1

Problem

Find the value of q and r :

5q = - 3r + 2 and 12r = - 20q + 8

2

Step

Given equations are

\$5x = 3y + 2\$ …​…​.equation (1)

\$12y = 20x + 8\$ …​…​.equation (2)

3

Formula

For a system of equations \$a_1 x + b_1 y + c_1 = 0; a_2 x + b_2 y + c_2 = 0\$ to have no solution, the condition to be satisfied is \$ (a_1)/(a_2) = (b_1)/(b_2) ne (c_1)/(c_2)\$.

4

Step

The given equation can be written as:

\$5x - 3y = 2\$…​..equation(3)

\$12y - 20x = 8\$…​.equation(4)

5

Hint

So, here \$a_1 = 5, a_2 = -20, b_1 = -3, b_2 = 12, c_1 = 2 and c_2 = 8\$.

6

Step

Now plug the values in the formula:

\$5/(-20) = (-3)/(12) = 2/8\$

\$(\cancel(5))/(\cancel(-20)^4) = (\cancel(-3))/(\cancel(12)^4) ne (\cancel(2))/(\cancel(8)^4)\$

7

Step

After simplifaction:

\$-1/4 = -1/4 ne 1/4\$

8

Solution

All these fractions satisfy the required condition, hence the system of equations has no solutions.

9

Sumup

Please summarize steps

Choices

10

Choice-A

This option is not correct because it does not satisfy the system of equation formula

Wrong Infinite solutions

11

Choice-B

This option is correct because it satisfies the system of equation formula

Correct No solutions

12

Choice-C

This option is not correct because it does not satisfy the system of equation formula

Wrong Finite solutions

13

Choice-D

This option is incorrect because it does not satisfy the system of equation formula

Wrong None of these

14

Answer

Option

B

15

Sumup

Please summarize choices