Question

How do you solve `(7x)/5+5/2=x/10+1/5` and check for extraneous solutions?

Answer 1

`x=-23/13`

Explanation:

To eliminate the fractions in this equation, multiply ALL terms on both sides by the `color(blue)"lowest common multiple"` (LCM) of the denominators, 2 , 5 and 10

The LCM of 2 ,5 and 10 is 10

`(cancel(10)^2xx(7x)/cancel(5)^1)+(cancel(10)^5xx5/cancel(2)^1)=(cancel(10)^1xxx/cancel(10)^1)+(cancel(10)^2xx1/cancel(5)^1)`

`rArr14x+25=x+2larrcolor(red)" no fractions"`

subtract x from both sides.

`14x-x+25=cancel(x)cancel(-x)+2`

`rArr13x+25=2`

subtract 25 from both sides.

`13xcancel(+25)cancel(-25)=2-25`

`rArr13x=-23`

To solve for x, divide both sides by 13

`(cancel(13) x)/cancel(13)=(-23)/13`

`rArrx=-23/13`

`color(blue)"As a check"`

Substitute this value into both sides of the equation and if the left side is equal to the right side then it is the solution.

`(7/5xx(-23/13))+5/2=-161/65+5/2=3/130`

`-23/130+26/130=3/130`

left side = right side

`rArrx=-23/13" is the solution"`

Extraneous solutions are solutions generated by the solving of the equation which do not satisfy the original equation. There are no extra solutions here.