Question

How do you solve `\frac { 2} { 3} x - 5= \frac { x } { 4} - 10`?

Answer 1

`x=-12`

Explanation:

Let's solve this one step at a time.

First, we need to get rid of the pesky fractions. Since there are 2 fractions, let's find the LCM (Lowest common multiple). Both 3 and 4 goes in 12, so let's multiply the equation by 12.
`2/3*12/1=24/3=8`

`x/4*12/1=(12x)/4=3x`
Therefore,
`8x-60=3x-120`
To get `x` on one side, let's add 60 to both sides.
`8x=3x-60`
Subtract `3x` from both sides to get the constant (-60) on its own.
`5x=-60`
Finally, divide both sides by 5 to isolate `x`
`x=-12`

Answer 2

`x=-12`

Explanation:

`"collect terms in x on the left side and numeric values on the"`
`"right side"`

`"Note " 2/3x=(2x)/3`

`"subtract " x/4" from both sides"`

`rArr(2x)/3-x/4-5=cancel(x/4)cancel(-x/4)-10`

`"add 5 to both sides"`

`(2x)/3-x/4cancel(-5)cancel(+5)=-10+5`

`rArr(2x)/3-x/4=-5`

`"before subtracting the fractions we require them to"`
`"have a "color(blue)"common denominator"`

`" the lowest common multiple of 3 and 4 is 12"`

`rArr(2x)/3xx4/4-x/4xx3/3=-5`

`rArr(8x)/12-(3x)/12=-5`

`rArr(5x)/12=-5`

`"multiply both sides by 12"`

`cancel(12)xx(5x)/cancel(12)=(-5xx12)`

`rArr5x=-60`

`"divide both sides by 5"`

`(cancel(5) x)/cancel(5)=(-60)/5`

`rArrx=-12`

`color(blue)"As a check"`

Substitute this value into the equation and if both sides are equal then it is the solution.

`"left side "=(2/3xx-12)-5=-8-5=-13`

`"right side "=(-12)/4-10=-3-10=-13`

`rArrx=-12" is the solution"`