Question

How do you solve the following linear system: `-x+3y=-9 , 8x-4y=32`?

Answer 1

`y=-2`
`x=3`

Explanation:

We have, `3y-x+9=0` and `4(2x-y)=4xx8`
Or, `3(y+3)=x` and `2x-y=8`
Substituting the value of `x` from Equation 1 in Equation 2,we get,
`2(3y+9)-y-8=0`
Or, `6y-y+18-8=0`
Or, `5y=-10`
Or, `y=-2`
Thus, `x` comes out to be `3xx(3-2)` or `3xx1` or `3`

Answer 2

`x=3`

`y=-2`

Explanation:

Given:

`-x+3y=-9" "......................Equation(1)`
`color(white)("d")8x-4y=32" ".........................Equation(2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "y" - in detail using first principles")`

`color(brown)("Consider "Eqn(1))`

`color(brown)("Multiply both sides by "(-1))`

`+x-3y=+9`

`color(brown)("Add "3y" to both sides")`

`x=9+3y" ".......Equation(1_a)`

`color(brown)("Using "Eqn(1_a)" substitute for "x" in "Eqn(2))`

`color(green)(8color(red)(x)-4y=32 color(white)("dddd")->color(white)("dddd")8(color(red)(9+3y))-4y=32 )`

`color(green)( color(white)("dddddddddddddd.")->color(white)("dddd")72+24y-4y=32 )`

`color(green)( color(white)("dddddddddddddd.")->color(white)("dddd")72+20ycolor(white)("dddd")=32 )`

`color(brown)("Subtract 72 from both sides")`

`color(green)( color(white)("dddddddddddddd.")->color(white)("dddddddd")20ycolor(white)("dddd")=-40 )`

`color(brown)("Divide both sides by 20")`

`color(green)( color(white)("dddddddddddddddd.")->color(white)("dddddddd")ycolor(white)("dddd")=-2 )`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "x" - not so much detail")`

`color(brown)("Substitute for "y" in "Eqn(1))`

`-x+3y=-9color(white)("dddd")->color(white)("dddd")-x+3(-2)=-9`

`color(white)("ddddddddddddddddddddddddd")->color(white)("dddd") -x=-3`

`x=3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

`-x+3y color(white)("d")->color(white)("d")-3+(3xx-2) ->9` as required

`8x-4y color(white)("d")->color(white)("d") 8(3)-4(-2) -> 24+8->32` as req'd