Question

How do you solve `\frac { 1} { 14} x + \frac { 1} { 7} = \frac { 22} { 14}`?

Answer 1

See a solution process below:

Explanation:

First, multiply each side of the equation by `color(red)(14)` to eliminate the fractions. `color(red)(14)` is the Lowest Common Denominator (LCD) of the three fractions:

`color(red)(14)(1/14x + 1/7) = color(red)(14) xx 22/14`

`(color(red)(14) xx 1/14x) + (color(red)(14) xx 1/7) = cancel(color(red)(14)) xx 22/color(red)(cancel(color(black)(14)))`

`(cancel(color(red)(14)) xx 1/color(red)(cancel(color(black)(14)))x) + (cancel(color(red)(14))2 xx 1/color(red)(cancel(color(black)(7)))) = 22`

`1x + 2 = 22`

`x + 2 = 22`

Now, subtract `color(red)(2)` to solve for `x` while keeping the equation balanced:

`x + 2 - color(red)(2) = 22 - color(red)(2)`

`x + 0 = 20`

`x = 20`

Answer 2

A lot of detail given. Once practised, a lot of steps will be done in your head and your solutions will be a lot shorter.

`x=20`

Explanation:

Multiply by 1 and you do not change the inherent value. However, 1 comes in many forms so you can change the way something looks without changing its inherent value.

`1/14x+[1/7color(red)(xx1)]=22/14`

`1/14x+[1/7color(red)(xx2/2)]=22/14`

`color(green)(1/14 x+2/14=22/14)`

Multiply `ul("all of both sides")` by `color(red)(14)` giving:

`color(green)([1/14color(red)(xx14)xx x]+[2/14color(red)(xx14)]=22/14color(red)(xx14))`

`color(white)(.)[1xx14/14xx x] +[2xx14/14]color(white)(.)=22xx14/14`

`" "[1xx1xx x] color(white)(..) +color(white)(.)[2xx1]" "=22xx1`

`" "x" "+" "2" "=22`

Subtract 2 from both sides

`x=20`