Question

Solve this equation ​

Answer 1

Question :-

Solve 6/(x + y) = {7/(x - y)} + 3, {1/2(x + y)} = {1/3(x - y)}

where x + y ≠ 0, x - y ≠ 0

Answer :-

• x = - 5/4

• y = - 1/4

Explanation :-

$\sf \dfrac{6}{x + y} = \dfrac{7}{x - y} + 3 \\ \\ \\ \sf \dfrac{1}{2(x + y)} = \dfrac{1}{3(x - y)}$

It can be written as

$\sf 6 \bigg(\dfrac{1}{x + y} \bigg) = 7 \bigg( \dfrac{1}{x - y} \bigg) + 3 \\ \\ \\ \sf \dfrac{1}{2} \bigg(\dfrac{1}{x + y} \bigg) = \dfrac{1}{3} \bigg(\dfrac{1}{x + y} \bigg)$

Substitute 1/(x + y) = a and 1/(x - y) = b we get the following pair of linear equations

6a = 7b + 3 ---eq(1)

(1/2) * a = 1/3 * (b)

a/2 = b/3 ---eq(2)

a = 2b/3

Substitute a = 2b/3 in eq(1)

$\sf 6 \bigg( \dfrac{2b}{3} \bigg) = 7b + 3$

6(2b) = 7b + 3

12b = 3(7b + 3)

12b = 21b + 9

- 9 = 21b - 12b

- 9 = 9b

- 9/9 = b

- 1 = b

b = - 1

Substitute b = - 1 in a = 2b/3

a = 2(-1)/3

a = - 2/3

But a = 1/(x + y) and b = 1/(x - y)

i) 1/(x + y) = a

1/(x + y) = - 2/3

x + y = - 3/2 ---eq(3)

ii) 1/(x - y) = b

1/(x - y) = - 1

x - y = - 1 --eq(4)

Adding eq(3) and eq(4)

(x + y) + (x - y) =( - 3/2) + (- 1)

2x = (-3/2) - 1

2x = (-3 - 2)/2

2x = - 5/2

x = (-5/2)/2

x = (-5/2) * (1/2)

x = - 5/4

Subsitute x = - 5/4 in eq(4)

(-5/4) - y = - 1

(-5/4) + 1 = y

(-5 + 4)/4 = y

- 1/4 = y

y = - 1/4

Answer 2

Step-by-step explanation:

HOPE IT HELPS YOU AND PLZZZ MARK AS THE BRAINLIEST MY MATE...