Question #f47ec

In this example multiplying one of the equations by `-1` will result in one positive `x` term and one negative `x` term.
When added they will equal `0x`

.Step `1.` multiply the first equation by `-1`

`-1( -x -y =11) = + x + y = -11`

Step `2` add the two equations to get a `0` value for the `x` term.

`+x + y = -11`
`ul(-x +3y = -13)" "` which gives
`0x + 4y = -24`

Step `3.` solve for the other variable. `(y)`

`(4y) /4 = (-24)/4` so

`y = -6`

step `4.` substitute the value for one variable into one of the equations and solve for the other variable.

`-x - (-6) = 11 " "` add `-6` to both sides

`-x + 6 - 6 = 11 - 6` which gives

`-x = 5` divide both sides by `-1`

`(-x)/-1 = 5/-1` so

`x = -5`

`-x - y= 11.`

This means:

`-x = 11+y`

Now we substitute `- x` in the second equation:

`- x + 3y = -13`

We have:

`11+ y+ 3y= -13`

`11+ 4y = -13`

`4y = -13 -11`

`y = -24/4 = -6`

Now we substitute `y` in either one of the two equations:

`- x -(-6) = 11`

`- x +6 = 11`

`- x = 11 - 6`

`x = - 5`

We can now substitute `x` and `y` in either equation to be sure that the answer is right.

The goal here is to solve for one variable by first eliminating one variable, hence, the elimination method.

Given:

`-x-y=11`

`-x+3y=-13`

We can eliminate the `x` variable since they have the same coefficient by subtracting both equations. Thus,

`" "cancel(-x)-y=11`
`-`
`" "cancel(-x)+3y=-13`

This yields:

`-4y=24`

Now we solve for the variable `y`

`cancel(-4)/cancelcolor(red)(-4)y=24/color(red)(-4)`

`y=-6`

The next step is to find the value for the variable `x` by substituting the value for `y` into one of the original equations. I will use the first equation.

We substitute `-6` for `y` in the first equation and solve for `x`

`-x-(color(red)(-6))=11`

`-x+6=11`

`-x+6color(red)(-6)=11color(red)(-6)`

`-x=5`

`x=-5`

We now have the solutions `x=-5` and `y=-6` or `(-5,-6)` but we must check if the solution checks out but plugging both values into both equations.

Equation 1:

`-(-5)-(-6)=11`

`5+6=11`

`11=11` TRUE

Equation 2:

`-(-5)+3(-6)=-13`

`5-18=13`

`13=13` TRUE

The values check out. The solution `(-5,-6)`

You can learn more about this method here: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L2_T2_text_final.html

The most important concept with the ELIMINATION method concerns ADDITIVE INVERSES.

Values which have the same number but opposite signs are called additive inverses. Their sum is `0`

`(+5)+(-5) = 0," " (-12)+(+12) =0`

To eliminate one of the variables, create additive inverses.
Other contributors have eliminated `x`, but let's look
at the `y`-terms instead

Notice that the `y`-terms already have opposite signs, but have different numbers. Create additive inverses.

`color(white)(xxxx)-xcolor(blue)(-y) =+11 " "A`
`color(white)(xxxx) -xcolor(blue)(+3y)=-13" "B`

`A xx3: -3xcolor(blue)(-3y)=+33" "C" "` now there are
`color(white)(xxx.xx) -xcolor(blue)(+3y)=-13" "B" "` additive inverses

Add the equations together:

`C+B:" "-4x " "= 20" "larr` only `x` values,
`color(white)(xxxxx.xxxx) x=-5`

That that you know the value for `x`, substitute to find `y`

`" "-x -y = 11`

`" "-(-5)-y = 11`

`5-11 = y`

`-6=y`