Question

Determine the values of a and b for which the following system of linear equations has infinite solutions.
$\begin{align}
  & 2x-(a-4)y=2b+1 \\
 & 4x-(a-1)y=5b-1 \\
\end{align}$

Answer 1

Hint: The pair of linear equations ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ and ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ has infinite solutions if $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$. Two linear equations have infinite solutions if they are one and the same, in other words, one of the equations must be a multiple of the other. In this question, we have been given two equations, so we just have to substitute the values in the condition and then evaluate the values of a and b.

Complete step by step answer:
For two linear equations to have infinite solutions, one of the equations must be a multiple of the other. One equation can be obtained by dividing a constant from the other.
$\begin{align}
  & {{a}_{1}}x+{{b}_{1}}y={{c}_{1}}\to (k{{a}_{2}}x+k{{b}_{2}}y=k{{c}_{2}})........(1) \\
 & {{a}_{2}}x+{{b}_{2}}y={{c}_{2}}..........................................(2) \\
\end{align}$
This leads us to the fact that linear equations with infinite solutions has the following property,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=k$
Where,
${{a}_{1}}$ is the coefficient of x in equation (1), ${{a}_{2}}$ is the coefficient of x in equation (2)
${{b}_{1}}$ is the coefficient of y in equation (1), ${{b}_{2}}$ is the coefficient of y in equation (2)
${{c}_{1}}$ is the constant in equation (1), ${{c}_{2}}$ is the constant in equation (2)
$\begin{align}
  & 2x-(a-4)y=2b+1...................(3) \\
 & 4x-(a-1)y=5b-1....................(4) \\
\end{align}$
Therefore, we have the following values from the question,
${{a}_{1}}=2,\,\,\,{{b}_{1}}=-(a-4),\,\,\,{{c}_{1}}=(2b+1)$ (From equation (3))
${{a}_{2}}=4,\,\,\,{{b}_{2}}=-(a-1),\,\,\,{{c}_{2}}=(5b-1)$ (From equation (4))
For getting the value of a, we have,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}$
Substituting the values, $\dfrac{2}{4}=\dfrac{-(a-4)}{-(a-1)}$
On cross-multiplying, we get,
$2(a-1)=4(a-4)$
$14=2a$
$\therefore a=7$
For getting the value of b, we have,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$
Substituting the values, $\dfrac{2}{4}=\dfrac{2b+1}{5b-1}$
On cross-multiplying, we get,
$2(5b-1)=4(2b+1)$
$10b-2=8b+4$
$2b=6$
$\therefore b=3$
Hence, the given linear equation has infinite solution when $a=7$and $b=3$.

Note:
The same question can be asked in terms of equations with no common solution and even one unique solution.
For the condition of equations with no common solution, we use,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$
For the condition of equations with one unique solution, we use,
$\dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}}$.
One must be careful while dealing with the values of coefficients ( not to mix up) and also while cross multiplying.