Question

3x+14y=5xy
21y-x=2xy.

Kindly solve this pair of linear equation

Answer 1

Step-by-step explanation:

Given:

$3x + 14y = 5xy \\ 21y - x = 2xy$

To find: Value of x and y

Solution:

Take xy from RHS to LHS

$\frac{3x + 14y}{xy} = 5 \\ \\ \frac{3x}{xy} + \frac{14y}{xy} = 5 \\ \\ \frac{3}{y} + \frac{14}{x} = 5 \\ \\ \frac{14}{x} + \frac{3}{y} = 5 \: \: ...eq1$

$21y - x = 2xy \\ \\ \frac{21y - x}{xy} = 2 \\ \\ \frac{21y}{xy} - \frac{x}{xy} = 2 \\ \\ \frac{21}{x} - \frac{1}{y} = 2 \: \: \: ...eq2$

Let

$\frac{1}{x} = a \\ \\ \frac{1}{y} = b$

substitute the assumption in eq1 and eq2

$14a + 3b = 5 \: \: \: ...eq 3\\ \\ 21a - b = 2 \: \: \: ..eq4$

Multiply eq 4 by 3 and add both eqs

$14a + 3b = 5 \: \: \: \\ \\ 63a - 3b = 6 \: \: \\ - - - - - \\ 77a = 11 \\ \\ a = \frac{11}{77} \\ \\\bold{a = \frac{1}{7}}$

Put the value of a in eq 3

$14 \times \frac{1}{7} + 3b = 5 \\ \\ 2 + 3b = 5 \\ \\ 3b = 5 - 2 \\ \\ 3b = 3 \\ \\ \bold{b =1 }$

Thus,

Value of

x= 7

y= 1

Hope it helps you.

To learn more in brainly:

1)Question 1 Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x+y=5...

https://brainly.in/question/1345026

2)2x+3y=9,3x+4y=5 substitution method

https://brainly.in/question/4360056

Answer 2

Answer:

Value of. x= 7 y= 1

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