Question

If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b represent coincident lines, then a and b satisfy the equation

(a) a + 5b = 0
(b) 5a + b = 0
(c) a − 5b = 0
(d) 5a − b = 0

Answer 1

The given system of equations are

For coincident lines , infinite number of solution

$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b2}=\frac{c_1}{c_2} \\ \Rightarrow \frac{2}{\left(a+b\right)}=\frac{-3}{-\left(a+b-3\right)}=\frac{7}{4a+b} \\ \Rightarrow \frac{2}{\left(a+b\right)}=\frac{3}{\left(a+b-3\right)}=\frac{7}{\left(4a+b\right)} \\ \Rightarrow 2\left(a+b-3\right)=3\left(a+b\right) \\ \Rightarrow 2a+2b-6=3a+3b \\ \Rightarrow 2a+2b-3a-3b=6 \\ \Rightarrow -a-b=6 \\ \Rightarrow a+b=-6---\left(i\right) \\ 3\left(4a+b\right)=7\left(a+b-3\right) \\ \Rightarrow 12a+3b=7a+7b-21 \\ \Rightarrow 5a-4b=-21---\left(ii\right) \\ \mathrm{multiply}\mathrm{equation}\left(i\right)\mathrm{by}5,\mathrm{we}\mathrm{get}5a+5b=-30---\left(iii\right) \\ \mathrm{subtract}\left(ii\right)\mathrm{from}\left(iii\right), \\ \left(5a+5b\right)-\left(5a-4b\right)=-30+21 \\ \Rightarrow 5a+5b-5a+4b=-9 \\ \Rightarrow 9b=-9 \\ \Rightarrow b=-1 \\ \mathrm{substitute}b=-1\mathrm{in}\mathrm{equation}\left(1\right) \\ a+\left(-1\right)=-6 \\ \Rightarrow a=-6+1=-5 \end{gathered}$$

Option A.:

$$\begin{gathered} a+5b=0 \\ -5+5\left(-1\right)=-5-5=-10\ne 0 \end{gathered}$$

Option B:

$$\begin{gathered} 5a+b=0 \\ 5\left(-5\right)+\left(-1\right)=-25-1=-26\ne 0 \end{gathered}$$

Option.C:

a - b = 0

-5 - (-1) = -4 $\ne$0

None of the option satisfies the values.