Solve the system of equation algerbraically. Show all steps y=x^2+2x
y=3x+20
To solve this system of equations algebraically, we can set the two equations equal to each other since they both equal y:
x^2 + 2x = 3x + 20
Next, we can simplify this equation by moving all terms to one side to set it equal to zero:
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0. We can solve this quadratic equation by factoring or using the quadratic formula. In this case, we can factor:
(x - 5)(x + 4) = 0
Setting each factor to zero gives us two possible solutions:
x - 5 = 0 or x + 4 = 0
x = 5 x = -4
So the solutions to the system of equations are x = 5 and x = -4.
To find the corresponding y values, we can substitute these x values back into either equation. Let's substitute into the first equation y = x^2 + 2x:
For x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
For x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
Therefore, the solutions to the system of equations are (5, 35) and (-4, 8).