Solve the following equations by substitution method.

In such questions , if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]

+ = 7

⇒ 3u + v = 7……(1)

= 6

⇒ 5u – 4v = 6 ……..(2)

From eq(1)

v = 7 – 3u…..(3)

Substituting the value of v in eq(2)

5u – 4(7 – 3u) = 6

⇒ 5u – 28 + 12u = 6

⇒ 17u = 6 + 28

⇒ 17u = 34

⇒ u = 34/17

⇒ u = 2

⇒ = 2

Substituting the value of u in eq(3)

v = 7 – 3×2

⇒ v = 1

⇒ = 1

⇒ y = 1

Hence, x = , y = 1