Question

NO LINKS!! Use the method of to solve the system. (if there's no solution, enter no solution). Part 2z​

Answer 1

smaller x-value: (-4, 18)

larger x-value: (3, 11)

Explanation:

Solving for x:

y = x^2 + 2

x + y = 14 ---> y = 14 - x

14 - x = x^2 + 2

0 = x^2 + x - 12

0 = (x + 4)(x - 3)

x = -4 or 3

Solving for y:

If x = -4

y = 14 + 4

y = 18

if x = 3

y = 14 - 3

y = 11

Answer 2

`(x,y)=\left(\; \boxed{-4,18} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{3,11} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}\phantom{bbbb}y=x^2+2\\x+y=14\end{cases}`

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

`\implies y=14-x`

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

`\begin{aligned}y=14-x \implies 14-x&=x^2+2\\x^2+2&=14-x\\x^2+2+x&=14\\x^2+x-12&=0\end{aligned}`

Factor the quadratic:

`\begin{aligned}x^2+x-12&=0\\x^2+4x-3x-12&=0\\x(x+4)-3(x+4)&=0\\(x-3)(x+4)&=0\end{aligned}`

Apply the zero-product property and solve for x:

`\implies x-3=0 \implies x=3`

`\implies x+4=0 \implies x=-4`

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

`\begin{aligned}x=3 \implies 3+y&=14\\y&=14-3\\y&=11\end{aligned}`

`\begin{aligned}x=-4 \implies -4+y&=14\\y&=14+4\\y&=18\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-4,18} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{3,11} \; \right)\quad \textsf{(larger $x$-value)}`