Mathematics, 01.09.2019 19:50 dhhdndnd

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|

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Answers

Answer from: amandaa25

(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 $\boxed{ \ 0 \ and \ \frac{8}{3} \ }$

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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 $\boxed{ \ 1 \ and \ \frac{13}{3} \ }$

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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 $\boxed{ \ \frac{21}{20} \ }$

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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.