# Help With Interval Inequality?

Solve the inequality; Put the answer in interval notation . type infinity . For more than one interval use a U to represent a " union ". Write () if there is no solution.

1) 55x^2 + 26x-56<0

2) (11x+4)(5x-4)<2511

3) (3x+2)(5x-10) >55

4) (x+11) / (x-8)(x-30) ≤ 0

5) (38x-18)(x-16) / (x-35)^2 > 0

6) x / (x-9) > 342 / 71

+1 vote

1). 55x2 + 26x - 56 < 0.

Now solve the equation using the factor method.

55x2 + 70x - 44x - 56 < 0.

Take out common term 5x and 4.

5x(11x + 14) - 4(11x + 14) < 0.

(5x - 4)(11x + 14) < 0.

(5x - 4) < 0 or (11x + 14) < 0.

Take (5x - 4) < 0

5x < 4

Divide each side by 5.

x < 4/5 = 0.8

And (11x + 14) < 0.

Subtract 14 from each side.

11x < -14

Divide each side by 11.

x < -14/11 = -1.272

Therefore x < 0.8 or x < -1.272

Graph the solution set on a number line.

Solution of 55x2 + 26x - 56 < 0 is {x | -14/11 < x < 4/5}.

+1 vote

2). (11x + 4)(5x - 4) < 2511

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

55x2 - 44x + 20x - 16 < 2511

55x2 - 24x - 16 < 2511

Subtract 2511 from each side.

55x2 - 24x - 2527 < 0

Now solve the equation using the factor method.

55x2 - 385x + 361x - 2527 < 0.

55x(x - 7) + 361(x - 7) < 0.

Take out common terms.

(x - 7)(55x - 361) < 0

x - 7 < 0 or 55x - 361 < 0.

x - 7 < 0 then x < 7

55x - 361 < 0

55x < 361

Divide each side by 55.

x < 361/55 = 6.56

Therefore x < 7 or x < 6.563

Graph the solution set on a number line.

The inequality (11x + 4)(5x - 4) < 2511 solution is {x | -361/55 < x < 7}.

3). (3x+2)(5x-10) >55

3x + 2 > 55 or 5x - 10 > 55

Take 3x + 2 > 55

Subtract 2 from each side.

3x > 53

Divide each side by 3.

x > 53/3 = 17.66

And 5x - 10 > 55

5x > 65.

Divide each side by 5.

x > 13.

Therefore x > 17.66 or x > 13.

Graph the solution set on a number line.

The inequality (3x+2)(5x-10) >55 solution  is {x| x > 3 or x < -5/3}.

4). (x+11) / [(x-8)(x-30)] ≤ 0

Multiply each side by [(x-8)(x-30)].

x + 11 ≤ 0

Subtract 11 from each side.

x ≤ -11.

Graph the solution set on a number line.

Solution of the inequality is

x ≤ -11

8 < x < 30.

+1 vote

6). x / (x-9) > 342 / 71

Cross multiplication.

71x > 342(x - 9)

Distribute terms using distributive property:  a( b + c) = ab + ac

71x > 342x - 3078

Subtract 342x from each side.

- 271x > -3078

Multiply each side by negative one and flip the symbol.

271x < 3078.

Divide each side by 271.

x < 3078/271

x < 11.357

Graph the solution set on a number line.

The solution of the inequality is 9 < x < 3078/271.

5)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 35.

• Step - 2

Solve the related equation

Solutions of related equation x   = 9/19 and x  = 16.

• Step-3

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

• Step-4

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

Test x  = -10

Above statement is true.

Test x = 10

Above statement is false.

Test x = 30

Above statement is true.

Above statement is true.

The statement is true for x  = -10, x = 30 and = 40.Therefore the solution is

Solution x  < 9/19

16 < < 35

x  > 35.

1) The equation 55x2 + 26x - 56 < 0

Related equation is 55x2 + 26x - 56 = 0

55x2 + 70x - 44x - 56 = 0

Take out common term 5x and 4.

5x(11x + 14) - 4(11x + 14) = 0

(11x + 14)(5x - 4) = 0

55x2 + 26x - 56 = (11x + 14)(5x - 4)

(11x + 14)(5x - 4) < 0

Now, there are two ways this product could be less than zero.One factor must be negative and one must be positive.

First situation: 11x + 14 < 0 and 5x - 4 > 0

11x < -14 and 5x > 4

x < -14/11 and x > 4/5

There are NO values for which this situation is true.

Second situation:

11x + 14 > 0 and 5x -4 < 0

11x > -14 and 5x < 4

x > -14/11 and x < 4/5

Solution  {x | -14/11 < x < 4/5}

Solution of 55x2 + 70x - 44x - 56 = 0 in interval notation (-∞, 4/5) U (-14/11, ∞)

The inequality solution in interval notation (-14/11, 4/5)

Solution on number line.

Observe the graph the open circle means -14/11 and 4/5 are not included in the solution set.

edited Jun 4, 2014 by david

2) The inequality (11x + 4)(5x - 4) < 2511

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

55x2 - 44x + 20x - 16 < 2511

55x2 - 24x - 16 < 2511

Subtract 2511 from each side.

55x2 - 24x - 2527 < 0

Now solve the related equation using the factor method.

55x2 - 24x - 2527 = 0

55x2 - 385x + 361x - 2527 = 0.

55x(x - 7) + 361(x - 7) = 0.

Take out common terms.

(x - 7)(55x + 361) = 0

55x2 - 24x - 2527 = (x - 7)(55x + 361)

Now, there are two ways this product could be less than zero.One factor must be negative and one must be positive.

First situation:  x - 7 > 0 55x + 361 < 0

x > 7 and 55x < -361

x > 7 and x < - 361/55

There are NO values for which this situation is true.

Second situation: x - 7 < 0 and 55x + 361 > 0

x < 7 and 55x < -361

x < 7 and x < -361/55

Solution  {x | -361/55 < x < 7}

Solution of (11x + 4)(5x - 4) < 2511 in interval notation (-∞, 7) U (-361/55, ∞)

The inequality solution in interval notation (-361/55, 7)

Solution on number line.

Observe the graph the open circle means -361/55 and 7 are not included in the solution set.

3) The inequality (3x + 2)(5x - 10) > 55

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

15x2 - 30x + 10x - 20 > 55

15x2 - 20x - 20 - 55 > 0

15x2 - 20x - 75 > 0

Factor the related equation 15x2 - 20x - 75 = 0

5(3x2 - 4x - 15) = 0

5[3x2 - 9x + 5x - 15] = 0

5 [3x(x - 3) + 5(x - 3)] = 0

5(x - 3)(3x + 5) = 0

15x2 - 20x - 75 = 5(x - 3)(3x + 5)

5(x - 3)(3x + 5) > 0

(x - 3)(3x + 5) > 0

Now, there are two ways that this product could be greater than zero (positive) - both factors are positive or both factors are negative.

Case 1: Both factors are positive

x - 3 > 0 and 3x + 5 > 0

x > 3 and 3x > -5

x > 3 and x > -5/3

The only condition that makes both true is x > 3.

Case 2 : Both factors are negative.

x - 3 < 0 and 3x + 5 < 0

x < 3 and 3x < -5

x < 3 and x < -5/3

The only condition that makes both true is x < -5/3

Solution {x| x > 3 or x < -5/3}

Solution of (3x + 2)(5x - 10) > 55 in interval notation (-∞, -5/3) U (3, ∞)

Solution on the number line.

Observe the graph the open circle means 3 and -5/3 are not include in the solution set.

4) 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 8, 30.

• Step - 2

Solve the related equation

Solution of related equation x   = -11.

• Step - 3

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

• 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

(-∞, -11)        x =  -20     True

(-11, 8)            x = 0                   False

(8, 30)             x = 10             True

(30, ∞)           x = 35                 False

Note that the original inequality contains a “” symbol, We inlude it into set of solutions at x = -11

Above statement is true.

x ≤ -11 is a solution of inequality.

The above conclude that the inequality is satisfied for all x - values in (-∞, -11) and (8, 30).

This implies that the solution  of  the  inequality is  the  interval (-∞, -11) and (8, 30) . as shown in Figure below. Note that the original inequality contains a “” symbol. This means that the solution set contain the endpoints of the test interval is (-∞, -11) .

Solution of the inequality

x ≤ -11

8 < x < 30

The interval notation form of inequality is (-∞, -11] U (8, 30).