Question 273009
I solved it by using the substitution method, if am right other wise explain in more detail what you looking for. HOPE i helped! ☺
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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