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

1 Answer
Jan 30, 2017

Answer:

#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.