Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Solve the system of equations algebraically. Show all of your steps.

y
=
x
2
+
2
x
y
=
3
x
+
20

3 answers

We can start by rearranging the first equation to solve for x:

y = x^2 + 2x

y - x^2 = 2x

y/x - x = 2

x = y/x - 2

Now we can substitute this expression for x into the second equation:

y = 3x + 20

y = 3(y/x - 2) + 20

y = 3y/x - 6 + 20

y = 3y/x + 14

y(x) = 3y + 14x

x = (y - 3y)/(14)

x = -2y/14

Simplifying further:

x = -y/7

Now we have expressions for both x and y in terms of the other variable. We can substitute either expression into one of the original equations to solve for the value of that variable. Let's use the first equation:

y = x^2 + 2x

y = (-y/7)^2 + 2(-y/7)

Multiplying both sides by 49 to get rid of the fraction:

49y = y^2 - 14y

y^2 - 63y = 0

y(y - 63) = 0

So either y = 0 or y = 63.

If y = 0, then from our expression for x:

x = -y/7 = 0

So one solution to the system is (0, 0).

If y = 63, then from our expression for x:

x = -y/7 = -9

So another solution to the system is (-9, 63).

Therefore, the system has two solutions: (0, 0) and (-9, 63).
The vertical formatting makes things so hard to read. I see
y = x^2 + 2x
y = 3x+20
so, substituting,
x^2 + 2x = 3x+20
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x = 5 or -4
Great, you correctly solved for x using the substitution method!

Now we just need to substitute these values of x back into one of the original equations to solve for y:

If x = 5:

y = x^2 + 2x

y = 5^2 + 2(5)

y = 35

So one solution is (5, 35).

If x = -4:

y = x^2 + 2x

y = (-4)^2 + 2(-4)

y = 8

So another solution is (-4, 8).

Therefore, the system has two solutions: (5, 35) and (-4, 8).