Question

3/3x-5-2/4x-7=6/7(4x-7)+7/9(3x-5)​

Answer 1

GIVEN :

The equation is $\frac{3}{3x-5}-\frac{2}{4x-7}=\frac{6}{7(4x-7)}+\frac{7}{9(3x-5)}$

TO SOLVE :

The given equation $\frac{3}{3x-5}-\frac{2}{4x-7}=\frac{6}{7(4x-7)}+\frac{7}{9(3x-5)}$ for solving x.

SOLUTION :

Given that the equation is $\frac{3}{3x-5}-\frac{2}{4x-7}=\frac{6}{7(4x-7)}+\frac{7}{9(3x-5)}$

Now solving for x to find its value :

$\frac{3}{3x-5}-\frac{2}{4x-7}=\frac{6}{7(4x-7)}+\frac{7}{9(3x-5)}$

$\frac{3}{3x-5}-\frac{2}{4x-7}-\frac{6}{7(4x-7)}-\frac{7}{9(3x-5)}=0$

$\frac{3}{3x-5}-\frac{7}{9(3x-5)}-\frac{2}{4x-7}-\frac{6}{7(4x-7)}=0$

$\frac{1}{3x-5}(3-\frac{7}{9})-\frac{1}{4x-7}(2+\frac{6}{7})=0$

$\frac{1}{3x-5}(\frac{27-7}{9})-\frac{1}{4x-7}(\frac{14+6}{7})=0$

$\frac{1}{3x-5}(\frac{20}{9})-\frac{1}{4x-7}(\frac{20}{7})=0$

By taking LCM we have that,

$\frac{7(4x-7)(1)(20)-20(1)(9)(3x-5)}{9(3x-5)(7)(4x-7)}=0$

$7(4x-7)(1)(20)-20(1)(9)(3x-5)=0$

$140(4x-7)-180(3x-5)=0$

By using the Distributive Law :

a(x+y)=ax+ay

$140(4x)+140(-7)-180(3x)-180(-5)=0$

$560x-980-540+900=0$

By adding the like terms we get,

$20x-80=0$

$20x=80$

$x=\frac{80}{20}$

x=4

∴ x=4

∴ the value of x in the given equation is 4.

Answer 2

Answer:

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