Question

What should be the value of $\lambda$ for the given equations to have infinitely many solutions?
$5x + \lambda y = 4$ and $15x + 3y = 12$

Answer 1

The given equation represents two lines in the Cartesian plane and for two lines to have infinitely many solutions in a two-dimensional plane they must overlap each other. And for two lines ${a_1}x + {b_1}y = {c_1}{\text{ and }}{a_2}x + {b_2}y = {c_2}$ the condition for them to be overlapping will be $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ . Use this condition to find the value of $\lambda$ .

Complete step-by-step answer:
Here in this problem, we are given with two linear equations, i.e. $5x + \lambda y = 4$ and $15x + 3y = 12$ in two variables $x$ and $y$ . The coefficient of the variable $y$ in the first equation is $\lambda$ , which is the unknown in this equation. We need to find the value of the unknown $\lambda$ for which these equations have infinitely many solutions.
As we know that the two lines present in a two-dimensional Cartesian plane can either be intersecting, overlapping, or parallel to each other. A linear equation in two variables can be represented by a straight line.
The given equations are:
$5x + \lambda y = 4$ ………(i)
$15x + 3y = 12$ …………(ii)
Since the equation of x-axis is $y = 0$ , so if we will put $y = 0$ in the equation we will get the points on x-axis where these lines cross the axis.
Therefore, for $y = 0$ , we get:
In equation (i) $5x + \lambda y = 5x + \lambda \times 0 = 4 \Rightarrow 5x = 4 \Rightarrow x = \dfrac{4}{5}$
Hence, the line (i) crosses the x-axis at point $\left( {\dfrac{4}{5},0} \right)$
In equation (ii) $15x + 3y = 15x + 3 \times 0 = 12 \Rightarrow 15x = 12 \Rightarrow x = \dfrac{{12}}{{15}} = \dfrac{4}{5}$
Hence, the line (ii) crosses the x-axis at point $\left( {\dfrac{4}{5},0} \right)$
Therefore, both the lines pass through a common point, i.e. $\left( {\dfrac{4}{5},0} \right)$
Now we know that both lines lie in the same plane and have a point of intersection. So we can conclude that for these equations to have infinitely many solutions, they must completely overlap each other.
For any two equations of lines, ${a_1}x + {b_1}y = {c_1}{\text{ and }}{a_2}x + {b_2}y = {c_2}$ the condition for them to be overlapping will be $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ .
Therefore, from the above condition, the given lines $5x + \lambda y = 4$ and $15x + 3y = 12$ will be overlapping when:
$\Rightarrow \dfrac{5}{{15}} = \dfrac{\lambda }{3} = \dfrac{4}{{12}}$
Now the above equation can be solved by expressing the fractions in its simplest form
$\Rightarrow \dfrac{5}{{15}} = \dfrac{\lambda }{3} = \dfrac{4}{{12}} \Rightarrow \dfrac{1}{3} = \dfrac{\lambda }{3} = \dfrac{1}{3}$
We can find the value for $\lambda$ by taking the first fraction and evaluating them as:
$\Rightarrow \dfrac{1}{3} = \dfrac{\lambda }{3} \Rightarrow \lambda = \dfrac{1}{3} \times 3 \Rightarrow \lambda = 1$
Therefore, for the given equations to have infinitely many solutions the value of $\lambda$ will be $\lambda = 1$

Note: In coordinate geometry, it is always important to analyze the questions properly before starting the solution. Notice that when you put $\lambda = 1$ in the equation $5x + \lambda y = 4$ , you will get an equation $5x + y = 4$ , which is the same as the second equation $15x + 3y = 12$ after dividing it by $3$ on both the sides, i.e. $\dfrac{{15x + 3y}}{3} = \dfrac{{12}}{3} \Rightarrow \dfrac{{15}}{3}x + \dfrac{3}{3}y = 4 \Rightarrow 5x + y = 4$