Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Solve. x + 12 / x + 14 = x / x +8

OpenStudy (anonymous):

divide n factorise

OpenStudy (anonymous):

\[\LARGE {x+12\over x+14}={x\over x+8}\] I guess this is what you're asking.. isn't it?

OpenStudy (anonymous):

yess

jhonyy9 (jhonyy9):

first of all restrictions will be x not can being equal -14 and -8 so because than the fractions will be undefined

OpenStudy (anonymous):

not 14 sorry it was 4

OpenStudy (anonymous):

\[\LARGE x\neq -4 \quad \quad , x\neq -8\] :P lol....

OpenStudy (anonymous):

its Solve. x + 12 / x + 4 = x / x +8

jhonyy9 (jhonyy9):

all ways when there are fractions you need push the restrictions

OpenStudy (anonymous):

ok... here we go... \[\LARGE {x+12\over x+4}={x\over x+8 }\] use cross multiply: \[\LARGE (x+12)(x+8)= x(x+4)\] \[\LARGE x^2+8x+12x+96= x^2+4x)\] \[\LARGE \cancel{x^2}+20x+96= \cancel{x^2}+4x\] can you do it?

OpenStudy (anonymous):

Cross multiply \[ (x+8) (x+12)=x (x+4)\\ x^2+20 x+96=x^2+4 x\\ 20 x+96=4 x\\ 16 x = -96\\ x=-6 \]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!