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

Question

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

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

Answer 1

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).

Answer 2

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

Answer 3

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).