Question

Please solve for x: |2x + 5| < 3|x - 1|


`|2x+5| \ \textless \ 3|x-1|`

Show your work!

Answer 1

x > 8 or x < -2/5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtract Property of Equality

Algebra I

• Solving inequalities
• |Absolute Values| - makes any negative number positive

Explanation:

Step 1: Define

|2x + 5| < 3|x - 1|

Step 2: Rewrite

When solving for absolute values, x can be ± since the absolute value makes all values positive.

Positive x equation: |2x + 5| < 3|x - 1|

Negative x equation: |2x + 5| < -3|x - 1|

Step 3: Solve for x

When solving, we can replace the absolute value with parenthesis because we have now rewritten the absolute value in dual form.

2x + 5 < 3(x - 1)

2x + 5 < -3(x - 1)

Positive Equation

1. [Division Property of Equality] Divide both sides by 3:
   `\displaystyle (2x)/(3) + (5)/(3) < x - 1`
2. [Addition Property of Equality] Add 1 on both sides:
   `\displaystyle (2x)/(3) + (8)/(3) < x`
3. [Subtraction Property of Equality] Isolate x terms:
   `\displaystyle (8)/(3) < (x)/(3)`
4. [Multiplication Property of Equality] Multiply 3 on both sides:
   `\displaystyle 8 < x`

Negative Equation

1. [Division Property of Equality] Divide both sides by -3:
   `\displaystyle (-2x)/(3) - (5)/(3) > x - 1`
2. [Addition Property of Equality] Add 1 on both sides:
   `\displaystyle (-2x)/(3) - (2)/(3) > x`
3. [Addition Property of Equality] Isolate x terms:
   `\displaystyle - (2)/(3) > (5x)/(3)`
4. [Division Property of Equality] Isolate x term:
   `\displaystyle - (2)/(5) > x`

We see that our inequality values are x < -2/5 and/or x > 8.

∴ We have solved your absolute value inequality problem.

Hope this helps! <3 :)

Answer 2

Hi!

The solution to this would be a set of values. To find that set, you would want to find the critical points of both cases (as you're only going to take out the absolute value of one side for now), which are the defining points of inequality of the solution, for example if you take the equation 5 < x < 9, the critical points would be 5 and 9.

To find them, you want to create two equations (just to check, you only really need one):

2x + 5 = 3 | x - 1 |

2x + 5 = -3 | x - 1 |

These are the equations you can use to figure out the critical points, as they calculate the exact boundaries for the zone of correct solutions.

The only thing you have left to do is solve.

2x + 5 = 3 | x - 1 |

2/3x + 5/3 = | x - 1 |

Take both the negative and positive cases of x - 1, as the equation equal to | x - 1 | can be either positive or negative.

2/3 x + 5/3 = x - 1

-1/3x = -8/3

x = 8

other case:

2/3x + 5/3 = -x + 1

5/3x = -2/3

x = -2/5

If you take the other equation, you will get the same thing.

Now to see if it is -2/5 < x < 8 or x < -2/5 OR 8 < x.

To do this, you can just take one inequality, I'll take -2/5 < x < 8, and test a value inside of it. I'll test 1.

|2 * 1 + 5| < 3 |1 - 1|

|3 + 5| < 3|0|

8 < 0

It doesn't work. So therefore, your solution to x is,

x < -2/5 OR x > 8