Question

Find the values of $a$ and $b$ for which the system of linear equations have infinite number of solutions:

$$\begin{gathered} 2x+3y=7 \\ (a+b+1)x+(a+2b+2)y=4(a+b)+1 \end{gathered}$$

Answer 1

Step 1: Recall the condition for an infinite number of solutions:

The system of linear equation

$\begin{array}{rcl}{a}_{1}x+b_{1}y+c_1& =& 0\\ a_{2}x+b_{2}y+c_2& =& 0\end{array}$

has an infinite number of solutions if:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

Step 2: Apply condition for an infinite solution

The system of linear equation

$$\begin{gathered} 2x+3y=7 \\ (a+b+1)x+(a+2b+2)y=4(a+b)+1 \end{gathered}$$

will have an infinite number of solutions if

$\frac{2}{a+b+1}=\frac{3}{a+2b+2}=\frac{7}{4(a+b)+1}$.

Now, on comparing

$\begin{array}{rcl}\frac{2}{a+b+1}& =& \frac{3}{a+2b+2}\\ 2a+4b+4& =& 3a+3b+3\\ a-b& =& 1.......\left(1\right)\end{array}$

$\begin{array}{rcl}\frac{3}{a+2b+2}& =& \frac{7}{4(a+b)+1}\\ 12a+12b+3& =& 7a+14b+14\\ 5a-2b& =& 11.........\left(2\right)\end{array}$

Step 3: Solve (1) and (2) by substitution method

From equation (1) we have, $a=1+b$

Substitute the above value in equation (2):

$\begin{array}{rcl}5(1+b)-2b& =& 11\\ 5+5b-2b& =& 11\\ 3b& =& 6\\ \mathit{b}& \mathbf{=}& \mathbf{2}\end{array}$

Then,

$$\begin{gathered} a=1+2 \\ \mathit{a}\mathbf{=}\mathbf{3} \end{gathered}$$

Hence, the required values of $a\text{and}b$ are $3\text{and}2$ respectively.