Question

How do I solve these three simultaneous equations?

Answer 1

subtract or substitute to eliminate one variable Then substitute into the third equation to solve for the third variable.

`r=4, s =-2 and t = +1`

Explanation:

Add the first and the third equations to eliminate the `r` variable and then solve for `t` in terms of `s`

`" "+4r -4s -6t = 18`
`+(-4r) +s +5t = -13`
`" " 0r -3s -t = 5`

Solve the resulting equation in terms of `t`

`-3s -t = 5` add `t-5` to both sides

`t = -3s - 5" "` now substitute this value into one of the equations.

`-2r -2s + 2 ( -3s -5) = -2` Solve this equation for r in terms of s

`-2r -2s + -6s -10 = -2` Combining like terms gives

`-2r - 8s -10 =-2` Add 10 to both sides

`-2r -8s -10 +10 = - 2 + 10` which gives

`-2r -8s = +8` add -8s to both sides

`-2r -8s + 8s = +8 + 8s` which gives

`-2r = 8 + 8s` divide both sides by -2

`( -2r)/-2= (8 + 8s)/-2` which equals

`r = -4s -4` Which can be substituted into an equation

`4 ( -4s -4) -4s -6 ( -3s -5) = 18" "` now solve for s

`-16s - 16 -4s + 18s + 30 = 18` combining like terms gives

`-2s + 14 = 18` subtracting 14 from both sides gives

`-2s + 14 -14 = 18 -14` which gives

`-2s = 4" "` Divide both sides by -2

`(-2s)/-2 = 4/-2`

`s = -2" "` Now this value can be used to solve for `r and t`

`r = -4(-2) -4` so

`r = 8-4 =4`

`t = -3( -2) -5`

`t = +6 -5`

`t = +1`

With three equations and three variables it is necessary to substitute to find values for two of the variables in terms of the third. Then substituting the values find the value for one of the variables which can then be used to find values for the other two variables.

Answer 2

`r =4 and s=-2 and t=1`

Explanation:

`color(white)(www)color(blue)( 4r)-4s -6t =18color(white)(www.ww.ww)A`
`" "-2r-2s+2t =-2 color(white)(wwwwwww)B`
`" "color(blue)(-4r)+s+5t=-13color(white)(wwwwwww)C`

Notice that in `A and C` there are additive inverses.
`color(blue)(+4r+(-4r) =0)`

Adding these two equations together will eliminate the `r` term.

`A+C:" " -3scolor(red)(-t) =5" "larr` solve for `t` by isolating it

`color(white)(wwxxxwww)color(red)(-3s -5 =t)" "larr` an expression for `t`

Make additive inverses with the terms in `r`

`Adiv2:" "color(blue)(2r)-2s-3t = 9color(white)(xxxxxxx)D`
`color(white)(xxx.xx)color(blue)(-2r)-2s+2t =-2 color(white)(w.www)B`

`D+B:color(white)(.......)-4s-t =7larr` solve for `t` by isolating it

`color(white)(wwwwwwww)color(red)(-4s -7 =t)" "larr` another expression for `t`

We now have two expressions for `t`, they have to be equal

`color(white)(wwwwwwwwww)color(red)( t" "=" "t)`
`color(white)(www.wwww)color(red)(-3s -5 = -4s-7`

`color(white)(www.ww.ww)4s-3s = 5-7`
`color(white)(www.wwwww..ww)s = -2" "larr` the value of `s`

Substitute to find `t:" " t = -3s -5`
`color(white)(wwwwwwwwwww.ww)t = -3(-2) -5`
`color(white)(wwwwwwwwwww.ww)t = 1" "larr` the value of `t`

Substitute the values for `s and t` in `C` to find `r`

`" "-4r+" "s+" "5t=-13color(white)(wwwwwww)C`
`" "-4r+(-2)+5(1)=-13`

`color(white)(xxxxxx.x)-2+5+13 = 4r`
`color(white)(xxxxxxxxxxxxxx.x)16 = 4r`
`color(white)(xxxxxxxxxxx.xxx.x)4 = r`

You can substitute the three values into any of the original equations to check that they are right.