Solve the system by the substitution method. Show your work.
2y - x = 5
x2 + y2 - 25 = 0
well, you know that x=2y-5
so, substitute that into the other equation, solve for y, and then you can get x.
2y - x = 5
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
4y² - 10y - 10y + 25
4y² - 20y + 25
4y² - 20y + 25 + y² - 25 = 0
4y² + y² - 20y = 0
4y² + y² - 20y = 0
y(4y + y - 20) = 0
y=0
x= -5
4y + y - 20 = 0
5y - 20 = 0
5y=20
y = 4
x = 3
or should I just do this instead of all my factors and simplifies
2y - x = 5
2y - 5 = x
so x = 2y - 5
(2y - 5)² + y² - 25 = 0
(2y - 5)(2y - 5)
To solve the system of equations using the substitution method, we can follow these steps:
Step 1: Solve one of the equations for one of the variables.
Let's solve the first equation, 2y - x = 5, for x:
Rearrange the equation to isolate x:
x = 2y - 5
Step 2: Substitute the expression found in Step 1 into the other equation.
Now, substitute the expression for x found in Step 1, which is x = 2y - 5, into the second equation, x^2 + y^2 - 25 = 0:
(2y - 5)^2 + y^2 - 25 = 0
Step 3: Simplify and solve the resulting equation.
Square the expression inside the parentheses, distribute, and combine like terms:
4y^2 - 20y + 25 + y^2 - 25 = 0
5y^2 - 20y = 0
Step 4: Factor the equation if possible or use the quadratic formula.
Here, we can factor out common terms:
y(5y - 20) = 0
So, either y = 0 or 5y - 20 = 0.
For the first case, if y = 0, substitute this value back into the first equation to find x:
2(0) - x = 5
0 - x = 5
x = -5
For the second case, if 5y - 20 = 0, solve for y:
5y - 20 = 0
5y = 20
y = 4
Substitute this value back into the first equation to find x:
2(4) - x = 5
8 - x = 5
-x = 5 - 8
-x = -3
x = 3
Hence, the solutions to the system of equations are (x,y) = (-5, 0) and (x,y) = (3, 4).