Answer:
No real solutions.
Step-by-step explanation:
[tex]y=x^2-3x-4[/tex]
[tex]x=y+8[/tex]
I'm going to subtract the second expression for y and plug it into the first equation.
So solving x=y+8 for y by subtracting 8 on both sides gives us y=x-8.
I'm going to insert this for the first y like so:
[tex]x-8=x^2-3x-4[/tex]
Now I'm going to move everything to one side.
I'm going to subtract x on both sides and add 8 on both sides.
[tex]0=x^2-3x-x-4+8[/tex]
Simplifying:
[tex]0=x^2-3x+4[/tex]
Now our job since the coefficient of x^2 is 1 is to find two numbers that multiply to be 4 and at the same time add up to be -3. I can't think of any such numbers.
Let's check the discriminant.
Compare [tex]x^2-3x+4[/tex] to [tex]ax^2+bx+c[/tex].
So [tex]a=1,b=-3,c=4[/tex].
The discriminant is [tex]b^2-4ac[/tex].
So plugging in our numbers we get [tex](-3)^2-4(1)(4)[/tex].
Time to simplify:
[tex](-3)^2-4(1)(4)[/tex]
[tex]9-16[/tex]
[tex]-7[/tex]
So since the discriminant is negative, then the solutions will not be real.
