Question

Find the equation of the line passing through the point of intersection of the line 3x+y=7 and 3y=4x-5 and parallel to the line 2x-4=3

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Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: equations \: are \\ 3x + y = 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)\\ 4x - 3y = 5 \: \: \: \: \: \: \: \: \: \: \: \: (2) \\ \\ applying \: now \: \\ 3 \times( 1) \\ we \: have \: then \\ \\ 9x + 3y = 21 \: \: \: \: \: \: \: \: \: \: (3) \\ \\ similarly \: applying \\ (2) + (3) \\ \\ 4x + 9x = 5 + 21 \\ \\ 13x = 26 \\ \\ x = 2$

$substituting \: value \: of \: x \: in \: (1) \\ we \: get \\ \\ y = 1$

$so \: here \: let \: the \: slope \: of \: the \: \: line \: \\ whose \: \: equation \: is \: to \: be \: formed \: \\ be \: m1 \\ \\ so \: here \: \\ another \: given \: line \: is \: \\ 2x - 4y = 3 \\ \\ comparing \: it \: with \\ ax + by = c \\ we \: have \\ \\ a = 2 \\ b = - 4 \\ c = 3 \\ \\ so \: we \: know \: that \\ \\ slope \: = \frac{ - a}{b} \\ \\ m2 = \frac{ - 2}{ - 4} \\ \\ = \frac{2}{4} \\ \\ m2= \frac{1}{2}$

$so \: as \: the se \: two\: lines \: are \: parallel \\ their \: slopes \: must \: be \: equal \\ \\ \\ thus \: then \\ m1 = m2 = \frac{1}{2} \\ \\ so \: here \: let \\ (x1,y1) = (2,1) \\ \\ we \: know \: that \\ equation \: of \: slope - point \: form \\ is \: given \: \: by \\ \\ y - y1 = m(x - x1) \\ \\ y - 1 = \frac{1}{2} (x - 2) \\ \\ 2(y - 1) = x - 2 \\ \\ 2y - 2 = x - 2 \\ \\ x = 2y \\ \\ or \\ \\ x - 2y = 0$

$hence \: required \: equation \: of \: straight \\ line \: is \: : \\ x = 2y$