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disa [49]
3 years ago
9

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
Mathematics
1 answer:
malfutka [58]3 years ago
4 0

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.

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