There are two ways to solve this.... Elimination or substitutionSubstitution6x+3y=133x-y=4
First we free one value...
-y=4-3x
y=-4+3x
now we use this solve for x
6x+3(-4+3x)=136x-12+9x=13
15x= 13+12
x= 25/15
Now that we have x we find y
y= -4+3(25/15)y=1
Now we check
6(25/15)+3(1)=1310+3= 13
13=13
3x-y=4
3(25/15)-1=4
5-1=4
4=4
yeah... so x= 25/15 and y= 1
now elimination (Yay... *note of sarcasm*)
6x+3y=133x-y=4
we need to sum and both equation in a way to eliminate one variable... so we multiply 3 to eliminate y so...
6x+3y=133x-y=4 (*3)
6x+3y=139x-3y=12
now we resolve
15x= 25
x= 15/25
now we could use x in any equation to find y
6(25/15)+3y=1310+3y=13
3y=13-10
y=3/3
y=1
3x-y=4
3(25/15) -y= 4
5-y= 4
-y=4-5
-y=-1
y=1
So now we check if it is correct
6(25/15)+3(1)=1310+3= 13
13=13
3x-y=4
3(25/15)-1=4
5-1=4
4=4
So yeah... we can now tell that x= 25/15 and y = 1
your answer is y= 1...
Hope it helped...
