Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
y=3x+20
3x + 20 = x^2 + 2x ... 0 = x^2 - x - 20 ... 0 = (x - 5)(x + 4) ... x = 5 , x = -4
Y = x^2 + 2x.
Y = 3x + 20.
In Eq1, replace Y with 3x + 20:
3x + 20 = x^2 + 2x,
x^2 + 2x - 3x -20 = 0,
x^2 - x - 20 = 0.
Solve the Eq by factoring.
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To solve the system of equations algebraically, we'll set the two equations equal to each other and solve for x.
x^2 + 2x = 3x + 20
Rearranging the equation, we get:
x^2 - x - 20 = 0
Factoring the quadratic equation, we find:
(x - 5)(x + 4) = 0
Setting each factor equal to zero, we have:
x - 5 = 0 or x + 4 = 0
Solving each equation, we find:
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 y-values.
For x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
For x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So, the solutions to the system of equations are: (5, 35) and (-4, 8).
To solve this system of equations algebraically, we can use the method of substitution.
Step 1: Set the two equations equal to each other.
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 we have two possible values for x: x = 5 and x = -4.
Step 6: Substitute the values of x into one of the original equations to solve for y.
For x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
So when x = 5, y = 35.
For x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So when x = -4, y = 8.
Therefore, the solutions to the system of equations are:
(x, y) = (5, 35) and (x, y) = (-4, 8).
To solve the system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously. Here are the steps to solve this system of equations:
Step 1: Set the two equations equal to each other since they both equal y.
x^2 + 2x = 3x + 20
Step 2: Rearrange the equation to bring all terms to one side to make it a quadratic equation.
x^2 + 2x - 3x - 20 = 0
Simplify:
x^2 - x - 20 = 0
Step 3: Factor the quadratic equation.
(x - 5)(x + 4) = 0
Step 4: Set each factor equal to zero and solve for x.
x - 5 = 0 ---> x = 5
x + 4 = 0 ---> x = -4
So, we have two possible values for x: x = 5 and x = -4.
Step 5: Substitute the values of x back into one of the original equations (either equation will work) to find the corresponding y-values.
For x = 5:
y = (5^2) + 2(5)
y = 25 + 10
y = 35
For x = -4:
y = (-4^2) + 2(-4)
y = 16 - 8
y = 8
Therefore, the solutions to the system of equations are (x, y) = (5, 35) and (x, y) = (-4, 8).