[matin_hawk]

3x-y=7
5x+2y=8

x+3y=10
3x+7y=24

3x-7y=-27
6x-5y=0

8x+3y=16
2x+y=4

3x+4y-11=0
6x-5y+4=0

y=-x-8
y=7-4x

0.3x-0.2y=0.6
0.4x+0.3y=2.5

x+3/2y=2
x-2/3y=-1/6

y=x+4
2x+3y=22

x+y+z=6
x+2y+3z=14
x+3y+7z=28

could you please write down how you did it

[urmod4u]

I'll show you how to do it for the first equation, then you should be able to do the others yourself:
3x-y=7
5x+2y=8
From the first line:
3x-y=7
we can calculate the value of y in function of x:
3x=7+y
or y+7=3x
or y=3x-7
We fill in this value in the second line which means that we replace y by (3x-7):
5x+2(3x-7)=8
or 5x+6x-14=8
or 11x-14=8
or 11x=14+8
or 11x=22
or x=2
This exact value of x can now be reintroduced in one of the equations to calculate y. It doesn't matter which one, so we take the first one:
3x-y=7
or 3*2-y=7
or 6-y=7
or 6=7+y
or y=6-7
or y=(-1)
****
If you have 3 variables, you first take one equation to express the value of one variable in function of the two other variables, and introduce that value in a second equation to eliminate the next variable.
x+y+z=6
x+2y+3z=14
x+3y+7z=28
from the first:
x=6-y-z remember this one for later, call it (1)
fill this in in the second equation:
6-y-z+2y+3z=14
or: 6+y+2z=14
or y+2z=8
or y=(8-2z) remember this one for later, call it (2)
Remembering (1), we replace y in there by (8-2z):
x=6-(8-2z)-z
or x=6-8+2z-z
or x=z-2 remember this one, call it (3).
(2) and (3) both show the values for x and y expressed in function of z.
We fill them in in the third equation:
x+3y+7z=28
(z-2)+3(8-2z)+7z=28
or z-2+24-6z+7z=28
2z+22=28
z=3
Filling in this value of z in (2) and (3) will give you the values of x and y:
x=3-2
or x=1
y=(8-6)
or y=2
Result: x=1 y=2 z=3

[littletalavera]

how do I solve y=3x+4
y=4x

[Harryo]

-[2z(7z+2)] = 2+(4z+6)

[Unknown008]

*sigh* you people cannot start another thread?

littletalavera, what's common in both equations? y, isn't it? Then, equate both equations, i.e. 3x + 4 = y = 4x, therefore, 3x + 4 = 4x. Can you solve now?

Harryo, just expand the terms, then solve, that's easier since it contains only one unknown term.