Question

question 31 solve it by reducing and elimination method answer is 2 and 1​

Answer 1

Given Equations :

$\mathsf{\bigstar\;\;\dfrac{1}{2(2x + 3y)} + \dfrac{12}{7(3x - 2y)} = \dfrac{1}{2}\;-----\;[1]}$

$\mathsf{\bigstar\;\;\dfrac{7}{(2x + 3y)} + \dfrac{12}{(3x - 2y)} = 4\;-----\;[2]}$

Take :

$\mathsf{\bigstar\;\;\dfrac{1}{(2x + 3y)} = P}$

$\mathsf{\bigstar\;\;\dfrac{1}{(3x - 2y)} = Q}$

New form of Equations :

$\mathsf{\bigstar\;\;\dfrac{P}{2} + \dfrac{12Q}{7} = \dfrac{1}{2}\;-----\;[1]}$

$\mathsf{\bigstar\;\;7P + 12Q = 4\;-----\;[2]}$

Divide Entire Equation [2] with 7

$\mathsf{\implies \dfrac{7P}{7} + \dfrac{12Q}{7} = \dfrac{4}{7}}$

$\mathsf{\implies P + \dfrac{12Q}{7} = \dfrac{4}{7}\;-----\;[3]}$

Subtracting Equation [1] from Equation [3], We get :

$\mathsf{\implies \left[P + \dfrac{12Q}{7}\right] - \left[\dfrac{P}{2} + \dfrac{12Q}{7}\right] = \dfrac{4}{7} - \dfrac{1}{2}}$

$\mathsf{\implies \left[P + \dfrac{12Q}{7} - \dfrac{P}{2} - \dfrac{12Q}{7}\right] = \dfrac{4(2) - 7}{14}}$

$\mathsf{\implies \left[P - \dfrac{P}{2}\right] = \dfrac{8 - 7}{14}}$

$\mathsf{\implies \dfrac{P}{2} = \dfrac{1}{14}}$

$\mathsf{\implies P = \dfrac{2}{14}}$

$\mathsf{\implies P = \dfrac{1}{7}}$

$\mathsf{\implies \dfrac{1}{(2x + 3y)} = \dfrac{1}{7}}$

$\mathsf{\implies 2x + 3y = 7\;----\;[4]}$

Substituting the value of P in Equation [3], We get :

$\mathsf{\implies \dfrac{1}{7} + \dfrac{12Q}{7} = \dfrac{4}{7}}$

$\mathsf{\implies \dfrac{12Q}{7} = \dfrac{4}{7} - \dfrac{1}{7}}$

$\mathsf{\implies \dfrac{12Q}{7} = \dfrac{4 - 1}{7}}$

$\mathsf{\implies \dfrac{12Q}{7} = \dfrac{3}{7}}$

$\mathsf{\implies 12Q = 3}$

$\mathsf{\implies Q = \dfrac{3}{12}}$

$\mathsf{\implies Q = \dfrac{1}{4}}$

$\implies \mathsf{\dfrac{1}{(3x - 2y)} = \dfrac{1}{4}}$

$\implies \mathsf{3x - 2y = 4\;-----\;[5]}$

Multiplying Equation [4] with 2, We get :

$\mathsf{\implies 4x + 6y = 14\;----\;[6]}$

Multiplying Equation [5] with 3, We get :

$\implies \mathsf{9x - 6y = 12\;-----\;[7]}$

Adding Equations [6] and [7], We get :

$:\implies$  (4x + 6y) + (9x - 6y) = 14 + 12

$:\implies$  (4x + 9x + 6y - 6y) = 26

$:\implies$  13x = 26

$\mathsf{:\implies x = \dfrac{26}{13}}$

$\mathsf{:\implies x = 2}$

Substituting value of x in Equation [5], We get :

$:\implies$  3(2) - 2y = 4

$:\implies$  6 - 2y = 4

$:\implies$  2y = 6 - 4

$:\implies$  2y = 2

$:\implies$  y = 1

Answers :

★  x = 2

★  y = 1