Question

Given below are three linear equations. Two of them have infinitely many solutions and two have a unique solution. State the pairs:
4x -5y =3, 8x - 10y = 6, 5x - 4y = 5

Answer 1

Step-by-step explanation:

Correct option is

A

x+y=2 & x−y=−4;

As x = -1, y = 3 is a point

So many lines can paas through a point, 

Therefore infinitely many pairs can possible

like (i)x−y=−4,x+y=2

(ii)  6x−2y=−12,3x+y=0

hope it helps

Answer 2

Answer:

$4x -5y =3, 8x - 10y = 6$ have infinitely many solutions

and $8x - 10y = 65,5x - 4y = 5$ & $4x -5y =3,5x - 4y = 5$ have unique solution.

Step-by-step explanation:

Given three linear equations:

$4x -5y =3$

$8x - 10y = 6$

$5x - 4y = 5$

For a given pair of linear equations, $a_1x +b_1y = c_1~ \&~ a_2x +b_2y = c_2$

• if $\dfrac{a_1}{a_2} \ne\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$ , then the equations have unique solution, and
• if $\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$, then the equations have infinitely many solutions.

Testing the conditions for the given equation,

$\left\begin{array}{ccc}4x -5y =3\\8x - 10y = 6\end{array}\right\}\dfrac{4}{8} =\dfrac{5}{10}=\dfrac{3}{6}$  

Taking out the common factors, all are equal to $\dfrac{1}{2}$, therefore the equations have infinitely many solutions.

$\left{\begin{array}{ccc}8x - 10y = 65\\5x - 4y = 5\end{array}\right\}\dfrac{8}{5} \ne\dfrac{10}{4}\ne\dfrac{65}{5}$

This pair of equations have unique solution.

$\left{\begin{array}{ccc}4x -5y =3\\5x - 4y = 5\end{array}\right\}\dfrac{4}{5} \ne\dfrac{5}{4}\ne\dfrac{3}{5}$

This pair of equations also have unique solution.

Therefore, $4x -5y =3, 8x - 10y = 6$ have infinitely many solutions

and $8x - 10y = 65,5x - 4y = 5$ & $4x -5y =3,5x - 4y = 5$ have unique solution.