Question

Hitunglah nilai x ( jika ada ) yang memenuhi persamaan nilai mutlak berikut . Jika tidak ada nilai x yang memenuhi , berikan alasanmu.
a. |4-3x|= |-4|
b. 2|3x-8|= 10C. 2x+|3x-8|= -4D. 5 |2x-3|=2|3-5x|
e. 2x + |8-3x| = |x-4|

Answer 1

(a). The solutions are 0 and ⁸/₃.

(b). The solutions are 1 and ¹³/₃.

(c). The equation has no solution.

(d). The only solution is ²¹/₂₀.

(e). The equation has no solution.

Further explanation

These are the problems with the absolute value of a function.

For all real numbers x,

$\boxed{ \ |f(x)|=\left \{ {{f(x), for \ f(x) \geq 0} \atop {-f(x), for \ f(x) < 0}} \right. \ }$

Problem (a)

|4 – 3x| = |-4|

|4 – 3x| = 4

Case 1

$\boxed{ \ 4 - 3x \geq 0 \ } \rightarrow \boxed{ \ 4\geq 3x \ } \rightarrow \boxed{ \ x\leq \frac{4}{3} \ }$

For 4 – 3x = 4

Subtract both sides by four.

-3x = 0

Divide both sides by -3.

x = 0

Since $\boxed{ \ 0\leq \frac{4}{3} \ }$, x = 0 is a solution.

Case 2

$\boxed{ \ 4 - 3x < 0 \ } \rightarrow \boxed{ \ 4 < 3x \ } \rightarrow \boxed{ \ x > \frac{4}{3} \ }$

For -(4 – 3x) = 4

-4 + 3x = 4

Add both sides by four.

3x = 8

Divide both sides by three.

$x = \frac{8}{3}$

Since $\boxed{ \ \frac{8}{3} > \frac{4}{3} \ }$, $\boxed{ \ x = \frac{8}{3} \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 0 \ and \ \frac{8}{3} \ }$  

————————

Problem (b)

2|3x - 8| = 10

Divide both sides by two.

|3x - 8| = 5  

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x - 8 = 5

Add both sides by eight.

3x = 13

Divide both sides by three.

$x = \frac{13}{3}$

Since $\boxed{ \ \frac{13}{3} \geq \frac{4}{3} \ }$, $\boxed{ \ x = \frac{13}{3} \ }$ is a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x – 8) = 5

-3x + 8 = 5

Subtract both sides by eight.

-3x = -3

Divide both sides by -3.

x = 1  

Since $\boxed{ \ 1 < \frac{8}{3} \ }$, $\boxed{ \ x = 1 \ }$ is a solution.

Hence, the solutions are $\boxed{ \ 1 \ and \ \frac{13}{3} \ }$  

————————

Problem (c)

2x + |3x - 8| = -4

Subtracting both sides by 2x.

|3x - 8| = -2x – 4

Case 1

$\boxed{ \ 3x - 8 \geq 0 \ } \rightarrow \boxed{ \ 3x\geq 8 \ } \rightarrow \boxed{ \ x\geq \frac{8}{3} \ }$

For 3x – 8 = -2x – 4

3x + 2x = 8 – 4

5x = 4

$x = \frac{4}{5}$

Since $\boxed{ \ \frac{4}{5} \ngeq \frac{8}{3} \ }$, $\boxed{ \ x = \frac{4}{5} \ }$ is not a solution.

Case 2

$\boxed{ \ 3x - 8 < 0 \ } \rightarrow \boxed{ \ 3x < 8 \ } \rightarrow \boxed{ \ x < \frac{8}{3} \ }$

For -(3x - 8) = -2x – 4

-3x + 8 = -2x – 4

2x – 3x = -8 – 4

-x = -12

x = 12

Since $\boxed{ \ 12 \nless \frac{8}{3} \ }$, $\boxed{ \ x = 12 \ }$ is not a solution.

Hence, the equation has no solution.

————————

Problem (d)

5|2x - 3| = 2|3 - 5x|  

Let’s take the square of both sides. Then,

[5(2x - 3)]² = [2(3 - 5x)]²

(10x – 15)² = (6 – 10x)²

(10x - 15)² - (6 - 10x)² = 0

According to this formula $\boxed{ \ a^2 - b^2 = (a + b)(a - b) \ }$

$[(10x - 15) + (6 - 10x)][(10x - 15) - (6 - 10x)]] = 0$

(-9)(20x - 21) = 0

Dividing both sides by -9.

20x - 21 = 0

20x = 21

$x = \frac{21}{20}$

The only solution is $\boxed{ \ \frac{21}{20} \ }$

————————

Problem (e)

2x + |8 - 3x| = |x - 4|

We need to separate into four cases since we don’t know whether 8 – 3x and x – 4 are positive or negative.  We cannot square both sides because there is a function of 2x.

Case 1

• 8 – 3x is positive  (or 8 - 3x > 0)
• x – 4 is positive  (or x - 4 > 0)

2x + 8 – 3x = x – 4

8 – x = x – 4

-2x = -12

x = 6

Substitute x = 6 into 8 – 3x ⇒ 8 – 3(6) < 0, it doesn’t work, even though when we substitute x = 6 into x - 4 it does work.

Case 2

• 8 – 3x is positive  (or 8 - 3x > 0)
• x – 4 is negative  (or x - 4 < 0)

2x + 8 – 3x = -(x – 4)

8 – x = -x + 4

x – x =  = 4 - 8

It cannot be determined.

Case 3

• 8 – 3x is negative (or 8  - 3x < 0)
• x – 4 is positive. (or x - 4 > 0)

2x + (-(8 – 3x)) = x – 4

2x – 8 + 3x = x - 4

5x – x = 8 – 4

4x = 4

x = 1

Substitute x = 1 into 8 - 3x, $\boxed{ \ 8 - 3(1) \nless 0 \ }$, it doesn’t work. Likewise, when we substitute x = 1 into x – 4, $\boxed{ \ 1 - 4 \not> 0 \ }$

Case 4

• 8 – 3x is negative (or 8 - 3x < 0)
• x – 4 is negative (or x - 4 < 0)

2x + (-(8 – 3x)) = -(x – 4)

2x – 8 + 3x = -x + 4

5x + x = 8 – 4

6x = 4

$\boxed{ \ x=\frac{4}{6} \rightarrow x = \frac{2}{3} \ }$

Substitute $x = \frac{2}{3} \ into \ 8-3x$, $\boxed{ \ 8 - 3 \bigg(\frac{2}{3}\bigg) \not< 0 \ }$, it doesn’t work. Even though when we substitute $x = \frac{2}{3} \ into \ x-4$, $\boxed{ \ \bigg(\frac{2}{3}\bigg) - 4 < 0 \ }$ it does work.

Hence, the equation has no solution.

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Keywords: hitunglah nilai x, the equation, absolute  value of the function, has no solution, case, the only solution

Answer 2

The solutions are as follows:

(a): $\fbox{\begin\\\ \math x=0\ \text{and}\ x=\frac{8}{3}\\\end{minispace}}$  

(b): $\fbox{\begin\\\ \math x=1\ \text{and}\ x=\frac{13}{3}\\\end{minispace}}$

(c): $\fbox{\begin\\\ No solution\\\end{minispace}}$

(d): $\fbox{\begin\\\ \math x=\frac{21}{20}\\\end{minispace}}$

(e): $\fbox{\begin\\\ No solution\\\end{minispace}}$

Further explanation:

The problem is based on the concept of modulus function.

Modulus function is defined as a function which always gives a positive output for all real value of $x$.

Part (a):

The equation in part (a) is as follows:

$|4-3x|=|-4|$

For the above equation two cases are formed.  

Case 1:  $(x>\frac{4}{3})\rightarrow|4-3x|=-(4-3x)$.  

$\begin{aligned}-(4-3x)&=4\\-4+3x&=4\\3x&=8\\x&=\frac{8}{3}\end{aligned}$  

This implies that if $x>\dfrac{4}{3}$ then $x=\frac{8}{3}$.  

Case 2:

$(x$.

$\begin{aligned}(4-3x)&=4\\-3x&=4-4\\-3x&=0\\x&=0\end{aligned}$  

This implies that if $x$ then $x=0$.  

Part (b):

The equation in part (b) is as follows:

$2|3x-8|=10$

Further solve the above equation.

$\begin{aligned}2|3x-8|&=10\\|3x=8|&=\frac{10}{2}\\|3x-8|&=5\end{aligned}$

For the above equation two cases are formed.

Case 1:  $(x>\frac{8}{3})\rightarrow|3x-8|=3x-8$.

$\begin{aligned}3x-8&=5\\3x&=5+8\\3x&=13\\x&=\frac{13}{3}\end{aligned}$

This implies that if $x>\frac{8}{3}$ then $x=\frac{13}{3}$.

Case 2: $(x$.

$\begin{aligned}-(3x-8)&=5\\-3x+8&=5\\-3x&=5-8\\-3x&=-3\\x&=1\end{aligned}$  

This implies that if $x$ then $x=1$.

Part (c):  

The equation in part (c) is as follows:  

$2x+|3x-8|=-4$  

For the above equation two cases are formed.  

Case 1: $(x>\frac{8}{3})\rightarrow|3x-8|=3x-8$.  

$\begin{aligned}2x+3x-8&=-4\\5x&=-4+8\\5x&=4\\x&=\frac{4}{5}\end{aligned}$  

The value of $x$ cannot be equal to $\frac{4}{5}$ because as assumed above $x>\frac{8}{3}$ and $\frac{4}{5}$ so due to contradiction the solution $x=\frac{4}{5}$ is discarded.  

Case 2: $(x$.  

$\begin{aligned}2x-(3x-8)&=-4\\2x-3x+8&=-4\\-x&=-12\\x&=12\end{aligned}$  

The value of $x$ cannot be equal to $12$ because as assumed above $x$ and $(12)>\farc{8}{3}$ so due to contradiction the solution $x=12$ is discarded.

Part (d):  

The equation in part (d) is as follows:  

$5|2x-3|=2|3-5x|$  

Square both the side in the above equation.  

$\begin{aligned}(5|2x-3|)2&=(2|3-5x|)2\\(10x-15)2&=(2(3-5x))2\\(10x-15)2&=(6-10x)2\\(10x-15)2-(6-10x)2&=0\\(10x-15+6-10x)(10x-15-6+10x)&=0\\(-9)(20x-21)&=0\\(20x-21)&=0\\20x&=21\\x&=\frac{21}{20}\end{aligned}$  

Part (e):  

The equation in part (e) is as follows:  

$2x+|8-3x|=|x-4|$ (4)

For the above equation four cases are formed.

Case 1: $(x>4 \rightarrow |x-4|=x-4)$ and $(x$.  

$\begin{aligned}2x+8-3x&=x-4\\-x+8&=x-4\\2x&=12\\x&=6\end{aligned}$  

The value of $x$ cannot be equal to $6$ because as assumed above $x$ so, due to contradiction the solution $x=6$ is discarded.  

Case 2: $x$ and $x$.  

$\begin{aligned}2x+8-3x&=-(x-4)\\-x+8&=-x+4\\-x+x&=-4\\0&=-4(\text {false})\end{aligned}$  

Case 2 leads to a false statement so, no solution for this case.  

Case 3: $x>4\rightarrow|x-4|=x-4$ and $x>\frac{8}{3}\rightarrow|8-3x|=-(8-3x)$.  

$\begin{aligned}2x-(8-3x)&=x-4\\2x-8+3x&=x-4\\4x&=4\\x&=1\end{aligned}$  

The value of $x$ cannot be equal to $1$ because as assumed above $x>4$ so, due to contradiction the solution $x=1$ is discarded.  

Case 4: $x$ and $(x>\frac{8}{3})\rightarrow|8-3x|=-(8-3x)$.  

$\begin{aligned}2x-(8-3x)&=-(x-4)\\2x-8+3x&=-x+4\\6x&=12\\x&=2\end{aligned}$  

The value of $x$ cannot be equal to $2$ because as assumed above $x>\dfrac{8}{3}$ so, due to contradiction the solution $x=2$ is discarded.  

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Answer details:

Grade: High school

Subject: Mathematics  

Chapter: Functions

Keywords: Functions, modulus function, absolute function, domain, range, intervals, equation, graph, curve, relation, |4-3x|=|-4|, 2x+|8-3x|=|x-4|, 5|2x-3|=2|3-5x|, 2x+|3x-8|=-4, solutions, Hitunglah nilai, memenuhi persmaan.