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

Mathematics

2 answers:

krek1111 [17]1 year ago

8 0

Answer:

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

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

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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.

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

Or alternatively

x^2-4x-5 = 0

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

x-5 = 0 or x+1 = 0

x = 5 or x = -1

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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)

Roman55 [17]1 year ago

6 0

Answer:

$(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:

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