Question

For value k =3 system x +ky =2 and 4x+12y =8 has

Answer 1

$\red{ ☆ UPSC-ASPIRANT ☆ }$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

For value k =3 system x +ky =2 and 4x+12y =8 has

$\huge\bigstar\huge\tt\underline\red{ᴀɴsᴡᴇʀ }\bigstar$

For value k= 3 the system have many solutions:-

see in the picture above attach for complete solution step by step:-

Hope it helps you..!!!

_________________

Thankyou:)

Answer 2

The system of equations $x+ky=2,k=3\\or, x+3y=2\\4x+12y=8$ has infinite solutions.

Given:

The system of equations is as follows.

$x+ky=2,k=3\\or, x+3y=2\\4x+12y=8$

To Find:

the solution to the system of equations.

Solution:

We can find the solution to this problem in the following way.

The system of equations depends upon the ratio of their respective coefficients and accordingly can have a unique solution or an infinite solution or no solution.

Let us suppose we have the following system of equations.

$a_{1} x+b_{1}y =c_{1} \\a_{2} x+b_{2}y =c_{2}$

We shall use all the relevant coefficient ratios $\frac{a_{1}}{a_{2}} ,\frac{b_{1}}{b_{2}} ,\frac{c_{1}}{c_{2}}$ to arrive at the solution to this problem.

We can compute the coefficient ratios $\frac{a_{1}}{a_{2}} ,\frac{b_{1}}{b_{2}} ,\frac{c_{1}}{c_{2}}$ as under

$\frac{a_{1}}{a_{2}} =\frac{1}{4} ,\\\frac{b_{1}}{b_{2}}=\frac{3}{12}=\frac{1}{4} ,\\\frac{c_{1}}{c_{2}}=\frac{2}{8}=\frac{1}{4}$

We know that when the coefficient ratios are equal $\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}$, the system of equations has infinite solutions.

Therefore, the required solution is that the system of equations $x+ky=2,k=3\\or, x+3y=2\\4x+12y=8$ has infinite solutions.

#SPJ3