Question

Combined with the equation −9x + 3y = 12 creates a system of linear equations with no solution? Select one: A. 18x - 6y = 20 B. 3x - y = -4 C. -16x + 9y = 30 D. 5x + 8y = -1

Answer 1

Answer:

The correct option is A. 18x - 6y = 20

Step-by-step explanation:

A system of two equations has no solution if the lines having these equations are parallel to each other.

Also, two lines $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ are parallel,

If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

While,

They coincide ( infinitely many solution ) if $\frac{a_1}{a_2}=\frac{ b_1}{b_2}=\frac{c_1}{c_2}$,

They are non parallel ( a unique solution ) if $\frac{a_1}{a_2}\neq \frac{ b_1}{b_2}\neq \frac{c_1}{c_2}$,

Here, the given equation,

-9x + 3y = 12

Since in lines -9x + 3y = 12 and 18x - 6y = 20,

$\frac{-9}{18}=\frac{3}{-6}\neq \frac{12}{20}$

Hence, system −9x + 3y = 12, 18x - 6y = 20 has no solution.

In lines -9x + 3y = 12 and 3x - y = -4,

$\frac{-9}{3}=\frac{3}{-1}= \frac{12}{-4}$

Hence, system −9x + 3y = 12, 3x - y = -4 has infinitely many solutions.

In lines -9x + 3y = 12 and  -16x + 9y = 30 ,

$\frac{-9}{-16}\neq \frac{3}{9}\neq \frac{12}{30}$

Hence, system −9x + 3y = 12,  -16x + 9y = 30  has a solution.

In lines -9x + 3y = 12 and 5x + 8y = -1,

$\frac{-9}{5}\neq \frac{3}{8}\neq \frac{12}{-1}$

Hence, system −9x + 3y = 12, 5x + 8y = -1 has a solution.