Question

What will be the value of $K$, if the following pair of linear equations have infinite solutions?
$2x - 3y = 7$
$(K + 1)x + (1 - 2K)y = (5K - 4)$

Answer 1

Hint: For solving such questions we need to use the equations using the coefficients. When the two linear equations are of the form: ${a_1}x + {b_1}y = {c_1}$ and ${a_2}x + {b_2}y = {c_2}$
When the equations have infinitely many equations, we can say that:
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$, Here $a_1,a_2, b_1,b_2,c_1 $ and $c_2$ are the coefficients of the linear equations.
We use the above logic for the given problem which results in an equation with the variable $K$. We will solve it to find the value of $K$.

Complete step by step answer:
The two given equations are: $2x - 3y = 7$ and $(K + 1)x + (1 - 2K)y = (5K - 4)$
Let’s write down all the coefficients of both the equations:
${a_1} = 2$ , ${b_1} = - 3$ , ${c_1} = 7$
${a_2} = K + 1$ , ${b_2} = 1 - 2K$, ${c_2} = 5K - 4$
Let’s use the relation mentioned above and write down the equation.
$\dfrac{2}{{K + 1}} = \dfrac{{ - 3}}{{1 - 2K}} = \dfrac{7}{{5K - 4}}$
Cross multiplying the first two terms in the equation, we get:
$2(1 - 2K) = - 3(K + 1)$
After solving the equations, we get:
$2 - 4K = - 3K - 3$
On simplification,
$-4K+3K=-3-2$
$\Rightarrow -K=-5$
Multiplying the above equation with $-1$ on both sides, we get
\[K=5\]
Therefore, when $K = 5$, the two equations have infinitely many solutions.

Note:
> We can check whether the $K$ value we got is correct or not by substituting it in the equation.
$\dfrac{{ - 3}}{{1 - 2K}} = \dfrac{7}{{5K - 4}}$
Substituting $K$ value as 5
$\dfrac{{ - 3}}{{1 - 2(5)}} = \dfrac{7}{{5(5) - 4}}$
$\dfrac{{ - 3}}{{ - 9}} = \dfrac{7}{{21}} = \dfrac{1}{3}$
Hence, our value of $K$ is right.

> When two linear equations in two variables are given, the relation between the coefficients will give us information about the solutions of the equations.
$\bullet $ $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ means infinitely many solutions.
$\bullet $ $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ means no solution.
$\bullet $ $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ means unique solution.
> There is an alternative method of solving this question by using matrices. We convert the equations into matrix forms A, B and X. If the value of $adjo\operatorname{int} (A) \times B = 0$ , then the two equations have infinitely many solutions.