QUESTION

If 2x-3y=7 and (a+b)x-(a+b-3)y=4a+b have infinite solutions. $(a,b)=$(a) (-5, -1)(b) (-5, 1)(c) (5, 1)(d) (5, -1)

Hint: The two linear equations have infinite solution when $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. We will substitute the respective coefficients of the two equations and then take first two ratios and then last two ratios and finally solve the simultaneous equations to get a and b.

Before proceeding with the question we should understand the concept of a system of two linear equations having infinite solutions.
To solve systems of an equation in two variables, first, we need to determine whether the equation is dependent, independent, consistent or inconsistent. If a pair of the linear equations have unique or infinite solutions, then the system of equations is said to be a consistent pair of linear equations. Thus, suppose we have two equations in two variables as follows:
${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$
${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$
The given equations are consistent and dependent and have infinitely many solutions, if and only if,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}......(1)$
For equation $2x-3y=7$, ${{a}_{1}}=2$, ${{b}_{1}}=-3$ and ${{c}_{1}}=7$.
For equation $(a+b)x-(a+b-3)y=4a+b$, ${{a}_{2}}=a+b$, ${{b}_{2}}=3-a-b$ and ${{c}_{2}}=4a+b$.
Using this information and putting all these values in equation (1) we get,
$\,\Rightarrow \dfrac{2}{a+b}=\dfrac{-3}{3-a-b}=\dfrac{7}{4a+b}......(2)$
Solving the first two ratios of equation (2) we get,
$\,\Rightarrow \dfrac{2}{a+b}=\dfrac{-3}{3-a-b}.........(3)$
Now forming an equation in terms of a and b from equation (3) we get,
\begin{align} & \,\Rightarrow \dfrac{2(3-a-b)}{1}=\dfrac{-3(a+b)}{1} \\ & \,\Rightarrow 6-2a-2b=-3a-3b \\ & \,\Rightarrow a+b=-6 \\ & \,\Rightarrow a=-6-b.........(4) \\ \end{align}
Solving the next two ratios of equation (3) we get,
$\,\Rightarrow \dfrac{-3}{3-a-b}=\dfrac{7}{4a+b}......(5)$
Now forming an equation in terms of a and b from equation (5) we get,
\begin{align} & \,\Rightarrow \dfrac{-3(4a+b)}{1}=\dfrac{7(3-a-b)}{1} \\ & \,\Rightarrow -12a-3b=21-7a-7b \\ & \,\Rightarrow 5a-4b=-21.........(6) \\ \end{align}
Now substituting value of a from equation (4) in equation (6) we get,
\begin{align} & \,\Rightarrow 5(-6-b)-4b=-21 \\ & \,\Rightarrow -30-5b-4b=-21 \\ & \,\Rightarrow 9b=-9 \\ & \,\Rightarrow b=-1 \\ \end{align}
Substituting value of b in equation (4) we get,
$\Rightarrow a=-6-(-1)=-5$
Hence $(a,b)=(-5,-1)$ is the answer.
So option (a) is the right answer.

Note: Remembering equation (1) is the key here to find a and b. From equation (2) we could have taken any two ratios at a time. We can commit a mistake in a hurry in solving the simultaneous equation so we need to be careful with every step.