Answer:
[tex](x,y)=\left(\; \boxed{-13,4} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{3,0} \; \right)\quad \textsf{(larger $x$-value)}[/tex]
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}\;\;\;\;\;\;\;y^2=3-x\\x+4y=3\end{cases}[/tex]
To solve by the method of substitution, rearrange the second equation to make x the subject:
[tex]\implies x=3-4y[/tex]
Substitute the found expression for x into the first equation and rearrange so that the equation equals zero:
[tex]\begin{aligned}x=3-4y \implies y^2&=3-(3-4y)\\y^2&=3-3+4y\\y^2&=4y\\y^2-4y&=0\end{aligned}[/tex]
Factor the equation:
[tex]\begin{aligned}\implies y^2-4y&=0\\y(y-4)&=0\end{aligned}[/tex]
Apply the zero-product property and solve for y:
[tex]\implies y=0[/tex]
[tex]\implies y-4=0 \implies y=4[/tex]
Substitute the found values of y into the second equation and solve for x:
[tex]\begin{aligned}y=0 \implies x+4(0)&=3\\x&=3\end{aligned}[/tex]
[tex]\begin{aligned}y=4 \implies x+4(4)&=3\\x+16&=3\\x&=-13\end{aligned}[/tex]
Therefore, the solutions are:
[tex](x,y)=\left(\; \boxed{-13,4} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y)=\left(\; \boxed{3,0} \; \right)\quad \textsf{(larger $x$-value)}[/tex]
