Question

solve this system of equations problem, both are togethor as the problem

2x-y=12
x+3y=13

Answer 1

2x-y=12
-2(x+3y=13)
------------
-7y=-14
y=2

2x-2=12
x=7

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x=13-3y
2(13-3y) - y = 12
26 - 6y - y = 12
-7y = -15
y = 2
x+3(2) = 13
x = 7

Answer 2

Solving a system of equations means finding their point of intersection. You can find this the most easily by rewriting both equations in slope-intercept form: y=mx+b and then setting them equal to each other.

2X-Y=12
2X-2X-Y=12-2X
-Y=12-2X
-Y=12-2X
---- -------
-1 -1
Y=-12+2X or Y=2X-12 <---This the equivalent of your first equation


X+3Y=13
X-X+3Y=13-X
3Y=13-X
3Y=13-X
--- ------
3 3
Y=4 1/3-1/3X or Y=-1/3X+4 1/3 <---This is the equivalent of the second

So now you set them equal to each other like this.

2X-12=4 1/3-1/3X
2X-12+12=4 1/3+12-1/3X
2X=16 1/3-1/3X
2X=16 1/3-1/3X
2X+1/3X=16 1/3-1/3X+1/3X
2 1/3X=16 1/3
X=7

This gives you the X-value of the point of intersection, and all you have to do now is plug this X-value into either of the original (or new) equations to get the Y-value for the coordinate. Sorry for the long answer!