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Question

# If 2x âˆ’ 3y = 7 and (a + b)x âˆ’ (a + b âˆ’ 3)y = 4a + b represent coincident lines, then a and b satisfy the equation (a) a + 5b = 0 (b) 5a + b = 0 (c) a âˆ’ 5b = 0 (d) 5a âˆ’ b = 0

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Solution

## The given system of equations are For coincident lines , infinite number of solution $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{b2}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’\frac{2}{\left(a+b\right)}=\frac{-3}{-\left(a+b-3\right)}=\frac{7}{4a+b}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’\frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’2\left(a+b-3\right)=3\left(a+b\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’2a+2b-6=3a+3b\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’2a+2b-3a-3b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’-a-b=6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’a+b=-6---\left(i\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3\left(4a+b\right)=7\left(a+b-3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’12a+3b=7a+7b-21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’5a-4b=-21---\left(ii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{multiply}\mathrm{equation}\left(i\right)\mathrm{by}5,\mathrm{we}\mathrm{get}5a+5b=-30---\left(iii\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{subtract}\left(ii\right)\mathrm{from}\left(iii\right),\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(5a+5b\right)-\left(5a-4b\right)=-30+21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’5a+5b-5a+4b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’9b=-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}â‡’b=-1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{substitute}b=-1\mathrm{in}\mathrm{equation}\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}a+\left(-1\right)=-6\phantom{\rule{0ex}{0ex}}â‡’a=-6+1=-5$ Option A.: Option B: Option.C: a - b = 0 -5 - (-1) = -4 0 None of the option satisfies the values.

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