Question

2/x+2/3y=1/6;3/x+2/y=0 solve by substitution method

Answer 1

2/x + 2/3y = 1/6 ---- (1)

3/x + 2/y = 0 ---- (2)

Let 1/x = a and 1/y = b

2 a + (2b/3) = 1/6

(6 a + 2 b)/3 = 1/6

6 (6 a + 2 b ) = 3

36 a + 12 b = 3

Dividing the whole equation by 3 we get

12 a + 4 b = 1 ------- (1)

3 a + 2 b = 0 ---------(2)

12 a + 4 b = 1

Multiply the second equation by 2 => 6 a + 4 b = 0

Subtracting the second equation from the first equation

12 a + 4 b = 1

6 a + 4 b = 0

- - -

-----------------

6 a = 1

a = 1/6


now we have to apply the value of a in either given equations to get the value of another variable b

Substitute a = 1/6 in the first equation we get

12(1/6) + 4 b = 1

2 + 4 b = 1

4 b = 1- 2

4 b = -1

b = -1/4

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