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Substitution: y=x2-6x+9 y+x=5 Elimination: y=x2-11x-36 y=-12x+36
Solution
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 = x2 – 6x + 9 where
y = -x + 5 and y = x2 – 6x + 9 such that
-x + 5 = x2 – 6x + 9………………Now we solve for x first by adding x to both sides such that
5 = x2 – 5x + 9………………Then subtracting 5 to both sides to get
0 = x2 – 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.
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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 = x2 – 11x – 36……………..First we add 12x to both sides such that
36 = x2 + x – 36………………Then we subtract 36 to both sides such that
0 = x2 + 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!