Question

2/x-1+3/y+1=2,3/x-1+2/y+1=13/6 solve for x and y using method of cross multiplication
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Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

$\sf= > \dfrac{2}{x - 1} + \dfrac{3}{y + 1} - 2=0......(1)$

$\sf= > \dfrac{3}{x - 1} + \dfrac{2}{y + 1} - \dfrac{13}{6} =0....(2)$

Let $\sf \dfrac{1}{x - 1} \:be\: 'u'$$\sf \dfrac{1}{y + 1} \:be \:'v'$

The new equation come :

$\sf= > 2u + 3v - 2 = 0...(3)$

$\sf= > 3u + 2v - \dfrac{13}{6} = 0....(4)$

General form of cross multiplication method is:

$\sf \dfrac{a}{b_{1}c_{2}- b_{2}c_{1}} = \dfrac{- b}{a_{1}c_{2} - a_{2}c_{1}} = \dfrac{1}{a_{1}b_{2} - a_{2}b_{1}}$

$\sf= > \sf \dfrac{a}{3 \times - \dfrac{13}{6} - 2 \times - 2} = \dfrac{ -b}{2 \times - \dfrac{13}{6} - 3 \times - 2 } = \dfrac{1}{2 \times 2 - 3 \times 3}$

$\sf= > \sf \dfrac{a}{ - \dfrac{13}{2} + 4} = \dfrac{ - b}{ - \dfrac{13}{3} + 6} = \dfrac{c}{4 - 9}$

$\sf= > \sf \dfrac{a}{ - \dfrac{5}{2} } = \dfrac{ -b}{ \dfrac{5}{3} } = \dfrac{c}{ - 5}$

$\sf = > \sf - \dfrac{2}{5} a = \dfrac{ - 3b}{5} = \dfrac{ 1}{ - 5}$

$\sf= > - 10a = -15b$

$\sf= > 2a = 3b$

$\sf= >2a = 3b....(5)$

$\sf = > \dfrac{ -3b}{5} = \dfrac{ 1}{ - 5}$

$\sf= >15b = 5$

$\sf = > \sf b = \dfrac{ 5}{15} = \dfrac{1}{3}$

Put b's value in equation (5)

$\sf = > 2a = 3 \times \dfrac{1}{3}$

$\sf = >2a = 1$

$\sf= > a = \dfrac{1}{2}$

Check from Equation (3) and (4) a which is used in place of u and b which is used in place of v

Now finding u and v value

$\sf\bold{= > a = u}$

$\sf= > \dfrac{1}{2} = \dfrac{1}{x - 1}$

$\sf= > x - 1 = 2$

$\sf = > x = 3$

$\sf\huge{x = 3}$

$\sf= > b = v$

$\sf= > b = \dfrac{1}{y + 1}$

$\sf= > \dfrac{1}{3} = \dfrac{1}{y + 1}$

$\sf= > y + 1 = 3$

$\sf= > y = 2$

$\sf\huge {y = 2}$