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).
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
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
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).
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).