Question

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

Answer 1

Question:

Solve the following simultaneous equations:

$\displaystyle{\sf\:\dfrac{5}{x\:-\:1}\:+\:\dfrac{1}{y\:-\:2}\:=\:2}$

$\displaystyle{\sf\:\dfrac{6}{x\:-\:1}\:-\:\dfrac{8}{y\:-\:2}\:=\:1}$

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:\dfrac{63}{17}\:,\:\dfrac{60}{7}\:\right)\:}}}$

Step-by-step-explanation:

The given simultaneous equations are

$\displaystyle{\sf\:\dfrac{5}{x\:-\:1}\:+\:\dfrac{1}{y\:-\:2}\:=\:2\:\qquad\cdots\:(\:1\:)}$

$\displaystyle{\sf\:\dfrac{6}{x\:-\:1}\:-\:\dfrac{8}{y\:-\:2}\:=\:1\:\qquad\cdots\:(\:2\:)}$

$\displaystyle{\sf\:Let\:\dfrac{1}{x\:-\:1}\:=\:a\:}$

$\displaystyle{\sf\:And\:\dfrac{1}{y\:-\:2}\:=\:b\:}$

By substituting these values in equation ( 1 ),

$\displaystyle{\sf\:\dfrac{5}{x\:-\:1}\:+\:\dfrac{1}{y\:-\:2}\:=\:2\:\qquad\cdots\:(\:1\:)}$

$\displaystyle{\implies\sf\:5\:\times\:\dfrac{1}{x\:-\:1}\:+\:1\:\times\:\dfrac{1}{y\:-\:2}\:=\:2}$

$\displaystyle{\implies\sf\:5a\:+\:b\:=\:2\:}$

$\displaystyle{\implies\sf\:b\:=\:2\:-\:5a}$

$\displaystyle{\implies\:\boxed{\sf\:b\:=\:-\:5a\:+\:2\:}\sf\:\qquad\cdots\:(\:3\:)}$

By substituting

$\displaystyle{\sf\:\dfrac{1}{x\:-\:1}\:=\:a\:}$

$\displaystyle{\sf\:And\:\dfrac{1}{y\:-\:2}\:=\:b\:}$ in equation ( 2 ),

$\displaystyle{\sf\:\dfrac{6}{x\:-\:1}\:-\:\dfrac{8}{y\:-\:2}\:=\:1\:\qquad\cdots\:(\:2\:)}$

$\displaystyle{\implies\sf\:6\:\times\:\dfrac{1}{x\:-\:1}\:-\:8\:\times\:\dfrac{1}{y\:-\:2}\:=\:1\:}$

$\displaystyle{\implies\sf\:6a\:-\:8b\:=\:1}$

$\displaystyle{\implies\sf\:6a\:-\:8\:(\:-\:5a\:+\:2\:)\:=\:1\:\qquad\cdots\:[\:From\:(\:3\:)\:]}$

$\displaystyle{\implies\sf\:6a\:+\:40a\:-\:16\:=\:1}$

$\displaystyle{\implies\sf\:46a\:=\:1\:+\:16}$

$\displaystyle{\implies\sf\:46a\:=\:17}$

$\displaystyle{\implies\:\boxed{\green{\sf\:a\:=\:\dfrac{17}{46}}}}$

By substituting this value in equation ( 3 ), we get,

$\displaystyle{\sf\:b\:=\:-\:5a\:+\:2\:\qquad\cdots\:(\:3\:)}$

$\displaystyle{\implies\sf\:b\:=\:-\:5\:\times\:\dfrac{17}{46}\:+\:2}$

$\displaystyle{\implies\sf\:b\:=\:\dfrac{-\:85}{46}\:+\:2}$

$\displaystyle{\implies\sf\:b\:=\:\dfrac{-\:85\:+\:2\:\times\:46}{46}}$

$\displaystyle{\implies\sf\:b\:=\:\dfrac{-\:85\:+\:92}{46}}$

$\displaystyle{\implies\:\boxed{\purple{\sf\:b\:=\:\dfrac{7}{46}}}}$

Now, by re-substituting the values of a and b, we get,

$\displaystyle{\sf\:\dfrac{1}{x\:-\:1}\:=\:a\:}$

$\displaystyle{\implies\sf\:\dfrac{1}{x\:-\:1}\:=\:\dfrac{17}{46}}$

$\displaystyle{\implies\sf\:46\:=\:17\:(\:x\:-\:1\:)}$

$\displaystyle{\implies\sf\:17x\:-\:17\:=\:46}$

$\displaystyle{\implies\sf\:17x\:=\:46\:+\:17}$

$\displaystyle{\implies\sf\:17x\:=\:63}$

$\displaystyle{\implies\:\boxed{\pink{\sf\:x\:=\:\dfrac{63}{17}}}}$

Now,

$\displaystyle{\sf\:\dfrac{1}{y\:-\:2}\:=\:b\:}$

$\displaystyle{\implies\sf\:\dfrac{1}{y\:-\:2}\:=\:\dfrac{7}{46}}$

$\displaystyle{\implies\sf\:46\:=\:7\:(\:y\:-\:2\:)}$

$\displaystyle{\implies\sf\:7y\:-\:14\:=\:46}$

$\displaystyle{\implies\sf\:7y\:=\:46\:+\:14}$

$\displaystyle{\implies\sf\:7y\:=\:60}$

$\displaystyle{\implies\:\boxed{\blue{\sf\:y\:=\:\dfrac{60}{7}}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:(\:x\:,\:y\:)\:=\:\left(\:\dfrac{63}{17}\:,\:\dfrac{60}{7}\:\right)\:}}}}$