Question

Easy algebra 2 h3lp?

Answer 1

\[\frac{ w ^{2}+2w-24 }{ w ^{2}+w-30 }+\frac{ 8 }{ w-5 }\]
you need to first get a common denominator
\[\frac{ w ^{2}+2w-24 }{ w ^{2}+w-30 }+\frac{ 8 }{ w-5 }X \frac{ w+6 }{ w+6 }\]
\[\frac{ w ^{2}+2w-24 }{ w ^{2}+w-30 }+\frac{ 8w+48 }{ w ^{2}+ w-30 }\]
\[\frac{ w ^{2}+2w-24+8w+48 }{ w ^{2}+w-30 }\]
\[\frac{ w ^{2}+10w+24}{ w ^{2}+w-30 }\]
then you cancel out
\[9w\frac{ 24 }{ -30 }\]

Answer 2

\[\Large \frac{ w^{2}+2w-24 }{ w^{2}+w-30}+\frac{ 8 }{ w-5 }=\frac{ w^{2}+2w-24 }{ (w+6)(w-5)}+\frac{ 8 }{ w-5 }\]
\[\Large \frac{ w^{2}+2w-24 }{ (w+6)(w-5) }=\frac{ (w+6)(w-4) }{ (w+6)(w-5) }=\frac{ w-4 }{ w-5 }\]
\[\Large \frac{ w-4 }{ w-5 }+\frac{ 8 }{ w-5 }=\frac{ w-4+8 }{ w-5 }=\frac{ w+4 }{ w-5 }\]

Answer 3

\[\frac{ w ^{2}+10w+24 }{ w ^{2}+w-30 }\]
factor both the top and bottom
\[\frac{(w+6)(w+4) }{ (w-5)(w+6) }\]
\[\frac{ w+4 }{ w-5 }\]

Answer 4

note that it's only valid as long as w is not equal to -6 though

Answer 5

Wow that was really easy i didnt think it would be like that.

\(\rm\Huge\color{pink}{thanks~so~much!!!}\)\(\Huge\bigstar\)