How do you solve the following system: `8x-7y=-39, 7x − 5y = −11`?

`8x-7y=-39`...(1)
`7x-5y=-11`...(2)

(1)*5:
`40x-35y=-195`...(3)

(2)*7:
`49x-35y=-77`...(4)

(4)-(3):
`9x=118`
`x=118/9`

Sub x=118/9 into (1)
`944/9-7y=-39`
`-7y=-1295/9`
`y=185/9`

`color(blue)("Preamble")`

I choose to answer this the traditional way.

Example of the principle used. When I realised this (long time ago) it opened the door for me to a whole new way of thinking.

Suppose we have `color(white)("d")color(green)(3x-6=5)`

Then it is also true that

`color(green)(color(white)("dddddddddd") color(red)(2xx)(3x-6)=color(red)(2xx)(5))`
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`color(blue)("Answering the question")`

Given:
`8x-7y=-39" "....................Equation(1)`
`7x-5y=-11" "......................Equation(2)`

If we change one or both of these then subtract it will eliminate one of the unknown. Then we have 1 equation with 1 unknown and thus solvable

Lets 'get rid' of the `y's`

Multiply `Eqn(1)` by 5 and `Eqn(2)` by 7

`color(white)("d")40x-35y=-195" ".........Equation(1_a)`
`color(white)("d")ul(49x-35y=-77" "............Equation(2_a)larr" Subtract")`
`-9x+color(white)("d")0y= -118`

Divide both side by -9

`x=+118/9`

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I will let you determine `y`. Just substitute this value into one of the equation.