# Which of the following equations would not be consistent with 3y - 4x = 7? a) -8x = 6y - 14 b)...

## Question:

Which of the following equations would not be consistent with {eq}3y - 4x = 7 {/eq}?

a) {eq}-8x = 6y - 14 {/eq}

b) {eq}12y = -9x + 28 {/eq}

c) {eq}-16x + 28 = 12y {/eq}

d) {eq}-4x - 3y - 7 = 0 {/eq}

e) {eq}-56 - 24y = -32x {/eq}

## Inconsistent Systems of Equations:

A system of equations may have one unique solution, when the determinant of the matrix of coefficients of this system is not zero. The system may have infinitely many solutions, when the two or more equations are linearly dependent, that is, one can be represented as a linear combination of several others. Finally, the system may have no solution, which means that two or more equations in the system are inconsistent with each other.

## Answer and Explanation:

The answer is **e)**:

Explanation: the two equations are inconsistent if the system of those two equations does not have a solution. In this case, we have the following system:

{eq}\begin{cases} 3y - 4x = 7\\ -56 - 24y = -32x \end{cases} {/eq}

or

{eq}\begin{cases} 3y - 4x = 7\\ 24y -32x = -56 \end{cases} {/eq}

Here we simply re-arranged terms in the second equation. Now if we divide both sides of the second equation by 8, we obtain:

{eq}\begin{cases} 3y - 4x = 7\\ 3y -4x = -7 \end{cases} {/eq}

It is clear now that from this two equations it follows that {eq}7 = -7 {/eq}, which is impossible. Therefore, the system has no solution and two equations are inconsistent;

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from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 9