Question

please
solve this problem ​

Answer 1

Answer:

x = 65/9 and y = - 10/9

Step-by-step explanation:

Given that -

$\sf \dfrac{7}{x + 2y} + \dfrac{8}{2x+y} =2 \\\\\sf \dfrac{14}{x + 2y}- \dfrac{24}{2x+y} =-1$

Let, $\sf \dfrac{1}{x+2y}=a \: and \: \dfrac{1}{2x+y}=b$

Substituting the above assumptions in the equation ;

Equation (i) :

⇒ $\sf \dfrac{7}{x + 2y} + \dfrac{8}{2x+y} =2$

⇒ 7a + 8b = 2

Equation (ii) :

⇒ $\sf \dfrac{14}{x + 2y} - \dfrac{24}{2x+y} =-1$

⇒ 14a - 24b = - 1

Multiplying equation (i) by 2 and then, subtracting the resulting equation from (ii).

⇒ 2(7a + 8b) = 2(2)

⇒ 14a + 16b = 4

Subtraction :

14a + 16b - (14a - 24b) = 2 - (-1)

⇒ 14a + 16b - 14a + 24b = 2 + 1

⇒ 40b = 3

⇒ b = $\dfrac{3}{40}$

Substituting the value of b in (i) :

⇒ 7a + 8b = 2

⇒ 7a + $8 *\dfrac{3}{40}$ =2

⇒ 7a + $\dfrac{3}{5}$ = 2

⇒ 7a = 2 - $\dfrac{3}{5}$

⇒ 7a = $\dfrac{10 - 3}{5}$

⇒ a = $\dfrac{7}{7*5}$

⇒ a = $\dfrac{1}{5}$

Now, substituting the value of a and b ;

$\sf \implies \dfrac{1}{x+2y}=a$

$\sf \implies \dfrac{1}{x+2y}= \dfrac{1}{5}$

$\sf \implies x + 2y = 5 \: \:... (iii)$

Also,

$\sf \implies \dfrac{1}{2x+y}=b$

$\sf \implies \dfrac{1}{2x+y}=\dfrac{3}{40}$

$\sf \implies 2x + y = \dfrac{40}{3} \: \:... (iv)$

Multiplying equation (iii) by 2 and then, subtracting the resulting equation from (iv).

⇒ 2(x + 2y) = 2 * 5

⇒ 2x + 4y = 10

Subtraction :

2x + 4y - 2x - y = 10 - $\dfrac{40}{3}$

⇒ 3y = $\dfrac{30 - 40}{3}$

⇒ 3y = - 10/3

⇒ 9y = - 10

⇒ y = - 10/9

Substituting the value of y in (iii) ;

⇒ x + 2y = 5

⇒ x - 20/9 = 5

⇒ 9x - 20/9 = 5

⇒ 9x - 20 = 45

⇒ 9x = 45 + 20

⇒ 9x = 65

⇒ x = 65/9

Hence, the value of x is 65/9 and y is - 10/9.