Math
posted by Chris .
How do you solve this by linear combinations?
4x+3y=8
3x3y=15

L1 : 4x+3y=8
L2 : 3x3y=15
Do L1+L2> 7x=7 so x=1
in L1 4*(1)+3y=8
3y=12
y=4
the couple of solution is(1,4) 
How did you get x=1? by pugging it in with the 8 right?

No. Add equation 1 (what mktintin labels L1) to equation 2, (L2) to obtain
7x +0y = 8+15
7x=7
x=1 
doing L1+L2>4x+3x+3y3y=8+15
>7x=7
x=1
do u understand? 
Oh i was looking at it all wrong alright i understand it a bit better now, doh was plugging in the wrong numbers haha. Sorry.