Answer:
[tex](x,y)=\left(\; \boxed{-3,2} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{1,0} \; \right)\quad \textsf{(larger $x$-value)}[/tex]
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}\;\;\;\;\;\;\;y^2=1-x\\x+2y=1\end{cases}[/tex]
To solve by the method of substitution, rearrange the second equation to make x the subject:
[tex]\implies x=1-2y[/tex]
Substitute the found expression for x into the first equation and rearrange so that the equation equals zero:
[tex]\begin{aligned}x=1-2y \implies y^2&=1-(1-2y)\\y^2&=1-1+2y\\y^2&=2y\\y^2-2y&=0\end{aligned}[/tex]
Factor the equation:
[tex]\begin{aligned}\implies y^2-2y&=0\\y(y-2)&=0\end{aligned}[/tex]
Apply the zero-product property and solve for y:
[tex]\implies y=0[/tex]
[tex]\implies y-2=0 \implies y=2[/tex]
Substitute the found values of y into the second equation and solve for x:
[tex]\begin{aligned}y=0 \implies x+2(0)&=1\\x&=1\end{aligned}[/tex]
[tex]\begin{aligned}y=2 \implies x+2(2)&=1\\x+4&=1\\x&=-3\end{aligned}[/tex]
Therefore, the solutions are:
[tex](x,y)=\left(\; \boxed{-3,2} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{1,0} \; \right)\quad \textsf{(larger $x$-value)}[/tex]
