Solve the following equations and inequalities: 1) −x2+2x&gt;1 2) |1−3x|=5 3) |4x−3|≤2

a)−x2+2x&gt;1 −x2+2x−1&gt;0 −(x2−2x+1)&gt;0 x2−2x+1&lt;0 (x−1)2&lt;0 Square of any number cannot be negative.Thus, there is no solution.

Consider the part (2): |1−3x|=5 ±(1−3x)=5 1−3x=5 −3x=4 x=−43 −(1−3x)=5 −1+3=5 3x=6 x=2 Hence, the solution is x=−43,2 Consider the part (3): |4−3x|≤2 ±(4−3x)≤2 4−3x≤2 −3x≤−2 3x≥2 x≥23 −(4−3x)≤2 −4+3x≤2 3x≤6 x≤2 Thus, the solution is x=[23,2]

a)−x2+2x&gt;1 −x2+2x−1&gt;0 −(x2−2x+1)&gt;0 x2−2x+1&lt;0 (x−1)2&lt;0 Square of any number cannot be negative.Thus, there is no solution.

Consider the part (2): |1−3x|=5 ±(1−3x)=5 1−3x=5 −3x=4 x=−43 −(1−3x)=5 −1+3=5 3x=6 x=2 Hence, the solution is x=−43,2 Consider the part (3): |4−3x|≤2 ±(4−3x)≤2 4−3x≤2 −3x≤−2 3x≥2 x≥23 −(4−3x)≤2 −4+3x≤2 3x≤6 x≤2 Thus, the solution is x=[23,2]

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trefoniu1

Answered question

2022-01-24

Solve the following equations and inequalities:
1)

- x^{2} + 2 x > 1

| 1 - 3 x | = 5

| 4 x - 3 | \leq 2

Answer & Explanation

terorimaox

Beginner2022-01-25Added 16 answers

a) $- x^{2} + 2 x > 1$
$- x^{2} + 2 x - 1 > 0$
$- (x^{2} - 2 x + 1) > 0$
$x^{2} - 2 x + 1 < 0$
$(x - 1)^{2} < 0$
Square of any number cannot be negative.Thus, there is no solution.

Aiden Cooper

Beginner2022-01-26Added 14 answers

Consider the part (2):

| 1 - 3 x | = 5

\pm (1 - 3 x) = 5

1 - 3 x = 5

- 3 x = 4

x = - \frac{4}{3}

- (1 - 3 x) = 5

- 1 + 3 = 5

3 x = 6

x = 2

Hence, the solution is

x = - \frac{4}{3}, 2

Consider the part (3):

| 4 - 3 x | \leq 2

\pm (4 - 3 x) \leq 2

4 - 3 x \leq 2

- 3 x \leq - 2

3 x \geq 2

x \geq \frac{2}{3}

- (4 - 3 x) \leq 2

- 4 + 3 x \leq 2

3 x \leq 6

x \leq 2

Thus, the solution is

x = [\frac{2}{3}, 2]

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