how do you solve 3x-y=9, 2x+y=6?

3x - y = 9; solve for y first. -y = 9 - 3x then y = -9 + 3x or y = 3x - 9.

Now that you have that, plug in 3x - 9 into the other equation where y is. it should look like this: 2x + (3x - 9) = 6
5x - 9 = 6
x = 15/5 = 3

now, use that as x and plug back into the first equation to get y:
3(3) - y = 9
y = 0

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3x - y = 9 2x + y = 6 Solve for Y in first equation... 3x - y = 9 3x - 9 = y Substitute for Y in second equation... 2x + y = 6 2x + 3x - 9 = 6 5x = 15 x = 3 Substitute for X in first equation... 3x - y = 9 (3*3) - y = 9 9 - y = 9 y = 0 Answer is (3,0)

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RE:
how do you solve 3x-y=9, 2x+y=6?

3x-y=9 2x+y=6
(3*x)-y=9 (2*x)+y=6
(3/3)=1x-y=(9/3)=x-y= 3 (2/2)*x+y=(6/2)
(x-y= 3) (x+y=3)

3x - y = 9 (exchange y & 9 position, when they're on the side, they become the opposite value like -y to +y)
hence y= 3x -9

2x+y = 6
2x + (3x -9) = 6 (substitute y's value from above here)
5x -9 = 6
5x = 6+9
5x = 15
x = 3

this is really an easy one.

so add the equations together to get rid of the y's

3x-y=9
2x+y=6
________________
5x=15
x=3

then substitute in x=3

3x3-y=9
9-y=9
y=0

6+y=6
y=0