Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

97,166 users

A few pre-calc questions? please?

0 votes

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


Thank you so much!!!!!

asked Mar 21, 2013 in PRECALCULUS by angel12 Scholar

5 Answers

+1 vote

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

Multiply each side by 15

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

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

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

15 =5x + 10 -3x +3

15 =2x +13

Subtract 13 to each side

15 - 13 = 2x +13 - 13

2 = 2x + 0

2 = 2x

Divide each side by 2

1 = x

Symetric property : a = b then b =a

x = 1.

 

answered Mar 23, 2013 by diane Scholar
0 votes

2.

|3x - 5| = 6

Case 1 :

3x  - 5 = 6

Add 5 to each side

3x -5 +5 = 6 + 5

3x + 0 = 11

3x = 11

Divide each side by 3

3x / 3 = 11 / 3

x = 11 / 3.

Case 2 :

3x - 5 = -6

Add 5 to each side

3x - 5 + 5 = -6 + 5

3x + 0 = -1

3x = -1

Divide each side by 3

3x  / 3 = -1 / 3

There fore x = -1 / 3.

 

answered Mar 23, 2013 by diane Scholar
0 votes

3) The inequality is image

  • 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 image

image

image

image

image

image

image

image

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.

 

answered May 31, 2014 by david Expert
0 votes

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  inequalityimageis  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 image is x  < 0 or x  > 1/2

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

 

answered May 31, 2014 by david Expert
0 votes

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

Add 2 to each side

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

Add 6 to each side.

x - 6 + 6 ≥ 8 + 6

x  ≥ 14

Solve the inequality : x - 6 ≤ - 8

Add 6 to each side.

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.

answered May 31, 2014 by david Expert
edited May 31, 2014 by david

Related questions

asked Sep 10, 2014 in PRECALCULUS by anonymous
asked Dec 8, 2014 in ALGEBRA 2 by anonymous
asked Feb 14, 2017 in ALGEBRA 2 by anonymous
asked Nov 12, 2014 in PRECALCULUS by anonymous
asked Jul 19, 2014 in PRECALCULUS by anonymous
asked May 13, 2014 in PRECALCULUS by anonymous
asked Feb 14, 2017 in GEOMETRY by anonymous
asked Feb 8, 2017 in ALGEBRA 2 by anonymous
...