Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 9z​

Answer 1

• (-3, 2)
• (1, 0)

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Given system

• y² = 1 - x
• x + 2y = 1

Rearrange the first equation

• x = 1 - y²

Substitute the value of x into second equation

• 1 - y² + 2y = 1
• y² - 2y = 0
• y(y - 2) = 0
• y = 0 and y = 2

Find the value of x

• y = 0 ⇒ x = 1 - 0² = 1
• y = 2 ⇒ x = 1 - 2² = -3

Answer 2

`(x,y)=\left(\; \boxed{-3,2} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{1,0} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}\;\;\;\;\;\;\;y^2=1-x\\x+2y=1\end{cases}`

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

`\implies x=1-2y`

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

`\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}`

Factor the equation:

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

Apply the zero-product property and solve for y:

`\implies y=0`

`\implies y-2=0 \implies y=2`

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

`\begin{aligned}y=0 \implies x+2(0)&=1\\x&=1\end{aligned}`

`\begin{aligned}y=2 \implies x+2(2)&=1\\x+4&=1\\x&=-3\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-3,2} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{1,0} \; \right)\quad \textsf{(larger $x$-value)}`