# Simple math questions I forget how to do? Thank You!?

1. solve for "x"

1=((x+2)/3)-((x-1)/5)

2. solve for "x"

|3x-5|=6

3. Solve the inequality. Write the solution in interval notation.

(5/3)-(1/4x)>(2/3x)-(1/6)

4. Solve the inequality. Write solution in interval notation.

1/2|x-6|-2 ≥ 2

1.

1 = ((x+2)/3) - ((x - 1) /5

Recall : Symmetric property a =b Then b = a where a, b are any numbers

((x+2)/3) - ((x - 1) /5 = 1

Multiply each side by 15

15[((x+2)/3) - ((x - 1) /5 = 1(15)]

Recall : Distributive property a(b + c) = ab + ac

15(x + 2)/3 - 15(x - 1)/5 =15

Simplify

5(x + 2) - 3(x - 1) = 15

Recall : Distributive property

5x + 10  -3x +3 = 15

5x -3x +10 + 3 = 15

2x + 13 = 15

Subtract 13 from each side

2x + 13 - 13 = 15 - 13

2x +0 = 2

2x = 2

Divide each side by 2

2x /2 = 2/2

x = 1.

2.Solve for "x"  |3x-5|=6.

|3x-5|=6.

Case1:

3x -5 = 6

3x -5 + 5 = 6 + 5

Simplify

3x +0 = 11

3x = 11

Divide each side by 3

3x/3 = 11 / 3

x = 11 / 3.

Case2 :

3x - 5 = -6

3x -5 + 5 = -6 + 5

3x + 0 = -1

3x = -1

Divide each side by 3

x = -1 / 3.

3.

(5/3)-(1/4x)>(2/3x)-(1/6)

(5/3)-(1/4x)+(1/4x) > (2/3x)-(1/6) + 1/4x

5/3 > 2/3x + 1/4x -(1/6)

Multiply each side by 12

(5/3)12 > [(2/3x) +1/4x - 1/6]12

20 > 8 /x +3/x - 2

20 + 2 > 8/x +3/x - 2 + 2

22 > 11/x - 0

22 > 11/x

Multiply each side by x

22x > [11/x]x

22x > 11

Divide each side by 22

22x/22 > 11/22

x > 1/2

The interval of x is (1/2 , ∞).

The solution of the inequality is x  < 0 or x  > 1/2

The interval notation from of solution is (-∞, 0) U (1/2, ∞).

4.

1/2|x-6| - 2  ≥ 2

1/2|x-6| - 2 + 2  ≥ 2 + 2

1/2|x-6| - 0 ≥  4

1/2|x-6|  ≥  4

Multiply each side by 2

2(1/2|x-6|)  ≥ 2( 4)

Simplify

|x-6|  ≥ 8

Case1:

x - 6 ≥ 8

x - 6 + 6 ≥ 8 + 6

x + 0 ≥ 14

x ≥ 14

x interval is[14,∞]

Case 2 :

x - 6 ≥ -8

x - 6 + 6 ≥ -8 + 6

x + 0 ≥ -2

x ≥ -2

x interval is [-2 ,∞]

Solution in interval : [-2,∞] or [14,∞]

Solution in interval : [-2,∞].

Solution set is {x Є R| x ≤ -2 or x ≥ 14}

Solution in interval notation form (-∞, -2] U [14, ∞).

3) The inequality is

• Step-1

State the exclude values,These are the values for which denominator is zero.

The exclude value of the inequality is 0.

• Step - 2

Solve the related equation

Solution of related equation x   = 1/2.

• Step - 3

Draw the vertical lines at the exclude value and at the solution to separate the number line into intervals.

Continuous...

• Step - 4

Now test  sample values in each interval to determine whether values in the interval satisify the inequality.

Test interval     x - value      Inequality                                          Conclusion

(-∞, 0)                      x =  -1         (5/3)-[1/4(-1)] > [2/3(-1)]-(1/6)⇒1.91>0.832         True

(0, 1/2)                     x = 0.1        (5/3)-[1/4(0.1)] > [2/3(0.1)]-(1/6) -0.83 > -0.10    False

(1/2, ∞)                   x = 0.8         (5/3)-[1/4(0.8)] > [2/3(0.8)]-(1/6)1.35 > 0.36       True

The above conclude that the inequality is satisfied for all x - values in (-∞, 0) and (1/2, ∞).

This implies that the solution  of  the  inequalityis  the  interval (-∞, 0) and (1/2, ∞) . as shown in Figure below. Note that the original inequality contains a “less than” symbol. This means that the solution set does not contain the endpoints of the test intervals are (-∞, 0) and(1/2, ∞)

The solution of the inequality is x  < 0 or x  > 1/2

The interval notation from of solution is (-∞, 0) U (1/2, ∞).

4) The absolute inequality is 1/2|x - 6| - 2 ≥ 2

1/2|x - 6| ≥ 4

Multiply each side by 2.

|x - 6| ≥ 8

|x| ≥ a can be written as x ≥ a or x ≤ - a

|x - 6| ≥ 8 can be written as x - 6 ≥ 8 or x - 6 ≤ - 8

Solve the inequality : x - 6 ≥ 8

x - 6 + 6 ≥ 8 + 6

x  ≥ 14

Solve the inequality : x - 6 ≤ - 8

x - 6 + 6 ≤ - 8 + 6

x ≤ - 2

Therefore the solution of the absolute inequality is x  ≥ 14 or x ≤ - 2

Solution set is {x Є R| x ≤ -2 or x ≥ 14}

Solution in interval notation form (-∞, -2] U [14, ∞).

Observe the graph , the closed circle means that -2 and 14 are the solutions of the inequality.