Friday oct 20th

In today's class we continued learning about subtitutions. We learned that it's possible to cancel out a varible, even when it has a coeificent. example:
L1: 5x-3y=-5
L2: 3x+2y=4
(these equations can be solved)

Notes:

1. we need to find a common multipule between ether the y-values or the x-values. example:
example: (common y-value between 3 and 2 is 6)

2. then we need to subtitute 6 into both L1 and L2 values.
L1:10x-6y=-10
L2:9x-6y=12

3. After that we can cancel out the y-variable and solve for x.
19x=2, so x=2/19

4. then substitute 2/19 for x into one of the equations. and solve for y.
3x+2y=4
3(2/19)+2y=4
y=35/19

5. so in conclusion the answer is: (x=2/19) and (y=35/19)

Monday's scribe is Marty.