Solve this system by substitution or addition...which is easier?

Question

I've done this equation both ways and been wrong, not sure why? Help would be greatly appreciated :)

2y - x = 3
x = 3y-5

Answer 1

OK here we go

2y - x = 3 and x = 3y -5

so if we substitute 3y - 5 for x in the first equation we have

2y -(3y - 5) = 3

distributed the negative

now we have

2y -3y + 5 = 3

combine like terms

-y + 5 = 3

subtract 5 from each side

-y = -2

Now change the signs on both sided

y = 2

Solve x

x = 3y - 5 and y = 2

x = 6 -5

x = 1

check

2y -x = 3

2(2) - 1 = 3

4 - 1 = 3

3=3

check

Answer 2

2y - x = 3
x = 3y - 5
Transfirm equations in terms of x

x = 2y - 3
x = 3y - 5

Take eq1 from eq2.

0 = y - 2 therefore y = 2

Sub y = 2 in eq 1 x = 6 - 3 = 3
Sub y = 2 in eq 1 x = 4 - 3 = 1

Sub x = 1 and y = 2 in eq 2
1 = 6 - 5 true

Subtraction is the easy way to do this. If you want to add there are two additional steps. You must multiply the first equation by 3 and the second by 2.

Do it the easiest way possible!

Answer 3

substitution is easyer in this case....

2y-(3y-5)=3 (dont forget to distribute the negative so u get -3y+5)
-y+5=3
-y=-2
y=2

then find x value

x=3y-5
x=3(2)-5
x=6-5
x= 1

check

2y-x=3
2(2)-1=3
4-1=3
3=3 check

x=3y-5
1=3(2)-5
1=6-5
1=1 check

Answer 4

Use sub. because you allready have the value for x.

2y-(3y-5) = 3

solve for y, then plug that in and get the value for x.

Answer 5

y=2
x=1