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Chapter 11: Problem 220

Determine the constant \(\mathrm{A}\) so that the lines \(3 \mathrm{x}-4 \mathrm{y}=12\) and \(\mathrm{Ax}+6 \mathrm{y}=-9\) are parallel.

Short Answer

Expert verified
The constant A is -\(\frac{9}{2}\).

Step by step solution

01

Convert the first equation to slope-intercept form

Rewrite the first equation, \(3x -4y = 12\), as follows: 1. Subtract 3x from both sides: \(-4y = -3x + 12\) 2. Divide both sides by -4: \(y = \frac{3}{4}x - 3\) The slope of the first line is \(\frac{3}{4}\).
02

Convert the second equation to slope-intercept form

Rewrite the second equation, \(Ax + 6y = -9\), as follows: 1. Subtract Ax from both sides: \(6y = -Ax - 9\) 2. Divide both sides by 6: \(y = -\frac{A}{6}x - \frac{3}{2}\) The slope of the second line is \(-\frac{A}{6}\).
03

Set the slopes equal and solve for A

Since the lines are parallel, their slopes are equal, so we set the slopes equal to each other and solve for A: \[\frac{3}{4} = -\frac{A}{6}\] 1. Multiply both sides by 4 to eliminate the fraction: \(3 = -\frac{2A}{3}\) 2. Multiply both sides by 3 to eliminate the fraction: \(9 = -2A\) 3. Divide both sides by -2: \[A = -\frac{9}{2}\] The constant A is -\(\frac{9}{2}\).

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Most popular questions from this chapter

If \(\mathrm{f}(\mathrm{x})=-2 \mathrm{x}-5\), find the (a) slope, (b) \(\mathrm{x}\) -intercept, and (c) \(\mathrm{y}\) -intercept, (d) Graph the function

Show that the graphs of \(3 x-y=9\) and \(6 x-2 y+9=0\) are parallel lines.

Find the point of intersection of the graphs of \(3 \mathrm{x}-\mathrm{y}=5\) and \(9 \mathrm{x}-3 \mathrm{y}=15\)

Find the slope, the \(\mathrm{y}\) -intercept, and the \(\mathrm{x}\) -intercept of the equation \(2 x-3 y-18=0\)

Graph the function \(\mathrm{y}=9-\mathrm{x} / 2\).
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