Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
y=3x+20
just equate the two expressions for y:
x^2+2x = 3x+20
x^2-x-20 = 0
(x-5)(x+4) = 20
x = 5, -4
So the solutions are (5,35) and (-4,8)
long foot
x^2+2x = 3x+20
x^2-x-20 = 0
(x-5)(x+4) = 20
x = 5, -4
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To solve the system of equations algebraically, we'll substitute one equation into the other to eliminate one of the variables.
Let's start by substituting the expression for y from the first equation into the second equation:
x^2 + 2x = 3x + 20
Now, let's rearrange the equation to set it equal to zero:
x^2 + 2x - 3x - 20 = 0
Combining like terms:
x^2 - x - 20 = 0
Next, we need to factor the quadratic equation. We're looking for two numbers that, when multiplied, give the constant term (-20) and, when added, give the coefficient of the linear term (-1). In this case, the numbers are -5 and 4:
(x - 5)(x + 4) = 0
To find the values of x that satisfy the equation, we set each factor equal to zero and solve for x:
x - 5 = 0 or x + 4 = 0
Solving for x in each equation:
x = 5 or x = -4
Now that we have the values of x, we can substitute them back into either of the original equations to find the corresponding values for y.
Using the first equation, let's find y when x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
Using the second equation, let's find y when x = -4:
y = 3(-4) + 20
y = -12 + 20
y = 8
So, the solution to the system of equations is x = 5 and y = 35, or x = -4 and y = 8.