Question

A rectangle has perimeter P of 58 cm. Its length l is one centimeter longer than 3 times its width w. Write and solve a system of linear equations using these variables to find the length and width of the rectangle?

Answer 1

`color(red)("The"` `color(red)("length"=22 and "width" = 7`

Explanation:

Width(w)=`w`

Length`(l)=3w+1`

Perimeter`(P)=58`

`2(l+w)=58`

`2(3w+1+w)=58`

`3w+w+1=58/2`

`4w+1=29`

`4w=29-1`

`4w=28`

`w=28/4`

`color(magenta)(w=7`

`l=3w+1`

`l=3xx7+1`

`l=21+1`

`color(magenta)(l=22`

`therefore` `color(red)("The"` `color(red)("length"=22 and "width" = 7`

~Hope this helps! :)

Answer 2

`L=22 and w=7`

Explanation:

I am using capital L instead of lower case because the lower case looks like the digit 1

Note that `2L+2w=P=58`

Its length is ............`color(white)("d")L=?`
one cm longer than `L=?+1`
3 times ....................`color(white)("d")L=3(?)+1`
its width .................`color(white)("d")L=(3w)+1`

So we have:
`2L+2w=58" "....................Equation(1)`
`L=3w+1" ".........................Equation(2)`

The trick to solving for an unknown is to end up with just 1 equation containing just 1 of that unknown type and a whole set of numbers in some format.

I am going to change `Equation(1)` so that it has just 1 unknown.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(ul("This part in a lot of detail using first principles"))`
`color(blue)("The shortcut methods you were taught are only")` `color(blue)("remembering the outcomes of first principles.")`

Consider `Eqn(2) ->color(red)(L=3w+1)`

Using `Eqn(2)` substitute for `L` in `Eqn(1)`

`color(green)(2color(red)(L)+2w=58 color(white)("dddd")->color(white)("dddd") 2(color(red)(3w+1))+2w=58)`

`color(green)(color(white)("ddddddddddddddd")->color(white)("ddddd")6w+2color(white)("d.")+2w=58)`

`color(green)(color(white)("ddddddddddddddd")->color(white)("ddddd")8w+2color(white)("d.")=58)`

Subtract `color(red)(2)` from both sides

`color(green)(8w+2=58 color(white)("dddd")->color(white)("dddd") 8w+2color(red)(-2)color(white)("d")=58color(red)(-2))`

`color(green)(color(white)("dddddddddddddd")->color(white)("dddd")8wcolor(white)("d")=56)`

Divide both sides by `color(red)(8)`

`color(green)(8w=56color(white)("ddddddd")->color(white)("dddd")8/color(red)(8) w=56/color(red)(8))`

`color(green)(color(white)("dddddddddddd.d")->color(white)("dddddd")w=7`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("This part not so detailed")`

Given that `w=7` and that we have: `L=3w+1 =>L=22`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check using "Equation(1))`

`2L+2w=58`

`2(22)+2(7) = 58`