Question

1/7x+1/6y=3 1/2x-1/3y=5 solve it by elimination method

Answer 1

(1/7)x + (1/6)y = 3.........(i)
(1/2)x - (1/3)y = 5..........(ii)

multiplying equation (i) with 42 and equation (ii) with 6 on both sides we get

6x + 7y = 126........(iii)
3x - 2y = 30.........(iv)

now (iii)-2(iv)

11y= 66
y= 6

substituting in (iv)
3x = 42
x= 14

solved :-)

Answer 2

Given: Two equations $\frac{1}{7}x + \frac{1}{6}y = 3 \ and \ \frac{1}{2}x - \frac{1}{3}y = 5$

To find: Values of $x \ and \ y$

Solution:

The given equations are:

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

⇒ $\frac{6x \ + \ 7y}{42} = 3$

⇒ $6x + 7y = 126$ ...(1)

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

⇒ $\frac{3x \ - \ 2y}{6} = 5$

⇒ $3x - 2y = 30$   ...(2)

Using elimination method to solve the equations:

Multiplying equation (2) by 2

⇒ $6x - 4y = 60$   ...(3)

Subtracting equation (3) from equation (1)

∵ $(6x + 7y) - (6x - 4y) = 126 - 60$

⇒ $6x + 7y - 6x + 4y = 66$

⇒ $11y = 66$

⇒ $y = 6$

Putting $y = 6$ in equation (3)

⇒ $6x - 4*6 = 60$

⇒ $6x = 60 + 24 = 84$

⇒ $x = 14$

Hence, $x = 14 \ and \ y = 6$