Solve the system of equations algebraically. Show all of your steps.
Y=x^2+2x
y=3x+20
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To solve the system of equations algebraically, we can set the two expressions for y equal to each other.
First, set the expressions for y equal to each other:
x^2 + 2x = 3x + 20
Next, rearrange the equation to bring all terms to one side:
x^2 + 2x - 3x - 20 = 0
Combine like terms:
x^2 - x - 20 = 0
To factor the quadratic expression, we need to find two numbers that multiply to -20 and add up to -1 (the coefficient of x). After some trial and error, we find that -5 and 4 satisfy these conditions:
(x - 5)(x + 4) = 0
Now, we can apply the zero product property:
x - 5 = 0 or x + 4 = 0
Solving these two equations, we find two possible values for x:
x = 5 or x = -4
To find the corresponding y-values, substitute these values back into one of the original equations. Let's use the first equation:
When x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
When x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
Therefore, the solution to the system of equations is (x, y) = (5, 35) and (x, y) = (-4, 8).
To solve the system of equations algebraically, we can substitute the expression for "y" from the second equation into the first equation.
Step 1: Substitute the expression for "y" from the second equation into the first equation:
x^2 + 2x = 3x + 20
Step 2: Simplify the equation:
x^2 + 2x - 3x - 20 = 0
Step 3: Combine like terms:
x^2 - x - 20 = 0
Step 4: Factor the quadratic equation:
(x - 5)(x + 4) = 0
Step 5: Set each factor equal to zero and solve for "x":
x - 5 = 0 or x + 4 = 0
For x - 5 = 0:
x = 5
For x + 4 = 0:
x = -4
So, the solutions for "x" are x = 5 and x = -4.
Step 6: Substitute the values of "x" back into either equation to solve for "y":
For x = 5:
y = 3(5) + 20
y = 15 + 20
y = 35
For x = -4:
y = 3(-4) + 20
y = -12 + 20
y = 8
Therefore, the solutions to the system of equations are (x,y) = (5, 35) and (x,y) = (-4, 8).