Answer :
[tex]y=2x-5 \\
4x-y=7 \\ \\
\hbox{substitute 2x-5 for y in the 2nd equation and solve for x:} \\
4x-(2x-5)=7 \\
4x-2x+5=7 \\
2x+5=7 \\
2x=7-5 \\
2x=2 \\
x=\frac{2}{2} \\
x=1 \\ \\
\hbox{substitute 1 for x in the 1st equation:} \\
y=2 \times 1-5=2-5=-3 \\ \\
\hbox{the answer:} \\
x=1 \\ y=-3[/tex]
This is straight forward:
You have already said: y = 2x - 5 ...........(i)
4x - y = 7............(ii)
So anywhere we see, y in equation (ii) we replace it with (2x - 5).
4x - (2x - 5) = 7
4x - 2x + 5 = 7
Note minus sign before the bracket, changes sign inside the bracket.
2x + 5 = 7
2x = 7 - 5.
2x = 2 Divide both sides by 2.
x = 2/2
x =1.
Remember from (i) y = 2x - 5, y = 2*1 - 5 = 2-5 = -3.
Therefore, x = 1, y = -3.
Those are the steps.
You have already said: y = 2x - 5 ...........(i)
4x - y = 7............(ii)
So anywhere we see, y in equation (ii) we replace it with (2x - 5).
4x - (2x - 5) = 7
4x - 2x + 5 = 7
Note minus sign before the bracket, changes sign inside the bracket.
2x + 5 = 7
2x = 7 - 5.
2x = 2 Divide both sides by 2.
x = 2/2
x =1.
Remember from (i) y = 2x - 5, y = 2*1 - 5 = 2-5 = -3.
Therefore, x = 1, y = -3.
Those are the steps.
