Question

2x/3- y/2+1/6=0 and x/2+2y/3=3

Answer 1

Question -

Solve the following pair of linear equations -

1. $\: \dfrac{2x}{3} - \dfrac{y}{2} + \dfrac{1}{6} = 0$

2. $\: \dfrac{x}{2} + \dfrac{2y}{3} = 3$

Solution -

$\sf{ Equation \: 1 \: - } \: \dfrac{2x}{3} - \dfrac{y}{2} + \dfrac{1}{6} = 0 \\ \\ \sf{ Taking \: LCM } \\ \\ => \dfrac{4x - 3y + 1}{6} = 0 \\ \\ = > 4x - 3y + 1 = 0..........(1)$

$\sf{ Equation \: 2 \: - } \: \dfrac{x}{2} + \dfrac{2y}{3} = 3 \\ \\ \\ \sf{ Taking \: LCM } \\ \\ => \dfrac{3x + 4y}{6} = 3 \\ \\ = > 3x + 4y = 18.........(2)$

$\sf{ Now \: Solving \: These \: Equations \: - }$

Multiply Equation 1 by 3 and Equation 2 by 4 -

$12x - 9y = -3 \\ \\ 12x + 16y = 72 \\ \\ Subtracting \: - \\ \\ -25y = -75 \\ \\ => y = 3 \\ \\ => x = 2$

Answer 2

=============================================

Question : 2x/3- y/2+1/6=0 and x/2+2y/3=3

Solve for x and y from these equations

Answer : x = 2 , y = 3  

Step-by-step explanation :

$\frac{2x}{3}-\frac{y}{2}+\frac{1}{6}=0\\\\=>\frac{8x-6y+2}{12} =0\\\\ =>8x-6y=-2...(1)\\\\\frac{x}{2} +\frac{2y}{3}=3\\\\ =>\frac{3x+4y}{6}=3\\\\ =>3x+4y=18...(2)$

Now add 2*(1) and 3*(2) , we get ,

$=>2(8x-6y)+3(3x+4y)=2(-2)+3(18)\\\\=>16x-12y+9x+12y=-4+54\\\\=>25x=50\\\\=>x=2$

sub. x=2 in (2) , we get ,

$=>3(2)+4y=18\\\\=>6+4y=18\\\\=>4y=12\\\\=>y=3$

=============================================

>>> Hope Helps You <<<