Question 345939: linear system.
2x+3y=8
-x+y=-4
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! 2x+3y=8_-x+y=-4
Since y does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting y from both sides.
2x+3y=8_-x=-y-4
Multiply each term in the equation by -1.
2x+3y=8_-x*-1=-y*-1-4*-1
Multiply -x by -1 to get x.
2x+3y=8_x=-y*-1-4*-1
Simplify the right-hand side of the equation by multiplying out all the terms.
2x+3y=8_x=y+4
Replace all occurrences of x with the solution found by solving the last equation for x. In this case, the value substituted is y+4.
2(y+4)+3y=8_x=y+4
Multiply 2 by each term inside the parentheses.
2y+8+3y=8_x=y+4
Since 2y and 3y are like terms, add 3y to 2y to get 5y.
5y+8=8_x=y+4
Since 8 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 8 from both sides.
5y=-8+8_x=y+4
Add 8 to -8 to get 0.
5y=0_x=y+4
Divide each term in the equation by 5.
(5y)/(5)=(0)/(5)_x=y+4
Simplify the left-hand side of the equation by canceling the common factors.
y=(0)/(5)_x=y+4
0 divided by any number or variable is 0.
y=0_x=y+4
Replace all occurrences of y with the solution found by solving the last equation for y. In this case, the value substituted is 0.
y=0_x=(0)+4
Remove the parentheses around the expression 0.
y=0_x=0+4
Remove the 0 from the polynomial; adding or subtracting 0 does not change the value of the expression.
y=0_x=4
This is the solution to the system of equations.
y=0_x=4
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