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Chapter 2: Problem 29

Solve for \(\mathrm{x}\) in \(|2 \mathrm{x}-6|=|4-5 \mathrm{x}|\)

Short Answer

Expert verified
The solution for x in the equation \(|2x - 6| = |4 - 5x|\) is \(x = \frac{10}{7}, x = \frac{-2}{3}\).

Step by step solution

01

Create cases for each absolute value expression.

To do this, we will create two separate cases for each absolute value expression. Keep in mind that when you have an absolute value expression equal to another expression, it can be either positive or negative. So for each expression, we will solve for both the positive and the negative cases separately. Case 1: \(2x - 6 = 4 - 5x\) Case 2: \(-(2x - 6) = 4 - 5x\)
02

Solve Case 1.

First, let's solve Case 1, which is \(2x - 6 = 4 - 5x\). To solve for x, we can simply get x terms on one side and constant terms on the other side: \(2x + 5x = 4 + 6\) Combine like terms: \(7x = 10\) Divide by 7: \(x = \frac{10}{7}\) From solving Case 1, we get one possible value for x: \(x = \frac{10}{7}\)
03

Solve Case 2.

Now, we will solve Case 2, which is \(-(2x - 6) = 4 - 5x\). First, distribute the negative sign on the left side: \(-2x + 6 = 4 - 5x\) Add 2x to both sides: \(6 = 4 - 3x\) Subtract 4 from both sides: \(2 = -3x\) Divide by -3: \(x = \frac{-2}{3}\) From solving Case 2, we get another possible value for x: \(x = \frac{-2}{3}\)
04

Present the final solution.

Now that we have solved both cases, we have the two possible values for x: 1. \(x = \frac{10}{7}\) 2. \(x = \frac{-2}{3}\) So the solution for x in the equation \(|2x - 6| = |4 - 5x|\) is \(\boxed{x = \frac{10}{7}, x = \frac{-2}{3}}\).

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Most popular questions from this chapter

$$ |(2 x-5) /(x-6)|<3 $$

Find a number \(\mathrm{M}\) when \(|(\mathrm{x}+2) /(\mathrm{x}-2)| \leq \mathrm{M}\), and if \(\mathrm{x}\) is restricted to the Interval \([1 / 2,3 / 2]\).

Find a positive number \(\mathrm{M}\) such that \(\left|\mathrm{x}^{3}-2 \mathrm{x}^{2}+3 \mathrm{x}-4\right| \leq \mathrm{M}\) for all values of \(\mathrm{x}\) in the interval \([-3,2]\).

Find all real numbers satisfying the inequality \(|3 \mathrm{x}+2|>5\).

Graph the following two inequalities and show where the two graphs coincide: \(2 \leq x<3\) and \(|y-2|<1 / 2\).
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