Question

pls answer this question in simultaneous linear equation ​

Answer 1

Question :

$\sf \dfrac{x+y}{2} + \dfrac{3x-5y}{4} = 2$

$\sf \dfrac{x}{14} + \dfrac{y}{18} = 1$

Answer :

$\sf \dfrac{x+y}{2} + \dfrac{3x-5y}{4} = 2 \: \pink{- (1)}$

$\sf \dfrac{x}{14} + \dfrac{y}{18} = 1 \: \pink{- (2)}$

From equation (1) :

$\sf : \implies \dfrac{x+y}{2} + \dfrac{3x-5y}{4} = 2$

$\sf : \implies \dfrac{2(x+y)}{2 \times 2} + \dfrac{3x-5y}{4} = 2$

$\sf : \implies \dfrac{2x+2y}{4} + \dfrac{3x-5y}{4} = 2$

$\sf : \implies \dfrac{(2x+2y) + (3x-5y)}{4} = 2$

$\sf : \implies \dfrac{2x+2y + 3x-5y}{4} = 2$

$\sf : \implies \dfrac{5x-3y}{4} = 2$

$\sf : \implies 5x-3y = 2\times 4$

$\sf : \implies 5x-3y = 8$

$\sf : \implies 5x = 8 + 3y$

$\sf : \implies x = \dfrac{8 + 3y}{5} \: \pink{-(3)}$

Put in equation (2) :

$\sf : \implies \dfrac{\pink{\dfrac{8 + 3y}{5}}}{14} + \dfrac{y}{18} = 1$

$\sf : \implies \dfrac{8 + 3y}{5} \times \dfrac{1}{14} + \dfrac{y}{18} = 1$

$\sf : \implies \dfrac{8 + 3y}{70} + \dfrac{y}{18} = 1$

By taking LCM :

$\sf : \implies \dfrac{18(8 + 3y)}{1260} + \dfrac{70y}{1260} = 1$

$\sf : \implies \dfrac{144 + 54y}{1260} + \dfrac{70y}{1260} = 1$

$\sf : \implies \dfrac{144 + 54y + 70y}{1260}= 1$

$\sf : \implies \dfrac{144 + 124y}{1260}= 1$

$\sf : \implies 144 + 124y= 1 \times 1260$

$\sf : \implies 144 + 124y= 1260$

$\sf : \implies 124y = 1260 - 144$

$\sf : \implies 124y = 1116$

$\sf : \implies y = \cancel{\dfrac{1116}{124}}$

$\sf : \implies y = 9$

$\large \underline{\boxed{\bf{\pink{y = 9}}}}$

Put value of y in equation (3) :

$\sf : \implies x = \dfrac{8 + 3(\pink{9})}{5}$

$\sf : \implies x = \dfrac{8 + 27}{5}$

$\sf : \implies x = \cancel{\dfrac{35}{5}}$

$\sf : \implies x = 7$

$\large \underline{\boxed{\bf{\pink{x = 7}}}}$

Hence, value of :

• $\bf \pink{x = 7}$
• $\bf \pink{y = 9}$