1

Problem

Find the value of q and r :

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

2

Step

Given equations are

\$5x = 3y + 2\$ …​…​.eq (1)

\$12y = 20x + 8\$ …​…​.eq (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\$…​..eq(3)

\$12y - 20x = 8\$…​.eq(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 is not correct because it does not satisfy the system of equation formula

Wrong Infinite solutions

11

Choice-B

This is correct because it satisfies the system of equation formula

Correct No solutions

12

Choice-C

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

Wrong Finite solutions

13

Choice-D

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

Wrong None of these

14

Answer

Option

B

15

Sumup

Please summarize choices