Question

Rational Equations and Partial Fractions
Solve the equation: (7a/3a+3)-(5/4a-4)=(3a/2a+2)

Answer 1

\[\left( \frac{ 7a }{ 3a }+3 \right)-\left( \frac{ 5 }{ 4 }a-4 \right)=\left( \frac{ 3a }{ 2a }+2 \right)\] this one or \[\left( \frac{ 7a }{ (3a+3) } \right)-\left( \frac{ 5 }{ (4a-4) } \right)=\frac{ 3a }{ (2a+2) }\]

Answer 2

I'm thinking second

Answer 3

Hey are you there @snowywinter123

Answer 4

Hello, I'm here but yes the second one is the equation I need help on

Answer 5

\[\frac{ 7a }{ (3a+3) }-\frac{ 5 }{ (4a-4) }=\frac{ 3a }{ (2a+2) }\] it is best if you multiply both sides by the denominator, so as following \[\color{red}{(3a+3)(4a-4)(2a+2)}\left( \frac{ 7a }{ (3a+3) } -\frac{ 5 }{ (4a-4) }\right)=\left( \frac{ 3a }{ (2a+2) } \right)\color{red}{(3a+3)(4a-4)(2a+2)}\]

Answer 6

Ah darn it got cut off on the right but the red part is same as the left

Answer 7

We're multiplying both sides by all the denominators

Answer 8

\[7a(4a-4)(2a+2) -5(3a+3)(2a+2)=3(3a+3)(4a-4) \] ignore "isth" but is this correct so far?

Answer 9

oh nevermind, it didn't show up the "isthe"

Answer 10

Yes! Good work, now see if you can simplify by distributing

Answer 11

For
\[ \frac{ 7a }{ (3a+3) } -\frac{ 5 }{ (4a-4) } =\frac{ 3a }{ (2a+2) } \]
I would first factor each of the denominators, then multiply both sides by 12 to
"clear" the numbers:
\[ \frac{7a}{3(a+1)}- \frac{5}{4(a-1)} = \frac{3a}{2(a+1)} \\
\frac{28a}{a+1}- \frac{15}{a-1} = \frac{18a}{a+1}
\]
we have two terms with the same denominator: add -18/(a+1) to both sides and simplify
\[ \frac{10a}{a+1} - \frac{15}{a-1}=0\]
Notice we can divide both sides by 5 (to make the numbers a bit smaller):
\[ \frac{2a}{a+1} - \frac{3}{a-1}=0\]
now we use a common denominator of (a+1)(a-1)
to get
\[ \frac{2a(a-1)}{(a+1)(a-1)}- \frac{3(a+1)}{(a+1)(a-1)}= 0 \]
to get zero , the numerator must be zero. We have
\[ 2a^2 -2a -3a-3=0 \\ 2a^2 -5a-3=0\]
Finally, factor the quadratic (or use the quadratic formula) to solve for a