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...
