Question

How do you solve the system of equations `-8x + 3y = 7` and `13x - 3y = - 1`?

Answer 1

`y=16.6/3 or 5.533333333`

`x=1.2`

Explanation:

`:.-8x+3y=7-------(1)`

`:.13x-3y=-1------(2)`

`:.(1)+(2)`

`:.5x=6`

`x=6/5` or `1.2`

substitute `x=1.2` in (1)

`:.-8(1.2)+3y=7`

`:.-9.6+3y=7`

`:.3y=7+9.6`

`:.3y=16.6`

`:.y=16.6/3` or `5.533333333`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

check:

substitute `y=5.533333333` and `x=1.2` in (1)

`:.-8(1.2)+3(5.533333333)=7`

`:.-9.6+16.6=7`

`:.7=7`

Answer 2

`x=6/5 =1.2`

`y= 5 8/15->83/15 larr" Exact value"`

Explanation:

`-8x+3y=7" "...........................Equation(1)`
`13x-3y=-1" "........................Equation(2)`

Although one is negative and the other is positive they both have `3y`

If we were to add these two equations together the `-3y` and `+3y` would cancel each other out and we would be left with only `x` as the unknown.

`-8x+3y=+7" "...........................Equation(1)`
`ul(color(white)("d")13x-3y=-1" ")........................Equation(2) larr" Add"`
`5x+0y=+6`

`x=6/5 =12/10=1.2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute for `x` in `Eqn(1)` giving:

`-8(6/5)+3y=7`

`3y=7+8(6/5)`

`y=7/3+8/cancel(3)^1(cancel(6)^2/5)`

`y= 5 8/15->83/15 larr" Exact value"`

`y~~15.53bar3`
where the bar over the three means the 3 goes on for ever.