Question

Solve the following pairs of equations
by reducing them to a pair of linear
equation
1/x +1/y=4
2/x-4/y=8​

Answer 1

$\tt{Let \: \dfrac{1}{x} \: be \: u \: and \: \dfrac{1}{y} \: be \:v}\\\\\\\\\bf \underline{According \: to \: the \: question,}\\\\\sf{u + v = 4 \quad \: \: \: \dashrightarrow [Equation\:1] }\\\\\sf{2u - 4v=8 \quad \dashrightarrow [Equation\:2]}\\\\\\\rm{Taking\:2\:as\:a\:common\:factor\:in\:Equation\:2}\\\\\sf{2(u-2v)=8}\\\\\\\implies \sf{u - 2v=\dfrac{8}{2}}\\\\\\\implies \sf{u-2v=4 \:\:\: \quad \dashrightarrow [Equation \: 3]}\\\\\\\sf{(Equation\:1)-(Equation\:3)}$

                  u + v = 4

         {-}    u - 2v = 4  

                      3v = 0

⇒ v = 0

Substitute value of v in Equation 1 to find the value of u

u + v = 4

⇒ u + 0 = 4

⇒ u = 4

v = 1 ÷ y

⇒ y = 1 ÷ 0

⇒ Value of y is not defined.

u = 1 ÷ x

⇒ x = 1÷4

$\implies \sf{x=\dfrac{1}{4}}$

Therefore, there is no solution for y and value of x is ¼

Answer 2

Solution:-

Given Equations :-

$\sf {\dfrac {1}{x}}+\dfrac {1}{y}=4\dots\dots (1)$

$\sf \dfrac {2}{x}+\dfrac {4}{y}=8 \dots\dots (2)$

• Let

$\qquad\quad {:}\longmapsto\sf \dfrac {1}{x}=u$

$\qquad\quad {:}\longmapsto\sf \dfrac {1}{y}=v$

• Now the new equations

$\qquad\quad {:}\longmapsto\sf u+v=4\dots\dots (3)$

$\qquad\quad {:}\longmapsto\sf 2u-4v=8$

• Make it simplified

$\qquad\quad {:}\longmapsto\sf \cancel {2} (u-2v)=\cancel {2} (4)$

$\qquad\quad {:}\longmapsto\sf u-2v=4\dots\dots (4)$

• There are now Four ways by which we can solve

___________________________

$\qquad\quad {:}\longmapsto\sf Substitution\:method$

$\qquad\quad {:}\longmapsto\sf Elimination \:method$

$\qquad\quad {:}\longmapsto\sf Cross\:multiplication\:method$

$\qquad\quad {:}\longmapsto\sf Cramer 's\: rule$

___________________________

• Let's use Elimination method

• By multiplying eq (3) by 2 and eq (4) by 1 we get

$\qquad\quad {:}\longmapsto\sf 2u+2v=8\dots\dots(5)$

$\qquad\quad {:}\longmapsto\sf u-2v=4\dots\dots (6)$

• Substract eq(6) from eq (5)
• we get

$\qquad\quad {:}\longmapsto{\underline{\boxed{\sf u=4 }}}$

• Substitute the value in eq (3)

$\qquad\quad {:}\longmapsto\sf u+v=4$

$\qquad\quad {:}\longmapsto\sf 4+v=4$

$\qquad\quad {:}\longmapsto\sf v=4-4$

$\qquad\quad {:}\longmapsto{\underline{\boxed{\sf v=0}}}$

____________________________

$\qquad\quad {:}\longmapsto\sf u={\dfrac {1}{x}} \\ \implies \sf \dfrac {1}{x}=4 \\ \implies x={\dfrac {1}{4}}$

$\qquad\quad {:}\longmapsto\sf v={\dfrac {1}{y}} \\ \implies\sf {\dfrac {1}{y}}=0 \\ \implies y=0$

$\therefore{\underline{\boxed{\sf (x,y)=({\dfrac {1}{4}}, 0)}}}$

$\Large\underbrace {Confused!!Let's\: verify}$

• We have to put the values of x and y in the equation instead of x and y . If the answer comes zero then our solution is correct.
• substitute the values in eq (1)

$\qquad\quad{:}\longrightarrow\sf \dfrac {1}{x}+{\dfrac {1}{y}}=4$

$\qquad\quad{:}\longrightarrow\sf \dfrac {1}{\dfrac {1}{4}}+{\dfrac {1}{0}}=4$

$\qquad\quad{:}\longrightarrow\sf 4+0=4$

$\qquad\quad{:}\longrightarrow\sf 4-4=0$

$\qquad\quad{:}\longrightarrow\sf 0=0$

$\therefore\underline{\sf Hence\:Verified}$