Question

$\frac { 5 } { x - 1 } + \frac { 1 } { y - 2 } = 2 \\ \cdot \frac { 6 } { x - 1 } - \frac { 8 } { y - 2 } = 1​$

Don't spam!​

Answer 1

$\sf \bigstar \: Correct \: Question :$

• Solve the following pair of equation by reducing them to a pair of linear equation.

$\rm \implies\dfrac{ 5 } { x - 1 } + \dfrac{ 1 } { y - 2 } = 2$

$\rm \implies \: \dfrac{ 6 } { x - 1 } - \dfrac{ 8 } { y - 2 } = 1$

$\sf \bigstar \: Correct \: Solution :$

$\rm \: Let's \: consider ,$

$\rm \implies \: \dfrac{1}{x - 1} = p$

$\rm \implies \: \dfrac{1}{y - 2} = q$

$\rm \: Substituting \: \dfrac{1}{x - 1} = p \: and \: \dfrac{1}{y - 2} = q \: in \: \: given \: eqution \: we \: get,$

$\sf \bigstar \: 5 \: p + q = 2 \: \: (1)$

$\sf \bigstar \: 6 \: p + 3 \: q = \: \: (2)$

$\rm \: Multiplying \: equation \: (1) \: \: by \: 3 \: and \: (2) \: by \: 1 \: ,we \: get$

$\sf \bigstar \: 15 \: p + 3 \: q = 6 \: \: \: \: (3)$

$\sf \bigstar \: 6 \: p - 3 \: q = 1 \: \: \: \: (4)$

$\rm \: By \: adding \: equation \: (3) \: \: and \: (4)$

$\sf \: \: \: \: \: \: \: \: \: 15 \: p + 3 \: q = 6$

$\sf \: \: \: \: \: \: \: \: \: 6 \: p - 3 \: q = 1$

____________________________________

$\sf \: \: \: \: \: \: \: \: \: 21\: p = 7$

$\rm \: p = \dfrac{7}{21} = \dfrac{1}{3}$

$\rm \: Substituting \:p = \dfrac{1}{3} \: in\: eqution \: (2)\: we \: get ,$

$\sf \implies \: 6 \: \times \dfrac{1}{3} - 3 \: q = 1$

$\sf \implies \: 6 \: \times \dfrac{1}{3} - 3 \: q = 1$$\sf \implies 2 - 3 \: q = 1$

$\rm \implies \: 3 \: q = 1$

$\rm \implies \: q = \dfrac{1}{3}$

$\rm\therefore \: \dfrac{1}{x - 1} = p$

$\rm\implies \: \dfrac{1}{x - 1} = \dfrac{1}{3}$

$\rm \: By \: cross \: \: multiplication :$

$\rm \implies \: x - 1 = 3$

$\rm \implies \: x= 3 + 1$

$\rm \implies \boxed{\: x= 4}$

$\rm\therefore \: \dfrac{1}{y- 2} = q$

$\rm\implies \: \dfrac{1}{y- 2} = \dfrac{1}{3}$

$\rm \: By \: cross \: \: multiplication :$

$\rm\implies \: y - 2 = 3$

$\rm \implies \: y = 3 + 2$

$\rm \implies \boxed{\: y= 5}$

$\sf \bigstar \: Final \: Answer :$

• Here solution of given pair of equations is

$\bf \implies \boxed{\: x= 4}$

$\bf\implies \boxed{\: y= 5}$