Linear Inequalities Exercise 1
Question:
Solve the inequality for real x. x+x/2+x/3<11
Answer:
 x + x/2 + x/3 < 11
 3x + 3x/2 < 33
 4.5x < 33
 x < 33/4.5
 x < 7.33
Question:
Solve the inequality and show the graph of the solution on number line: 3(1−x)<2(x+4)
Answer:
Step 1: Subtract 2x from both sides of the inequality. 3(1−x)  2x < 2(x+4)  2x
Step 2: Simplify the left side of the inequality. 3  3x  2x < 2x + 8
Step 3: Simplify the right side of the inequality. x < 10
Step 4: Solve for x. x > 10
The graph of the solution is shown on the number line below:
10 ————> x —————>
Question:
Solve −12x>30, when (i) x is a natural number. (ii) x is an integer
Answer:
(i) When x is a natural number:
Step 1: Divide both sides by 12 to get x on one side.
12x > 30 x < 2.5
Step 2: Since natural numbers are positive integers, x must be greater than 0 and less than 2.5. Therefore, there is no solution when x is a natural number.
(ii) When x is an integer:
Step 1: Divide both sides by 12 to get x on one side.
12x > 30 x < 2.5
Step 2: Since integers include negative numbers, x can be any number less than 2.5. Therefore, the solution set is x < 2.5.
Question:
Solve 3x+8>2, when (i) x is an integer. (ii) x is a real number
Answer:
(i) When x is an integer, 3x+8>2 can be rearranged to 3x>6. The solution set for this inequality is x>2.
(ii) When x is a real number, 3x+8>2 can be rearranged to 3x>6. The solution set for this inequality is x>2.
Question:
5x−3≥3x−5
Answer:
 5x  3 ≥ 3x  5
 5x  3x ≥ 3  5
 2x ≥ 2
 x ≥ 1
Question:
To receive Grade A in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunitas marks in first four examinations are 87,92,94 and 95, find minimum marks that Sunita must obtain in the fifth examination to get grade A in the course.
Answer:
Step 1: Calculate the average marks Sunita has obtained in the first four examinations.
87 + 92 + 94 + 95 = 368
Average = 368/4 = 92
Step 2: Calculate the total marks Sunita needs to obtain in the five examinations to receive Grade A.
90 marks x 5 examinations = 450 marks
Step 3: Calculate the minimum marks Sunita must obtain in the fifth examination.
450  368 = 82
Therefore, Sunita must obtain a minimum of 82 marks in the fifth examination to get Grade A in the course.
Question:
A man wants to cut three lengths from a single piece of board of length 91cm.The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?
Answer:

Let x = the length of the shortest board.

The second board is 3 cm longer than the shortest, so it has a length of x + 3 cm.

The third board is twice as long as the shortest, so it has a length of 2x cm.

The third board must be at least 5 cm longer than the second, so 2x must be greater than or equal to x + 8 cm.

Therefore, the shortest board must have a length of x ≥ 5 cm.
Question:
Solve the inequality for real x. 4x<(5x−2)/3−(7x−3)/5
Answer:
 4x < (5x−2)/3−(7x−3)/5
 20x < 5x−2−21x+9
 16x < 11
 x > 11/16
Question:
Solution set of inequality 4x+3<5x+7 is (−4,∞).
Answer:

Rearrange the inequality: 4x  5x < 7  3

Simplify: x < 4

Divide both sides by 1: x > 4

Solution set of inequality 4x+3<5x+7 is (4,∞).
Question:
Solve: 37−(3x+5)≥9x−8(x−3)
Answer:
 37  (3x + 5) ≥ 9x  8(x  3)
 37  3x  5 ≥ 9x  8x + 24
 32  3x ≥ 9x + 24
 3x ≥ 9x + 56
 4x ≥ 56
 x ≤ 14
Question:
Solve the inequality for real x. (2x−1)/3≥(3x−2)/4−(2−x)/5
Answer:
Step 1: Multiply both sides of the inequality by 15.
15(2x−1)/3 ≥ 15(3x−2)/4 − 15(2−x)/5
Step 2: Simplify the left side of the inequality.
10x − 5 ≥ 9x − 6 − 6 + 3x
Step 3: Simplify the right side of the inequality.
10x − 5 ≥ 6x
Step 4: Subtract 6x from both sides of the inequality.
4x − 5 ≥ 0
Step 5: Add 5 to both sides of the inequality.
4x ≥ 5
Step 6: Divide both sides of the inequality by 4.
x ≥ 5/4
Question:
Solve the inequality for real x. 1/2(3x/5+4)≥1/3(x−6)
Answer:
 1/2(3x/5+4) ≥ 1/3(x−6)
 3/5x + 4 ≥ x/3 − 6/3
 10/5x + 12 ≥ x − 6
 10/5x ≥ x − 18
 10/5x − x ≥ −18
 5/5x ≥ −18
 x ≥ −18*5/5
 x ≥ 36
Question:
Find all pairs of consecutive odd positive integers both which are smaller than 10 such that their sum is more than 11.
Answer:
 The smallest possible consecutive odd positive integers are 1 and 3.
 The sum of 1 and 3 is 4, which is not more than 11.
 The next consecutive odd positive integers are 3 and 5.
 The sum of 3 and 5 is 8, which is not more than 11.
 The next consecutive odd positive integers are 5 and 7.
 The sum of 5 and 7 is 12, which is more than 11.
 Therefore, the pair of consecutive odd positive integers both which are smaller than 10 such that their sum is more than 11 is (5, 7).
Question:
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23
Answer:
Pair 1: 6 and 8. Sum = 14 Pair 2: 10 and 12. Sum = 22
Question:
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length in cm. of the shortest side.
Answer:
Let x = the length of the shortest side.
Longest side = 3x Third side = 3x  2
Perimeter = 3x + 3x  2 = 6x  2
6x  2 ≥ 61 6x ≥ 63 x ≥ 63/6 x ≥ 10.5
Therefore, the minimum length of the shortest side is 10.5 cm.
Question:
Solve 24x<100, when (i) x is a natural number. (ii) x is an integer
Answer:
(i) x = 4
(ii) x = 4
Question:
Solve 5x−3<7, when (i) x is an integer. (ii) x is a real number.
Answer:
(i) 5x  3 < 7 5x < 10 x < 2
(ii) 5x  3 < 7 5x < 10 x < 2.0
Question:
Solve for x: x/2=x/3+1.
Answer:
Step 1: Multiply both sides of the equation by 2.
2x/2 = x/3 + 1*2
Step 2: Simplify the left side of the equation.
2x/2 = 2x/3 + 2
Step 3: Simplify the right side of the equation.
2x/2 = 2(x/3 + 1)
Step 4: Subtract x/3 from both sides of the equation.
2x/2  x/3 = 2 + x/3
Step 5: Simplify the left side of the equation.
x/2 = 2 + x/3
Step 6: Subtract x/2 from both sides of the equation.
0 = 2 + x/3  x/2
Step 7: Simplify the right side of the equation.
0 = x/6 + 2
Step 8: Subtract 2 from both sides of the equation.
2 = x/6
Step 9: Multiply both sides of the equation by 6.
12 = x
Question:
Solve:3(2−x)≥2(1−x) when x is an integer.
Answer:
 3(2−x)≥2(1−x)
 3(2x) ≥ 2  2x
 6  3x ≥ 2  2x
 4 ≥ x
 x ≤ 4
Question:
Solve: 3(x−2)/5≤5(2−x)/3
Answer:
 Multiply both sides by 15: 3(x−2) ≤ 5(2−x)
 Simplify: 3x  6 ≤ 10  5x
 Add 5x to both sides: 8x  6 ≤ 10
 Add 6 to both sides: 8x ≤ 16
 Divide both sides by 8: x ≤ 2
Question:
Solve the following quadratic equation. 3x^2−2x−1=0
Answer:
Step 1: Rewrite the equation in the form ax^2 + bx + c = 0. In this case, a = 3, b = 2 and c = 1.
Step 2: Calculate the discriminant (b^24ac). In this case, the discriminant is (2)^2  4(3)(1) = 4+12 = 16.
Step 3: Calculate the two solutions using the quadratic formula. The two solutions are x = (2 ± √16)/6
Step 4: Simplify the solutions. The two solutions are x = 1 ± 2√2/3
Question:
Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Answer:

Let the marks obtained in the third test be ‘x’.

Average = (70+75+x)/3

60 = (70+75+x)/3

60*3 = 70+75+x

180 = 70+75+x

1807075 = x

x = 35
Therefore, Ravi should obtain at least 35 marks in the third test to have an average of at least 60 marks.
Question:
Solve the inequalities for real x. 3(x−1)≤2(x−3)
Answer:
 3(x−1) ≤ 2(x−3)
 3x  3 ≤ 2x  6
 3x ≤ 2x  3
 3x  2x ≤ 3
 x ≤ 3
Question:
Solve the inequality and show the graph of the solution on number line:
Answer:
2x + 4 ≤ 8
Step 1: Subtract 4 from both sides: 2x ≤ 4
Step 2: Divide both sides by 2: x ≤ 2
The graph of the solution on the number line is:
————— 3 2 1 0 1 2
Question:
x/2≥(5x−2)/3−(7x−3)/5
Answer:
 x/2 ≥ (5x  2)/3  (7x  3)/5
 3x/2 ≥ 5x/3  7x/5
 3x/2  5x/3 + 7x/5 ≥ 0
 (3x  10x + 15x)/10 ≥ 0
 18x/10 ≥ 0
 x ≥ 0
Question:
Solve: 3x−7>5x−1
Answer:

Add 7 to both sides: 3x > 5x + 6

Subtract 5x from both sides: 3x  5x > 6

Simplify: x > 6