Question

Solve the following systems of equations:

$\frac{1}{7x}+\frac{1}{6y}=3$

$\frac{1}{2x}-\frac{1}{3y}=5$

Answer 1

Step1: Reduce the linear equations.

Give $\frac{1}{7x}+\frac{1}{6y}=3$

$\frac{1}{2x}-\frac{1}{3y}=5$

Let p = $\frac{1}{x}$ ; q = $\frac{1}{y}$

$\frac{p}{7}$ + $\frac{q}{6}$ = 3

6p + 7q = 126 ___(i) [$\because$ LCM of 6, 7 is 42]

$\frac{p}{2}-\frac{q}{3}=5$

3p - 2q = 30 __(ii) [$\because$ LCM of 3, 2 is 6]

Step2: Solve equation i and ii by using elimination method

equation ii multiple with '2'

2(3p - 2q = 30)

6p - 4q = 60 ___(iii)

Subtrat i and iii

(6p + 7q) - (6p - 4q) = 126 - 60

$7q+4q=66$

11q = 66

$\therefore$ q = 6

substitute q = 22 in equation i

6p + 7(6) = 126

6p = 126 - 42

6p = 84

$\therefore$ p =14

Step3: Find x, and y values

14 = $\frac{1}{x}$

x = $\frac{1}{14}$

6 = $\frac{1}{y}$

y = $\frac{1}{6}$

Hence the value of x and y is $\frac{1}{14}$, $\frac{1}{6}$