Question

Consider a pair of equations as shown. 5/(x+2) + 7/(y+2)= 31/12 & 4/(x+2) + 3/(y+2)= 17/12 What is the value of x and y?

x = 4 , y = 2
x = 2 , y = 4
x = 16 , y = 14
x = 14 , y = 16​

Answer 1

Answer:

Correct answer is x=4 and y=2...

Step-by-step explanation:

Take 1/x+2 and 1/y+2 as p and q...

Make two equations in variable p and q...

Eliminate the two equation...

Thus, find the value of both the variables...

Hope it helps you....If u understand...Then please make me as brilliant....

Answer 2

Given:

A pair of equations 5/(x+2) + 7/(y+2)= 31/12 and 4/(x+2) + 3/(y+2)= 17/12.

To find:

The value of x and y.

Solution:

Let the value of 1/(x+2) = p and 1/(y+2) = q.

So, after putting these values in the given pair of equations. We have,

$5p + 7q = \frac{31}{12} \: \: (i)$

and

$4p + 3q = \frac{17}{12} \: \: \: (ii)$

After multiplying equation (i) by 4 and equation (ii) by 5, we have

$20p + 28q = \frac{124}{12} \: \: (iii)$

and,

$20p + 15q = \frac{85}{12} \: \: (iv)$

Now, subtract (iv) from (iii), we have

$13q = \frac{39}{12}$

$q = \frac{39}{12 \times 13} = \frac{3}{12} = \frac{1}{4}$

$q = \frac{1}{4}$

Now, put the value of q in equation (iii), we have

$20p + 28( \frac{1}{4} ) = \frac{124}{12}$

$20p + 7 = \frac{124}{12}$

$20p = \frac{124}{12} - \frac{7 \times 12}{12}$

$20p = \frac{40}{12}$

$p = \frac{40}{12 \times 20}$

$p = \frac{2}{12} = \frac{1}{6}$

$p = \frac{1}{6}$

Now, p = 1/(x+2) and q = 1/(y+2).

So, put the value of p = 1/6 and q = 1/4 in 1/(x+2) and 1/(y+2) respectively.

$\frac{1}{x + 2} = \frac{1}{6}$

$x + 2 = 6$

$x = 4 \: \: \:$

Also,

$\frac{1}{y + 2} = \frac{1}{4}$

$y + 2 = 4$

$y = 2$

Hence, the correct option is x = 4, y = 2.