Question

Find the value of 'a' and 'b' for which the following system of linear equations has an infinite number of solutions 2x + 3y =7, (a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1.

Answer 1

Hint: For solving this question two linear equations are given. First, we write down the coefficient of both the equations. It is given that both the linear equations have infinite solutions. Using this information, the ratio of their coefficients is equal. So, we can easily calculate the value of a and b.

Complete step-by-step answer:
As per given in the question, the two equations are: 2x + 3y = 7 and (a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1. Now, we identify the coefficients as
2x + 3y -7 = 0
The coefficients of the above equation are:
${{a}_{1}}=2,{{b}_{1}}=3,{{c}_{1}}=-7$
(a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1
(a + b + 1) x + (a + 2b + 2) y - (4 (a + b) + 1) = 0
Similarly, the coefficients of the equation are:
$\Rightarrow {{a}_{2}}=\left( a+b+1 \right),{{b}_{2}}=\left( a+2b+2 \right),{{c}_{2}}=-\left[ 4\left( a+b \right)+1 \right]$
It is given that both the equations have infinite solutions. Therefore, the ratio of their coefficients is equal.
$\Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$
We know all the values, so putting in the above expression, we get:
$\Rightarrow \dfrac{2}{\left( a+b+1 \right)}=\dfrac{3}{\left( a+2b+2 \right)}=\dfrac{-7}{-\left[ 4\left( a+b \right)+1 \right]}$
Now, using these equalities one at a time, we get
$\begin{align}
  & \dfrac{2}{\left( a+b+1 \right)}=\dfrac{3}{\left( a+2b+2 \right)} \\
 & \Rightarrow 2\left( a+2b+2 \right)=3\left( a+b+1 \right) \\
 & \Rightarrow 2a+4b+4=3a+3b+3 \\
 & \Rightarrow 4-3=3a-2b+3b-4b \\
 & \Rightarrow 1=a-b \\
 & \Rightarrow a-b=1...\left( 1 \right) \\
\end{align}$
And,
$\begin{align}
  & \dfrac{2}{\left( a+b+1 \right)}=\dfrac{-7}{-\left[ 4\left( a+b \right)+1 \right]} \\
 & \Rightarrow 2\left[ 4\left( a+b \right)+1 \right]=7\left( a+b+1 \right) \\
 & \Rightarrow 8a+8b+2=7a+7b+7 \\
 & \Rightarrow 8a-7a+8b-7b=7-2 \\
 & \Rightarrow a+b=5...\left( 2 \right) \\
\end{align}$
Adding both the equations (1) and (2), we get:
\[\begin{array}{*{35}{l}}
   \Rightarrow a-b+a+b=1+5 \\
   \Rightarrow 2a=6 \\
   \Rightarrow a=3 \\
\end{array}\]
By substituting the value of a in equation (1),
\[\begin{array}{*{35}{l}}
   \Rightarrow a-b=1 \\
   \begin{align}
  & \Rightarrow 3-b=1 \\
 & \Rightarrow b=3-1 \\
 & \Rightarrow b=2 \\
\end{align} \\
\end{array}\]
Hence, the value of a and b is 3 and 2 respectively.

Note: Students must be aware of the condition of the relationship between coefficients of an equation having infinite solution. For an infinite solution, the two lines are coincident. Hence, the ratio of coefficients is equal.