Question

1 ÷ x+1÷y= 4,2÷x+3÷y=7 find by substitution or elimination method​

Answer 1

Answer:-

x = 1/5

y = -1

Explanation:-

Given:

$\dfrac{1}{x} +\dfrac{1}{y} = 4$

$\dfrac{2}{x} +\dfrac{3}{y} = 7$

To Find:

Values of x and y

Solution:

Substitution Method

$\dfrac{1}{x} +\dfrac{1}{y} = 4....</p><p>..(1)$

$\dfrac{2}{x} +\dfrac{3}{y} = 7.....(2)$

From eq. (1)

$\dfrac{1}{x} +\dfrac{1}{y} = 4$

$\dfrac{1}{x} = 4 - \dfrac{1}{y}$

$\dfrac{1}{x} = \dfrac{4y - 1}{y}.....(3)$

Put eq.(3) in (2)

$2(\dfrac{4y-1}{y}) +\dfrac{3}{y} = 7$

$\dfrac{8y-2}{y} = 7- \dfrac{3}{y}$

$\dfrac{8y-2}{\cancel{y}} = \dfrac{7y-3}{\cancel{y}}$

$8y-2= 7y-3$

$8y-7y = -3+2$

$\bold{y = -1}$

Put y = -1 in eq (3)

$\dfrac{1}{x} = \dfrac{4(-1)- 1}{-1}$

$\dfrac{1}{x} = \dfrac{-4- 1}{-1}$

$\dfrac{1}{x} = \dfrac{-5}{-1}$

$\bold{x = \dfrac{1}{5}}$

Elimination Method

$\dfrac{1}{x} +\dfrac{1}{y} = 4....</p><p>..(1)$

$\dfrac{2}{x} +\dfrac{3}{y} = 7.....(2)$

(2) - 3 × (1)

$\dfrac{-1}{x} = -5$

$\bold{x = \dfrac{1}{5} }$

Put x value in eq(1)

$\dfrac{5}{1} +\dfrac{1}{y} = 4$

$\dfrac{1}{y} = 4-5$

$\dfrac{1}{y} = -1$

$\bold{y = -1}$

Answer 2

Answer :-

• x = 1/5

• y = - 1

Explanation :-

Given

Pair of Linear equations

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

(2/x) + (3/y) = 7--eq(2)

I. Substitution method :

Finding the value of x in terms of y from eq(1)

⇒ (1/x) + (1/y) = 4

⇒ 1/x = 4 - (1/y)

⇒ 1/x = (4y - 1)/y

⇒ x = y/(4y - 1)

Substitute x = y/(4y - 1) in eq(2)

⇒ (2/x) + (3/y) = 7

⇒ 2/{y/(4y - 1)} + (3/y) = 7

⇒ 2(4y - 1)/y + (3/y) = 7

⇒ (8y - 2)/y + (3/y) = 7

⇒ (8y - 2 + 3)/y = 7

⇒ (8y + 1)/y = 7

⇒ (8y/y) + (1/y) = 7

⇒ 8 + (1/y) = 7

⇒ 1/y = 7 - 8

⇒ 1/y = - 1

⇒ y = - 1

Substitute y = - 1 in x = y/(4y - 1)

⇒ x = y/(4y - 1)

⇒ x = - 1/{4(-1) - 1}

⇒ x = - 1/(-4 - 1)

⇒ x = - 1/-5

⇒ x = 1/5

II. Elimination method :

To eliminate first we need to make any one one coefficients equal

Multiply eq(1) by 2

⇒ (2/x) + (2/y) = 8 --eq(3)

Subtracting eq(1) from eq(3)

⇒ (2/x) + (2/y) - { (2/x) + (3/y) } = 8 - 7

⇒ (2/x) + (2/y) - (2/x) - (3/y) = 1

⇒ (2/y) - (3/y) = 1

⇒ (2 - 3)/y = 1

⇒ - 1/y = 1

⇒ 1/y = - 1

⇒ y = - 1

Substitute y = - 1 in eq(1)

⇒ (1/x) + (1/y) = 4

⇒ (1/x) + (1/-1) = 4

⇒ (1/x) + (-1) = 4

⇒ (1/x) - 1 = 4

⇒ 1/x = 4 + 1

⇒ 1/x = 5

⇒ 1/x = 5

⇒ x = 1/5

∴ x = 1/5 and y = -1.