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Solve the following absolute value equation:
|2x-1|=|x-8|
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First look in this plot.
Plot y = |2x-1| (red) and y = |x-8| (green)
Second, below is the solution and the procedure on how you should treat this equation problem.
Divide the entire number line in these subsets:
a) < x < ; b) <= x <= and c) < x < .
In each of this "interval" the given equation becomes a linear equation (different on different subsets).
The linear equation you can easily solve.
But in addition, you must check that the solution you found for the specific equation belongs to the corresponding interval.
a) Interval < x < .
In this interval (2x-1) is negative, so |2x-1| = -(2x-1);
(x-8) is negative too, so |x-8| = -(x-8).
Therefore the original equation becomes
-(2x-1) = -(x-8)
in this interval. Simplify and solve it:
-(2x-1) = -(x-8) ---> 2x-1 = x-8 ---> 2x-x = -8+1 ---> x = -7.
The number "-7" belongs to the interval (,), so this value really is the solution of the original equation.
b) Interval <= x <= .
In this interval (2x-1) is positive, so |2x-1| = (2x-1);
(x-8) is negative, so |x-8| = -(x-8).
Therefore the original equation becomes
(2x-1) = -(x-8)
in this interval. Simplify and solve it:
(2x-1) = -(x-8) ---> 2x-1 = -x+8 ---> 2x+x = 8+1 ---> 3x = 9 ---> x = 3.
The number "3" belongs to the interval (,), so this value really is the solution of the original equation.
b) Interval < x < .
In this interval (2x-1) is positive, so |2x-1| = (2x-1);
(x-8) is positive too, so |x-8| = (x-8).
Therefore the original equation becomes
(2x-1) = (x-8)
in this interval. Simplify and solve it:
(2x-1) = (x-8) ---> 2x-1 = x - 8 ---> 2x-x = -8+1 ---> x = -7.
But the number "-7" does not belong to the interval (,), so this value really is NOT the solution of the original equation in this interval.
Answer. The solutions are x = -7 and x = 3.
Compare the answer with the plot.
What I described here is a STANDARD method on solving similar problems.
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Lesson to learn from this solution:
A strategy on solving of absolute value equations is to break up the entire set of real numbers into sub-domains (ranges)
where the absolute value of linear terms is a linear function, and then to solve the corresponding linear equations in each sub-domain (range).
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My other lessons on Absolute Value equations in this site are
- 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
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".