Rational equations problem help?

**homeylova223** · December 7th 2010, 03:45 PM

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.

**yeKciM** · December 7th 2010, 04:07 PM

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)

**homeylova223** · December 7th 2010, 04:52 PM

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.

**skeeter** · December 7th 2010, 05:13 PM

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 ...

**homeylova223** · December 7th 2010, 05:36 PM

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.

**HallsofIvy** · December 8th 2010, 02:38 AM

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 $28a^2- 28a- 15a- 15= 3a$ .

**homeylova223** · December 8th 2010, 02:36 PM

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).

**skeeter** · December 8th 2010, 05:18 PM

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 ...

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

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

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

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