24. solve the system of equations algerbraically. show all of your steps. (2 points)
y=x^(2)+2x
y=3x+20
I'm just having trouble understanding.
Please help
Since you have two values for y, just set them equal to each other:
x^2+2x = 3x+20
x^2-x-20 = 0
(x-5)(x+4) = 0
and you're home free...
To solve the system of equations algebraically, we can use the method of substitution or elimination. Let's use the substitution method to solve this system.
Step 1: Start by isolating one variable in one of the equations. We can choose either equation to begin with. Let's solve the first equation for y to get:
y = x^2 + 2x .............(Equation 1)
Step 2: Now substitute the value of y from Equation 1 into the second equation:
x^2 + 2x = 3x + 20 ...........(Equation 2)
Step 3: Simplify Equation 2 by moving all the terms to one side:
x^2 + 2x - 3x - 20 = 0
Step 4: Combine like terms:
x^2 - x - 20 = 0 ...........(Equation 3)
Step 5: Now, we can solve Equation 3 by factoring, completing the square, or using the quadratic formula. In this case, let's factor Equation 3:
(x - 5)(x + 4) = 0
Step 6: Set each factor equal to zero and solve for x:
x - 5 = 0 --> x = 5
x + 4 = 0 --> x = -4
Step 7: Now that we have the values of x, we can substitute them back into either equation to solve for y. Let's substitute x = 5 into Equation 1:
y = (5^2) + 2(5)
y = 25 + 10
y = 35
So one solution to the system of equations is x = 5 and y = 35.
Similarly, substituting x = -4 into Equation 1:
y = (-4^2) + 2(-4)
y = 16 - 8
y = 8
So another solution to the system of equations is x = -4 and y = 8.
Hence, the system of equations has two solutions: (5, 35) and (-4, 8).