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.
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 \displaystyle \frac {12}{t} + t - 8 = 0 $
so multiply by "t" to loose fraction
$\displaystyle \displaystyle 12 + t^2 -8t = 0$
and that's
$\displaystyle t^2-8t+12 = 0 $
same as that you have
$\displaystyle ax^2+bx+c = 0$
solutions would be
$\displaystyle \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)
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.
$\displaystyle \displaystyle \frac{7a}{3(a+1)} - \frac{5}{4(a-1)} = \frac{3a}{2(a+1)}$
common denominator is $\displaystyle 12(a+1)(a-1)$ ...
$\displaystyle \displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}$
try it again ...
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 $\displaystyle \displaystyle \frac{7a \cdot 4(a-1)}{12(a+1)(a-1)} - \frac{5 \cdot 3(a+1)}{12(a+1)(a-1)} = \frac{3a \cdot 6(a-1)}{12(a+1)(a-1)}$, 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 $\displaystyle 28a^2- 28a- 15a- 15= 3a$.