Question

How do you solve the system of equations `-6x + 5y = 8` and `x + 5y = - 13`?

Answer 1

`x=-3 and y = -2`

Explanation:

Let,
`-6x + 5y = 8` ----- (1)

`x + 5y = - 13`----(2)

(2)`=> x = -13 -5y`

Substitute this value of `x` in (1):

(1)`=> -6(-13 -5y) + 5y = 8`

`=> 78 + 30y +5y =8`

`=> 35y = 8-78`

`=> y = -70/35 = -2`

`therefore y = -2`

Now substitute this value in the expression of `x`:

`=> x = -13 -5y`

`=> x = -13 -5(-2)`

`=> x = -13 +10`

`therefore x = -3`

Answer: `x= -3 and y = -2`

Answer 2

`x =-3 and y=-2`

Explanation:

Notice that the `y` terms both have the same coefficient.

Subtracting the equations will therefore eliminate the `y` terms.

`" "x" "color(red)(+" "5y)" " =-13" "............................ A`
`-6x" "color(red)(+" "5y)" " =8" ".......................... B`

Subtract `A - B" "` (change the signs of `B`)

`" "x" "+" "5y" " =-13" "color(white)(xxxxxxxxx)A`
#ul(color(blue)(+)6x" "color(blue)(-)" "5y" " =color(blue)(-)8)" "color(white)(xxxxxxxxx)B#
#" "7xcolor(white)(xxxxxxxxx) =-21" "color(white)(xxxxxxxxxx)C" "larr div 7#
`" "xcolor(white)(xxxxxxxxx) =-3`

Substitute `-3 " for " x` in `A`

`-3+5y =-13`
`" "5y = -13+3`
`" "5y=-10`
`" "y=-2`

An alternative method is to equate the `y` terms.
Transpose the equations to isolate `5y` in each:

`5y = -13-x" and "5y =8+6x`

We know that `" "5y=5y`

Therefore: `" "6x+8 = -13-x" "larr` now solve for `x`

`6x+x = -13-8`
`" "7x = -21`
`" "x =-3`

Then proceed as described above to get `y=-2`

Check in `B`
`-6x+5y`
`=-6(-3)+5(-2)`
`=18-10`
`=8" "larr` the answer is the same as the RHS