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
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2
x
y
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3
x
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20
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).
To solve the system of equations algebraically, we need to find the values of x and y that satisfy both equations.
Given:
y = x^2 + 2x
y = 3x + 20
We can solve this system of equations through substitution or elimination. Let's use the substitution method:
Step 1: Solve the first equation for y in terms of x.
y = x^2 + 2x
Step 2: Substitute the value of y from the first equation into the second equation.
x^2 + 2x = 3x + 20
Step 3: Simplify the equation.
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Step 4: Factorize or use the quadratic formula to find the values of x.
This quadratic equation can be factored as (x - 5)(x + 4) = 0. So, either x - 5 = 0 or x + 4 = 0.
If x - 5 = 0, then x = 5.
If x + 4 = 0, then x = -4.
Step 5: Substitute the values of x into either of the original equations to find the corresponding values of y.
Using the first equation, when x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35.
So, one possible solution is x = 5 and y = 35.
Using the first equation, when x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8.
So, another possible solution is x = -4 and y = 8.
Therefore, the system of equations can be solved algebraically, and the possible solutions are (x = 5, y = 35) and (x = -4, y = 8).
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).