(a). The solutions are 0 and βΈ/β.
(b). The solutions are 1 and ΒΉΒ³/β.
(c). The equation has no solution.
(d). The only solution is Β²ΒΉ/ββ.
(e). The equation has no solution.
Further explanation
These are the problems with the absolute value of a function.
For all real numbers x,
Problem (a)
|4 β 3x| = |-4|
|4 β 3x| = 4
Case 1
For 4 β 3x = 4
Subtract both sides by four.
-3x = 0
Divide both sides by -3.
x = 0
Since , x = 0 is a solution.
Case 2
For -(4 β 3x) = 4
-4 + 3x = 4
Add both sides by four.
3x = 8
Divide both sides by three.
Since , is a solution.
Hence, the solutions are Β
ββββββββ
Problem (b)
2|3x - 8| = 10
Divide both sides by two.
|3x - 8| = 5 Β
Case 1
For 3x - 8 = 5
Add both sides by eight.
3x = 13
Divide both sides by three.
Since , is a solution.
Case 2
For -(3x β 8) = 5
-3x + 8 = 5
Subtract both sides by eight.
-3x = -3
Divide both sides by -3.
x = 1 Β
Since , is a solution.
Hence, the solutions are Β
ββββββββ
Problem (c)
2x + |3x - 8| = -4
Subtracting both sides by 2x.
|3x - 8| = -2x β 4
Case 1
For 3x β 8 = -2x β 4
3x + 2x = 8 β 4
5x = 4
Since , is not a solution.
Case 2
For -(3x - 8) = -2x β 4
-3x + 8 = -2x β 4
2x β 3x = -8 β 4
-x = -12
x = 12
Since , is not a solution.
Hence, the equation has no solution.
ββββββββ
Problem (d)
5|2x - 3| = 2|3 - 5x| Β
Letβs take the square of both sides. Then,
[5(2x - 3)]Β² = [2(3 - 5x)]Β²
(10x β 15)Β² = (6 β 10x)Β²
(10x - 15)Β² - (6 - 10x)Β² = 0
According to this formula
(-9)(20x - 21) = 0
Dividing both sides by -9.
20x - 21 = 0
20x = 21
The only solution is
ββββββββ
Problem (e)
2x + |8 - 3x| = |x - 4|
We need to separate into four cases since we donβt know whether 8 β 3x and x β 4 are positive or negative. Β We cannot square both sides because there is a function of 2x.
Case 1
8 β 3x is positive Β (or 8 - 3x > 0)x β 4 is positive Β (or x - 4 > 0)
2x + 8 β 3x = x β 4
8 β x = x β 4
-2x = -12
x = 6
Substitute x = 6 into 8 β 3x β 8 β 3(6) < 0, it doesnβt work, even though when we substitute x = 6 into x - 4 it does work.
Case 2
8 β 3x is positive Β (or 8 - 3x > 0)x β 4 is negative Β (or x - 4 < 0)
2x + 8 β 3x = -(x β 4)
8 β x = -x + 4
x β x = Β = 4 - 8
It cannot be determined.
Case 3
8 β 3x is negative (or 8 Β - 3x < 0)x β 4 is positive. (or x - 4 > 0)
2x + (-(8 β 3x)) = x β 4
2x β 8 + 3x = x - 4
5x β x = 8 β 4
4x = 4
x = 1
Substitute x = 1 into 8 - 3x, , it doesnβt work. Likewise, when we substitute x = 1 into x β 4,
Case 4
8 β 3x is negative (or 8 - 3x < 0)x β 4 is negative (or x - 4 < 0)
2x + (-(8 β 3x)) = -(x β 4)
2x β 8 + 3x = -x + 4
5x + x = 8 β 4
6x = 4
Substitute , , it doesnβt work. Even though when we substitute , it does work.
Hence, the equation has no solution.
Learn moreThe inverse of a function linkThe piecewise-defined functions linkThe composite function link
Keywords: hitunglah nilai x, the equation, absolute Β value of the function, has no solution, case, the only solution