How do solve the following linear system?: $8 x + 2 y = 3, 6 x + 4 y = - 1$ $8x+2y=3 , 6x+4y=-1$ ?

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Answer 1 · 2018-03-07T01:11:21

x=0.5
y=-0.5

8x+2y=3 (1)
6x+4y=1 (2)

We now need to make 1 value in each eqaul. So lets do y. we need to multiply all in (1) by 2 to make the y value the same as (2).

16x+4y=6 (3)

we can now take (2) from (3).

16x+4y=6
-6x+4y=1
:10x+0=5
:10x=5
therefore x=0.5 (4)

sub (4) into (1):
8*0.5+2y=3
4+2y=3
2y=3-4
2y=-1
y=-0.5 (5)

Solve w/2

60.5+4-0.5=1

3+-2=1
3-2=1
OK!

x=0.5
y=-0.5

How do solve the following linear system?: 8x+2y=3 , 6x+4y=-1 8x+2y=3,6x+4y=−1 8x+2y=3 , 6x+4y=-1 ?