Question

find the value of x and y in 2/x+5/y=1/4,3/x+6/y=1/3​

Answer 1

Answer:

Step-by-step explanation:

2*1/x+5*1/y=1/4(eq1),3*1/x+6*1/y=1/3(eq2)

lets take;1/x=p,1/y=q

now multiply eq1 with 3 &eq2 with 2

2p+5q=1/4

3p+6q=1/3

solve (eq1)&(eq2)

6p+15q=3/4

6p+12q=2/3

3p=1/12

p=1/12*1/3

p=1/36

2(1/36)+5q=1/4

1/18+5q=1/4

5q=1/4-1/18

5q=9/36-2/36

7/36=5q

q=7/36*1/5

q=7/180

Answer 2

Answer

The values are

x = 18 and y = 36

Given

The equations are :

$\frac{2}{x} + \frac{5}{y} = \frac{1}{4} - - - - (1)$

And

$\frac{3}{x} + \frac{6}{y} = \frac{1}{3} - - - - (2)$

To Find

The value of x and y

Solution

Let us consider

• 1/x = p
• 1/y = q

So that the equation (1) and (2) becomes :

$2p + 5q = \frac{1}{4} \\ \implies 4(2p + 5q) = 1 \\ \implies8p + 20q = 1 - - - (3)$

And

$3p + 6q = \frac{1}{3} \\ \implies3(3p + 6q) = 1\\ \implies9p + 18q = 1 - - - (4)$

Subtracting (3) from (4) we have

$9p + 18q - 8p - 20q = 1 - 1 \\ \implies p - 2q = 0 \\ \implies p = 2q - - - - (5)$

Using the above value of p in (3) we have

$8(2q) + 20q = 1 \\ \implies16q + 20q = 1 \\ \implies36q = 1 \\ \implies q = \frac{1}{36}$

And putting the value of q in (5) we have

$p = 2 \times \frac{1}{36} \\ \implies p = \frac{1}{18}$

Since p = 1/x and q = 1/y so

$\frac{1}{x } = \frac{1}{18} \\ \implies x = 18$

And

$\frac{1}{y} = \frac{1}{36} \\ \implies y = 36$