Check my answer-
Solve the system of equations algebraically. Show all of your steps.
Y=x^2+2x
Y=3x+20
My answer:
X^2+2x+3x+20
X^2-x-20
(x-5),(x+4)=0
x=5, x=-4
y=35, y=8
Thank you!
No, you might have come up mysteriously with the answer but it does not follow from what you did.
You start off with two lines are are not equations,
X^2+2x+3x+20
X^2-x-20 ----> this does not follow from the previous line,
then in the third line all of a sudden you have
.... = 0
proper way:
x^2 + 2x = 3x + 20 , since Y = Y
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x = 5 or x =-4
if x = 5, y = 3(5) + 20 = 35
if x = -4, y = 3(-4) + 20 = 8
To solve the system of equations algebraically, we need to find the values of x and y that satisfy both equations simultaneously.
The given equations are:
1. Y = x^2 + 2x
2. Y = 3x + 20
To start, we can set the right sides of both equations equal to each other since they both equal Y:
x^2 + 2x = 3x + 20
Now, we have a quadratic equation. To solve it, we move all terms to one side:
x^2 - x - 20 = 0
To factor this quadratic equation, we need to find two numbers that multiply to -20 and add up to -1. After some trial and error, we find that -5 and 4 satisfy these conditions:
(x - 5)(x + 4) = 0
Now, we set each factor equal to zero and solve for x:
x - 5 = 0 or x + 4 = 0
Solving these equations gives us two possible values for x:
x = 5 or x = -4
Now, we substitute these values back into either of the original equations to find the corresponding y-values:
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:
x = 5, y = 35
x = -4, y = 8