Question

If there is no solution of the linear equation system kx-5y=2, 6x+2y=7 then find the value of k?

Answer 1

Hint: In this question it is given that we have two linear equations kx-5y=2 and 6x+2y=7, which have no solution, and we have to find the value of k which is the coefficient of x. So for this let’s just consider two linear equations,
$${}a_{1}x+b_{1}y=c_{1}$$ …………...equation(1)
$${}a_{2}x+b_{2}y=c_{2}$$ …………...equation(2)
If equation (1) and equation(2) have no solution then we can write,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$..........equation(3)

Complete step-by-step solution:
First of all we are going to compare kx-5y=2 with equation(1) and 6x+2y=7 with equation(2).
So by comparing the coefficients we can write,
$$a_{1}=k,\ b_{1}=-5,\ c_{1}=2$$
$$a_{2}=6,\ b_{2}=2,\ c_{2}=7$$

Now putting the values which we have obtained, in equation(3), we get,
$$\dfrac{a_{1}}{a_{2}} =\dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$$
$$\Rightarrow \dfrac{k}{6} =\dfrac{-5}{2} \neq \dfrac{2}{7}$$
Now taking the first two part of the above equation, we get,
$$\dfrac{k}{6} =\dfrac{-5}{2}$$
$$\Rightarrow 2k=-6\times 5$$ [by cross multiplication]
$$\Rightarrow 2k=-30$$
$$\Rightarrow k=\dfrac{-30}{2}$$
$$\Rightarrow k=-15$$
So our required solution is k = - 15.

Note: To solve this type of equation we have to know that if two linear equations have no solution then two equations must be parallel to each other, and if two linear equations parallel then the ratio of the coefficient of x is equal to the ratio of y coefficient.