.
This inequality has two linear functions under the absolute value sign each.
This "absolute value" sign transform a linear function into non-linear, which is not so simple to solve.
Therefore, the solution strategy is to divide the entire number line into separate intervals / segments
in a way that at each interval/segment an absolute value function is LINEAR.
Then the solution is doable and simple.
Below is how I implement this idea.
In this case we have two critical points, x= and x= , where the functions change their linear behavior.
These points divide the entire number line in 3 non-intersecting intervals/segments
1) x < ; 2) <= x <= ; and 3) x > .
Let's analyze each interval separately.
1) If x < , then | 2x+1 | = -(2x+1) and | 4-3x | = 4-3x.
therefore, the original inequality takes the form
(-3)*(-(2x+1)) + 2*(4-3x) < 7.
Simplify and solve it step by step
6x + 3 + 8 - 6x < 7
11 < 7.
This inequality is FALSE.
It means that the interval x < is NOT the solution to the original inequality.
2) If <= x <= , then | 2x+1 | = 2x+1 and | 4-3x | = 4-3x.
therefore, the original inequality takes the form
(-3)*(2x+1) + 2*(4-3x) < 7.
Simplify and solve it step by step
-6x - 3 + 8 - 6x < 7
-12x < 2.
x > =
It means that in the interval <= x <= is the partial solution to the original inequality.
2) If x > , then | 2x+1 | = 2x+1 and | 4-3x | = -(4-3x).
therefore, the original inequality takes the form
(-3)*(2x+1) + 2*(-(4-3x)) < 7.
Simplify and solve it step by step
-6x - 3 - 8 + 6x < 7
-11 < 7.
This inequality is TRUE.
It means that in the interval x >= is the partial solution to the original inequality.
Thus, after completing analyses of all 3 cases/intervals we come to this conclusion
ANSWER. The given inequality has the solution set x >= .
See the plot below, which visually confirms the found solution.
Plot y = 2*|4-3x| - 3*|2x+1| (red) and y = 7 (green).
Solved.
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To see many other similar solved problems, look into the lessons
- Absolute Value equations
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3
- HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1
- HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2
- OVERVIEW of lessons on Absolute Value equations
Read them attentively and become an expert in this area.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Solving Absolute values equations".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.