|
Question 273009: solve the following systems of equations.
x+2y=3
2x-y=1
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! I solved it by using the substitution method, if am right other wise explain in more detail what you looking for. HOPE i helped! ☺
--------------------------------------------------------------------------------
x+2y=3_2x-y=1
Since 2y does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2y from both sides.
x=-2y+3_2x-y=1
Replace all occurrences of x with the solution found by solving the last equation for x. In this case, the value substituted is -2y+3.
x=-2y+3_2(-2y+3)-y=1
Multiply 2 by each term inside the parentheses.
x=-2y+3_(-4y+6)-y=1
Since -4y and -y are like terms, subtract y from -4y to get -5y.
x=-2y+3_-5y+6=1
Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.
x=-2y+3_-5y=-6+1
Add 1 to -6 to get -5.
x=-2y+3_-5y=-5
Divide each term in the equation by -5.
x=-2y+3_-(5y)/(-5)=-(5)/(-5)
Simplify the left-hand side of the equation by canceling the common terms.
x=-2y+3_y=-(5)/(-5)
Simplify the right-hand side of the equation by simplifying each term.
x=-2y+3_y=1
Replace all occurrences of y with the solution found by solving the last equation for y. In this case, the value substituted is 1.
x=-2(1)+3_y=1
Multiply -2 by each term inside the parentheses.
x=-2+3_y=1
Add 3 to -2 to get 1.
x=1_y=1
This is the solution to the system of equations.
1st Answer: x=1
2nd Answer: y=1
|
|
|
| |
