solve the systems of equations algebraically. please show all your steps.
y=x^2+2x
y=3x+20
since both are expressed as y = ...
we can equate them:
x^2 + 2x = 3x + 20
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
x=5, y = 35
x = -4, y = 8
check:
http://www.wolframalpha.com/input/?i=solve+y%3Dx%5E2%2B2x,++y%3D3x%2B20
Well, well, well, looks like we've got some equations in need of a solution! Let's get cracking, shall we?
First, let's set the two equations equal to each other:
x^2 + 2x = 3x + 20
Now, let's rearrange the equation and get rid of those pesky x's on both sides:
x^2 - x - 20 = 0
Now, let's factor this quadratic equation (if it can be factored):
(x - 5)(x + 4) = 0
Mamma mia, we did it! We've got two possible solutions:
x - 5 = 0 OR x + 4 = 0
So, either x = 5 or x = -4. Phew, that was half the battle!
Now, let's substitute these values of x back into one of the original equations to find the corresponding values of y. Let's use the first equation y = x^2 + 2x:
For x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
So, one solution is x = 5, y = 35.
For x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So, another solution is x = -4, y = 8.
Voila! We've successfully solved the system of equations! Remember to double-check your work and make sure that those answers indeed satisfy both equations. Happy solving!
To solve the system of equations algebraically, we'll set the two equations equal to each other and then solve for x.
y = x^2 + 2x (equation 1)
y = 3x + 20 (equation 2)
Setting equation 1 equal to equation 2, we get:
x^2 + 2x = 3x + 20
To solve this quadratic equation, we'll move all the terms to one side to obtain a quadratic expression equal to zero.
x^2 + 2x - 3x - 20 = 0
Combining like terms:
x^2 - x - 20 = 0
Now we'll factor the quadratic expression, if possible. Otherwise, we'll use the quadratic formula.
(x - 5)(x + 4) = 0
Setting each factor equal to zero, we get:
x - 5 = 0 or x + 4 = 0
Solving for x in each equation:
x = 5 or x = -4
Now, we'll substitute the values of x into either equation 1 or 2 to solve for y.
First, let's substitute x = 5 into equation 1:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
So, when x = 5, y = 35.
Next, let's substitute x = -4 into equation 1:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So, when x = -4, y = 8.
Therefore, the solution to the system of equations is (x, y) = (5, 35) and (-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 the system:
Step 1: Set the two equations equal to each other:
x^2 + 2x = 3x + 20
Step 2: Rearrange the equation to get all the terms on one side:
x^2 + 2x - 3x - 20 = 0
Step 3: Combine like terms:
x^2 - x - 20 = 0
Step 4: Factor the quadratic equation if possible. If factoring is not possible or not easily visible, you can use the quadratic formula:
(x - 5)(x + 4) = 0
Step 5: Set each factor equal to zero and solve for x:
x - 5 = 0 or x + 4 = 0
Solving for x, we find:
x = 5 or x = -4
Step 6: Substitute the values of x back into either of the original equations to find the corresponding values of y.
For x = 5:
Using the first equation: y = 5^2 + 2(5) = 25 + 10 = 35
Therefore, when x = 5, y = 35.
For x = -4:
Using the first equation: y = (-4)^2 + 2(-4) = 16 - 8 = 8
Therefore, when x = -4, y = 8.
So the solution to the system of equations is (x, y) = (5, 35) and (-4, 8).