Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 11z​

Answer 1

`(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}`

Step-by-step explanation:

Given system of equations:

`\begin{cases}y=x^2-9\\y=4x-4\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^2-9&=4x-4\\x^2-4x-9&=-4\\x^2-4x-5&=0\end{aligned}`

Factor the quadratic:

`\begin{aligned}x^2-4x-5&=0\\x^2-5x+x-5&=0\\x(x-5)+1(x-5)&=0\\(x+1)(x-5)&=0\end{aligned}`

Apply the zero-product property and solve for x:

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

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

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

`\begin{aligned}x=-1 \implies y&=4(-1)-4\\y&=-4-4\\y&=-8\end{aligned}`

`\begin{aligned}x=5 \implies y&=4(5)-4\\y&=20-4\\y&=16\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-1,-8} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,16} \; \right)\quad \textsf{(larger $x$-value)}`

Answer 2

• smaller x value: -1,-8
• larger x value: 5,16

The parenthesis part is already taken care of by the teacher.

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

Step-by-step explanation:

y is equal to x^2-9 and also 4x-4. We can equate those two right hand sides and get everything to one side like this

x^2-9 = 4x-4

x^2-9-4x+4 = 0

x^2-4x-5 = 0

Then we can use the quadratic formula to solve that equation for x.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-4)\pm√((-4)^2-4(1)(-5)))/(2(1))\\\\x = (4\pm√(36))/(2)\\\\x = (4\pm6)/(2)\\\\x = (4+6)/(2) \ \text{ or } \ x = (4-6)/(2)\\\\x = (10)/(2) \ \text{ or } \ x = (-2)/(2)\\\\x = 5 \ \text{ or } \ x = -1\\\\`

Or alternatively

x^2-4x-5 = 0

(x-5)(x+1) = 0

x-5 = 0 or x+1 = 0

x = 5 or x = -1

------------------------------

After determining the x values, plug them into either original equation to find the paired y value.

Let's plug x = 5 into the first equation:

y = x^2-9

y = 5^2-9

y = 25-9

y = 16

Or you could pick the second equation:

y = 4x-4

y = 4(5)-4

y = 20-4

y = 16

We have x = 5 lead to y = 16

One solution is (x,y) = (5,16)

This is one point where the two curves y = x^2-9 and y = 4x-4 intersect.

If you repeat the same steps with x = -1, then you should find that y = -8 for either equation.

The other solution is (x,y) = (-1,-8)