Find the value of q and r:
5q = - 3r + 2 and 12r = - 20q + 8

Given equations are

\$5x = 3y + 2\$ …​.equation (1)
\$12y = 20x + 8\$ …​.equation (2)

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\$.

The given equation can be written as

\$5x - 3y = 2\$ …​.equation(3)
\$12y - 20x =8\$…​.eequation (4)

So, here \$a_1 = 5 , a_2 = -20\$

\$ b_1 = -3 , b_2 = 12\$,

\$ c_1 = 2 and c_2 = 8\$

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\$

After simplifaction

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

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

Choice A is incorrect. This does not satisfy the system of equation formula

Infinite solutions

Choice B is correct. This satisfies the system of equation formula

No solutions

Choice C is incorrect. This does not satisfy the system of equation formula

Finite solutions

Choice D is incorrect. This does not satisfy the system of equation formula

None of these

Option

B

Can you summarize what you’ve understood in the above steps?