Question

solve the simultaneous equations by t method of elimination
$\frac{x}{2} + \frac{y}{3} = 1 \\ \frac{x}{3} + \frac{y}{2} = 1$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm \: \dfrac{x}{2} + \dfrac{y}{3} = 1 - - - (1)$

and

$\rm \: \dfrac{x}{3} + \dfrac{y}{2} = 1 - - - (2)$

So, equation (1) can be rewritten as

$\rm \: \dfrac{x}{2} + \dfrac{y}{3} = 1$

$\rm \: \dfrac{3x + 2y}{6} = 1$

$\rm \: 3x + 2y = 6 - - - (3)$

Now, Equation (2) can be rewritten as

$\rm \: \dfrac{2x + 3y}{6} = 1$

$\rm \: 2x + 3y = 6 - - - (4)$

Now, multiply equation (3) by 2 and (2) by 3, we get

$\rm \: 6x + 4y = 12 - - - (5)$

$\rm \: 6x + 9y = 18 - - - (6)$

On Subtracting equation (5) from (6), we get

$\rm \: 5y = 6$

$\rm\implies \:y = \dfrac{6}{5} - - - (7)$

On substituting the value of y in equation (3) we get

$\rm \: 2x + 3 \times \dfrac{6}{5} = 6$

$\rm \: 2x + \dfrac{18}{5} = 6$

$\rm \: 2x = 6 - \dfrac{18}{5}$

$\rm \: 2x = \dfrac{30 - 18}{5}$

$\rm \: 2x = \dfrac{12}{5}$

$\rm\implies \:\rm \: x = \dfrac{6}{5}$

So, required solution is

$\rm \: x \: = \: \dfrac{6}{5}$

and

$\rm \: y \: = \: \dfrac{6}{5}$

Verification

Consider equation (1)

$\rm \: \dfrac{x}{2} + \dfrac{y}{3} = 1$

On substituting the values of x and y, we get

$\rm \: \dfrac{6}{2 \times 5} + \dfrac{6}{3 \times 5} = 1$

$\rm \: \dfrac{3}{5} + \dfrac{2}{5} = 1$

$\rm \: \dfrac{3 + 2}{5}= 1$

$\rm \: \dfrac{5}{5}= 1$

$\rm \: 1= 1$

Hence, Verified

Answer 2

Answer:

x = 6/5

y = 6/5

Step-by-step explanation:

Given,

x/2 + y/3 = 1 ----- (i)

x/3 + y/2 = 1​ ----- (ii)

Equating (i) and (ii),

x/2 + y/3 = x/3 + y/2

(3x + 2y)/6 = (2x + 3y)/6

3x + 2y = 2x + 3y

x = y ----- (iii)

Substituting the value of x in (i)

y/2 + y/3 = 1

(3y + 2y)/6 = 1

5y = 6

y = 6/5

x = 6/5