# Thread: Rational equations problem help?

1. ## Rational equations problem help?

I am having difficulty solving these equations

(7A/3A+3)-(5/4A-4)=(3A)/(2A+2)

(2Q/(2Q+3)-(2Q/2Q-3)=1

and

(12/t)+t-8=0
Can anyone help me solve these and show me how to as I am stuck on rational equations I am not good at it.

2. Originally Posted by homeylova223
I am having difficulty solving these equations

(7A/3A+3)-(5/4A-4)=(3A)/(2A+2)

(2Q/(2Q+3)-(2Q/2Q-3)=1

and

(12/t)+t-8=0
Can anyone help me solve these and show me how to as I am stuck on rational equations I am not good at it.

you have ...
$\displaystyle \frac {12}{t} + t - 8 = 0$

so multiply by "t" to loose fraction

$\displaystyle 12 + t^2 -8t = 0$

and that's

$t^2-8t+12 = 0$

same as that you have

$ax^2+bx+c = 0$

solutions would be

$\displaystyle x_{1,2} = \frac {-b \pm \sqrt{b^2-4ac}}{2a}$

and for first two you can do the same thing ... with what you should multiply (the whole ) equation to lose fractions ? (if fractions are thing that you have problems with)

3. Right now I am having trouble figuring out (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)
I think you can use 24a^2+12a^2-12a-24 as a common factor but then I get stuck when trying to solve it. This is what I usually get when I am done
56a^3+4a+30a^2+60a+3=36a^3-36a but I am not sure if this is right.

4. Originally Posted by homeylova223
Right now I am having trouble figuring out (7A/3A+3)-(5/4A-4)=(3A)/(2A+2)
I think you can use 24a^2+12a^2-12a-24 as a common factor but then I get stuck when trying to solve it. This is what I usually get when I am done
56a^3+4a+30a^2+60a+3=36a^3-36a but I am not sure if this is right.
$\displaystyle \frac{7a}{3(a+1)} - \frac{5}{4(a-1)} = \frac{3a}{2(a+1)}$

common denominator is $12(a+1)(a-1)$ ...

$\displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}$

try it again ...

5. After doing that I get
28A/12(A+1)-15/12(a-1)= 18A/12(a+1)

Now from the back of the book I know my final answer are (-1/2 and 3). I am just no sure how to get there.

6. Skeeter's point is that after you have the same denominator for every fraction, you can multiply through by that denominator and it disappears!

From $\displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}$, multiply each part by 12(a+1)(a-1) and you have 7a(4)(a-1)- 5(3)(a+1)= 3a which is the same as $28a^2- 28a- 15a- 15= 3a$.

7. Then I get 28a^2-46a-15
I am not sure how to factor this to get to -1/2 and 3 unless I have to use quadratic formula. Maybe I have to multiply the 3a(6)(a-1).

8. Originally Posted by homeylova223
Then I get 28a^2-46a-15 ... no.
I am not sure how to factor this to get to -1/2 and 3 unless I have to use quadratic formula. Maybe I have to multiply the 3a(6)(a-1).
correction ...

$10a^2 - 25a - 15 = 0$

$5(2a^2 - 5a - 3) = 0$

$5(2a + 1)(a - 3) = 0$