Question

8/3x-2 + 45/4y+3 =5 ; 12/3x-2 - 30/4y+3=1​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given :

$\dfrac{8}{3x-2}+\dfrac{45}{4y+3}=5\\\\\dfrac{12}{3x-2}-\dfrac{30}{4y+3}=1$

To find : x= ? and y= ?

Solution :

Let $\dfrac{1}{3x-2}$ = u

Let $\dfrac{1}{4y+3}$ = v

So, it becomes :

$8u+45v=5\\\\12u+30v=1$

Multiply $8u+45v=5$ by 3 and $12u+30v=1$ by 2

So, we get :

$24u+135v=15\\\\24u-60v=2\\\\----------------------------------------\\\\\text{Subtracting both the equations}\\\\195v=13\\\\v=\dfrac{13}{195}=\dfrac{1}{15}$

So, the value of u would be:

$24u-60\times \dfrac{1}{15}=2\\\\24u-4=2\\\\24u=2+4\\\\24u=6\\\\u=\dfrac{6}{24}=\dfrac{1}{4}$

So, it becomes :

$\dfrac{1}{4}=\dfrac{1}{3x-2}\\\\3x-2=4\\\\3x=4+2\\\\3x=6\\\\x=\dfrac{6}{3}\\\\x=2$

and

$\dfrac{1}{5}=\dfrac{1}{4y+3}\\\\5=4y+3\\\\5-3=4y\\\\2=4y\\\\\dfrac{2}{4}=y\\\\y=\dfrac{1}{2}$

Hence, the value of x = 2 and y = $\dfrac{1}{2}$