-1 = 0$

Combine like terms and multiply by $-1$ to get a coefficient of $1$ for $y^2$ .

$y^2-2y = 0$

This can be factored easily, so we will solve by factoring

$y(y-2) = 0$

$=> y = 0 or y-2 = 0$

$=> y = 0 or y = 2$

Substitute the possible solutions for $y$ back into $x = 1-y^2$ to find the $x$ that goes with them.

$y = 0 => x = 1-0^2 = 1$

$y = 2 => x = 1-2^2 = -3$

Thus, our potential solutions are

$(x, y) = (1, 0) or (x, y) = (-3, 2)$

Checking these answers by substituting them into the original equations, we find they both satisfy the system. Thus, our final result is:

$(x, y) in {(1, 0), (-3, 2)}$