That looks like a pretty basic "High School Algebra" question. First, of course, it is two equations:
2x−y=73(2x+y) and
73(2x+y)=6x−7y+4;
(The third equation,
2x−y=6x−7y+4, is implied but gives no new information.)
The first thing I would do is simplify both equations. Multiply both sides of the first equation by 7:
7(2x- y)= 14x- 7y= 3(2x+ y)= 6x+ 3y
14x- 7y= 6x+ 3y
Subtract 6x from both sides and add 7y to both sides:
8x= 10y. Divide both sides by 2: 4x= 5y.
That's about as simple as you can get it.
Multiply both sides of the second equation by 7: 3(2x+ y)= 6x+ 3y= 7(6x- 7y+ 4)= 42x- 49y+ 28
6x+ 3y= 42x- 49y+ 28
Subtract 6x from both sides and add 49y to both sides:
52y= 36x+ 28
Divide both sides by 4: 13y= 9x+ 7
That's about as simple as you can get that.
We can divide both sides of 4x= 5y by 4 to get
x=45y.
Replace x by that in 13y= 9x+7:
13y=445y+7
so that we have a single equation in the single unknown, y.
Subtract
445y from both sides.
13=1344=452 so
13y−445y=(452−445)y=47y=7.
Multiply both sides by
74:
y=4.
Since
x=45y from before,
x=45(4)=5.
x= 5, y= 4.
And, of course, check those answers in the original equations,
2x−y=73(2x+y)=6x−7y+4.
If x= 5, y= 4 then
2x−y=2(5)−4=10−4=6
73(2x+y)=73(2(5)+4)=73(14)=6
6x−7y+4=6(5)−7(4)+4=30−28+4=2+4=6.
Yes, those are all equal so x= 5, y= 4 satisfy the equation.