Question

solve by linear equation in two variable method
please don't write wrong answers​

Answer 1

Answer:

$\tt \dfrac{8}{3x - 2} + \dfrac{45}{4y + 3} = 5; \: \: \: \dfrac{12}{3x - 2} + \dfrac{30}{4y + 3} = 1$

$: \implies \tt \dfrac{8}{3x - 2} + \dfrac{45}{4y + 3} = 5 \: \: \bigg \lgroup \bf equation \: (1) \bigg \rgroup$

$: \implies \tt \dfrac{12}{3x - 2} + \dfrac{30}{4y + 3} = 1\: \: \bigg \lgroup \bf equation \: (2) \bigg \rgroup$

$\bf Put \: \dfrac{1}{3x-2}=m \: \: and \: \: \dfrac{1}{4y +3} = n</p><p></p><p>$

$: \implies \tt 8m + 45n = 5\: \: \bigg \lgroup \bf equation \: (3) \bigg \rgroup$

$: \implies \tt 12m - 30n = 1\: \: \bigg \lgroup \bf equation \: (4) \bigg \rgroup$

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$\bf Multiplying \: equation \: (3) \: by \: 2 \: and \: equation \: (4) \: by \: 2$

$: \implies \tt 24m + 135n = 15\: \: \bigg \lgroup \bf equation \: (5) \bigg \rgroup$

$: \implies \tt 24m - 60n = 2\: \: \bigg \lgroup \bf equation \: (6) \bigg \rgroup$

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$\bf Substracting \: equation \: (6) \: from \: equation \: (5) \: we \: get,$

$\tt 24m \: + \: 135n \: = \: 15 \\ \tt \dfrac{ \: 24m \: - \: 60n = \: 2 \qquad}{ 195n \: = \: 13} \\ \tt n \: = \dfrac{13}{195} \\ \tt n \: = \:\dfrac{1}{15}$

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$\bf Put \: the \: value \: of \: n \: = \: \dfrac{1}{15} \: in \: equation \: (4) \: we \: get,</p><p>$

$: \implies \tt 12m - 30n = 1$

$: \implies \tt 12m - 30 \times \dfrac{1}{15} = 1$

$: \implies \tt 12m - 2 = 1$

$: \implies \tt 12m = 1 + 2$

$: \implies \tt 12m = 3$

$: \implies \tt m = \dfrac{3}{12}$

$: \implies \tt m = \dfrac{1}{4}$

$\bf Substitute \: m \: = \: \dfrac{1}{3x-2} \: and \: n \: = \: \dfrac{1}{4y+3}$

$: \implies \tt m = \dfrac{1}{4}$

$: \implies \tt \dfrac{1}{3x-2} = \dfrac{1}{4}$

$: \implies \tt 3x-2 = 4$

$: \implies \tt 3x = 4 + 2$

$: \implies \red{\tt x = 6 }$

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$: \implies \tt n = \dfrac{1}{15}$

$: \implies \tt \dfrac{1}{4y+3}= \dfrac{1}{15}$

$: \implies \tt 4y+3= 15$

$: \implies \tt 4y= 15 - 3$

$: \implies \red{ \tt y= 3}$