Tim T. answered • 05/14/19

Math: K-12th grade to Advanced Calc, Ring Theory, Cryptography

Greetings! Lets solve this shall we ?

So, after reviewing this problem I see we cannot use elimination for the second part. We must use substitution for both problems such that

y + x = 5 and y = x^{2} - 6x + 9 where

y = -x + 5 and y = x^{2} - 6x + 9 such that

-x + 5 = x^{2} - 6x + 9..................Now we solve for x first by adding x to both sides such that

5 = x^{2} - 5x + 9..................Then subtracting 5 to both sides to get

0 = x^{2} - 5x + 4.................We can factor this into

0 = (x-1)(x-4)...................Then use the Zero-Product Property of Equality such that

x - 1 = 0 and x - 4 = 0..................Then the x-values are

x = 1 and 4

Now we plug these values into the equation such that

y = -(1) + 5

y = 4

and

y = -(4) + 5

y = 1

So, the solutions are **(1,4) and (4,1)** for the first system of equations.

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

Then, we must solve the second system of equations such that elimination will not work due to it having an x squared term. So, we must use substitution method. Then,

-12x + 36 = x^{2} - 11x - 36.................First we add 12x to both sides such that

36 = x^{2} + x - 36..................Then we subtract 36 to both sides such that

0 = x^{2} + x - 72...................We see this polynomial is factorable so we find factors that will add to give us

0 = (x+9)(x-8)................Then

x = -9,8

So, we plug these values into the equation y = -12x + 36 to get the solutions **(-9,144) and (8, -60).**

I hope this helped!