Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 10z​

Answer 1

`(x,y)=\left(\; \boxed{-2,13} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{1,10} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}\phantom{bbbb}y=x^2+9\\x+y=11\end{cases}`

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

`\implies y=11-x`

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

`\begin{aligned}y=11-x \implies 11-x&=x^2+9\\x^2+9&=11-x\\x^2+9+x&=11\\x^2+x-2&=0\end{aligned}`

Factor the quadratic:

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

Apply the zero-product property and solve for x:

`\implies x-1=0 \implies x=1`

`\implies x+2=0 \implies x=-2`

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

`\begin{aligned}x=1 \implies 1+y&=11\\y&=11-1\\y&=10\end{aligned}`

`\begin{aligned}x=-2 \implies -2+y&=11\\y&=11+2\\y&=13\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-2,13} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{1,10} \; \right)\quad \textsf{(larger $x$-value)}`

Answer 2

• (-2, 13)
• (1, 10)

=====================

Given system

• y = x² + 9
• x + y = 11

Substitute the value of y into second equation

• x + x² + 9 = 11
• x² + x - 2 = 0
• x² +2x - x - 2 = 0
• x(x + 2) - (x + 2) = 0
• (x + 2)(x - 1) = 0
• x + 2 = 0 and x - 1 = 0
• x = - 2 and x = 1

Find the value of y

• x = -2 ⇒ y = 11 - (-2) = 13
• x = 1 ⇒ y = 11 - 1 = 10