Question

asked Mar 24, 2023 211k views

2 Answers

Oy · Answer 1 · 2023-03-25T23:39:46+0000

Answer:

$(x,y)=\left(\; \boxed{-13,4} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{3,0} \; \right)\quad \textsf{(larger $x$-value)}$

Explanation:

Given system of equations:

$\begin{cases}\;\;\;\;\;\;\;y^2=3-x\\x+4y=3\end{cases}$

To solve by the method of substitution, rearrange the second equation to make x the subject:

$\implies x=3-4y$

Substitute the found expression for x into the first equation and rearrange so that the equation equals zero:

$\begin{aligned}x=3-4y \implies y^2&=3-(3-4y)\\y^2&=3-3+4y\\y^2&=4y\\y^2-4y&=0\end{aligned}$

Factor the equation:

$\begin{aligned}\implies y^2-4y&=0\\y(y-4)&=0\end{aligned}$

Apply the zero-product property and solve for y:

$\implies y=0$

$\implies y-4=0 \implies y=4$

Substitute the found values of y into the second equation and solve for x:

$\begin{aligned}y=0 \implies x+4(0)&=3\\x&=3\end{aligned}$

$\begin{aligned}y=4 \implies x+4(4)&=3\\x+16&=3\\x&=-13\end{aligned}$

Therefore, the solutions are:

$(x,y)=\left(\; \boxed{-13,4} \; \right)\quad \textsf{(smaller $x$-value)}$

$(x,y)=\left(\; \boxed{3,0} \; \right)\quad \textsf{(larger $x$-value)}$

Joel Cox · Answer 2 · 2023-03-31T03:58:15+0000

Answer:

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