Question

NO LINKS!! Use the method of to solve the system. (if there's no solution, enter no solution). Part 1z​

Answer 1

-2,-9

5,12

Explanation:

Set the equations equal to each other

x^2 - 13 = 3x - 3

x^2 - 10 = 3x

x^2 - 3x - 10 = 0

(x-5)(x+2) = 0

x = 5 x = -2

Now choose on the equations to plug into

y = 3x - 3

y = 3(-2) - 3 = -9

y = 3(5) - 3 = 12

-2,-9

5,12

Answer 2

`(x,y)=\left(\; \boxed{-2,-9} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,12} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}y=x^2-13\\y=3x-3\end{cases}`

To solve by the method of substitution, substitute the first equation into the second equation and rearrange so that the equation equals zero:

`\begin{aligned}x^2-13&=3x-3\\x^2-3x-13&=-3\\x^2-3x-10&=0\end{aligned}`

Factor the quadratic:

`\begin{aligned}x^2-3x-10&=0\\x^2-5x+2x-10&=0\\x(x-5)+2(x-5)&=0\\(x+2)(x-5)&=0\end{aligned}`

Apply the zero-product property and solve for x:

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

`\implies x-5=0 \implies x=5`

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

`\begin{aligned}x=-2 \implies y&=3(-2)-3\\y&=-6-3\\y&=-9\end{aligned}`

`\begin{aligned}x=5 \implies y&=3(5)-3\\y&=15-3\\y&=12\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-2,-9} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,12} \; \right)\quad \textsf{(larger $x$-value)}`