Question

find the answer of attached question
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Answer 1

Answer:

${\sf{x = {\dfrac{1}{2}} , \ y = {\dfrac{1}{3}}}}$

Step-by-step explanation:

Given :

1st equation : ${\sf{\ {\dfrac{2}{x}} + {\dfrac{3}{y}} = 13}}$

2nd equation : ${\sf{\ {\dfrac{5}{x}} - {\dfrac{4}{y}} = - 2}}$

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To Find : x & y

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Solution :

Let ${\sf{{\dfrac{1}{x}}}}$ = X and ${\sf{{\dfrac{1}{y}}}}$ = Y

New equations are :

• 2X + 3Y = 13 ...(1)

• 5X - 4Y = - 2 ...(2)

• Multiplying (1) by 4 and (2) by 3.

→ 4(2X + 3Y) = 4(13)

→ 3(5X - 4Y) = 3(- 2)

• On further solving, we get

→ 8X + 12Y = 52 ...(3)

→ 15X - 12Y = - 6 ...(4)

• Adding (3) and (4), we get

→ (8X + 12Y) + (15X - 12Y) = (52) + (- 6)

→ 8X + 12Y + 15X - 12Y = 52 - 6

→ 23X = 46

→ X = ${\sf{{\dfrac{46}{23}}}}$

→ X = 2

• Putting this value in (1), we get

→ 2(2) + 3(Y) = 13

→ 4 + 3Y = 13

→ 3Y = 13 - 4

→ 3Y = 9

→ Y = ${\sf{{\dfrac{9}{3}}}}$

→ Y = 3

• Now, from assumption :

• ${\sf{{\dfrac{1}{x}}}}$ = X

• Putting the value of X, we get

→ ${\sf{{\dfrac{1}{x}}}}$ = 2

→ ${\boxed{\sf{x = {\dfrac{1}{2}}}}}$

• ${\sf{{\dfrac{1}{y}}}}$ = Y

• Putting the value of Y, we get

→ ${\sf{{\dfrac{1}{y}}}}$ = 3

→ ${\boxed{\sf{y = {\dfrac{1}{3}}}}}$