Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 15 Linear Inequations

Exercise 15(A)

Question 1. If the replacement set is the set of natural numbers, solve:
(i) \( x - 5 < 0 \)
(ii) \( x + 1 \leq 7 \)
(iii) \( 3x - 4 > 6 \)
(iv) \( 4x + 1 \geq 17 \)
Answer:
(i) For \( x - 5 < 0 \):
Adding \( 5 \) on both sides of the inequality:
\( x - 5 + 5 < 0 + 5 \)
\( \implies x < 5 \)
Since the replacement set is the set of natural numbers \( \{1, 2, 3, 4, 5, \dots\} \), the solution set is:
\( \{1, 2, 3, 4\} \)

(ii) For \( x + 1 \leq 7 \):
Subtracting \( 1 \) from both sides:
\( x + 1 - 1 \leq 7 - 1 \)
\( \implies x \leq 6 \)
The natural numbers satisfying this condition are:
\( \{1, 2, 3, 4, 5, 6\} \)

(iii) For \( 3x - 4 > 6 \):
Adding \( 4 \) to both sides:
\( 3x - 4 + 4 > 6 + 4 \)
\( \implies 3x > 10 \)
Dividing by \( 3 \) on both sides:
\( \frac{3x}{3} > \frac{10}{3} \)
\( \implies x > 3\frac{1}{3} \)
The natural numbers greater than \( 3\frac{1}{3} \) are:
\( \{4, 5, 6, \dots\} \)

(iv) For \( 4x + 1 \geq 17 \):
Subtracting \( 1 \) from both sides:
\( 4x + 1 - 1 \geq 17 - 1 \)
\( \implies 4x \geq 16 \)
Dividing both sides by \( 4 \):
\( \frac{4x}{4} \geq \frac{16}{4} \)
\( \implies x \geq 4 \)
The natural numbers that are greater than or equal to \( 4 \) are:
\( \{4, 5, 6, \dots\} \)
In simple words: To solve these inequalities when the variable must be a natural number, find the range of x. Then, write down all the counting numbers (starting from 1) that fit into that range.

Exam Tip: Always pay attention to the replacement set (like natural numbers, whole numbers, or integers), as it determines the final set of values for your answer.

Question 2. If the replacement set = \(\{-6, -3, 0, 3, 6, 9\}\); find the truth set of the following:
(i) \( 2x - 1 > 9 \)
(ii) \( 3x + 7 \leq 1 \)
Answer:
(i) For \( 2x - 1 > 9 \):
Add \( 1 \) to both sides:
\( 2x - 1 + 1 > 9 + 1 \)
\( \implies 2x > 10 \)
Divide both sides by \( 2 \):
\( \frac{2x}{2} > \frac{10}{2} \)
\( \implies x > 5 \)
From the given replacement set, the values greater than \( 5 \) are \( 6 \) and \( 9 \). Thus, the solution set is:
\( \{6, 9\} \)

(ii) For \( 3x + 7 \leq 1 \):
Subtract \( 7 \) from both sides of the inequality:
\( 3x + 7 - 7 \leq 1 - 7 \)
\( \implies 3x \leq -6 \)
Divide both sides by \( 3 \):
\( \frac{3x}{3} \leq \frac{-6}{3} \)
\( \implies x \leq -2 \)
From the given replacement set, the values less than or equal to \( -2 \) are \( -6 \) and \( -3 \). Thus, the solution set is:
\( \{-6, -3\} \)
In simple words: Find the range for x first. Then look at the list of numbers given in the replacement set and pick only those numbers that fit inside your range.

Exam Tip: Only select values from the specified replacement set. Do not write general natural numbers or integers unless that is the set specified.

Question 3. Solve \( 7 > 3x - 8 \); \( x \in N \)
Answer:
Given the inequality:
\( 7 > 3x - 8 \)
Subtract \( 3x \) from both sides:
\( 7 - 3x > 3x - 3x - 8 \)
\( \implies 7 - 3x > -8 \)
Subtract \( 7 \) from both sides:
\( 7 - 7 - 3x > -8 - 7 \)
\( \implies -3x > -15 \)
Divide by \( -3 \) on both sides. Remember that dividing or multiplying by a negative value reverses the inequality sign:
\( \frac{-3x}{-3} < \frac{-15}{-3} \)
\( \implies x < 5 \)
Since \( x \) belongs to the natural numbers (\( N \)), the values of \( x \) are:
\( \{1, 2, 3, 4\} \)
In simple words: When you divide both sides of an inequality by a negative number, the inequality sign flips direction (from greater-than to less-than).

Exam Tip: The most common error in inequations is forgetting to reverse the inequality sign when multiplying or dividing by a negative number. Keep this rule in mind!

Question 4. Solve \( -17 < 9y - 8 \); \( y \in Z \)
Answer:
Given the inequality:
\( -17 < 9y - 8 \)
Add \( 8 \) to both sides:
\( -17 + 8 < 9y - 8 + 8 \)
\( \implies -9 < 9y \)
Divide both sides by \( 9 \):
\( \frac{-9}{9} < \frac{9y}{9} \)
\( \implies -1 < y \)
Since \( y \) belongs to the set of integers (\( Z \)), the values of \( y \) greater than \( -1 \) are:
\( \{0, 1, 2, 3, \dots\} \)
In simple words: Solve for y to find that it must be larger than -1. Since y is an integer, the solutions start from 0 and go up infinitely.

Exam Tip: Integers include negative numbers, zero, and positive numbers. Since \( y > -1 \), zero is included in the solution set.

Question 5. Solve \( 9x - 7 \leq 28 + 4x \); \( x \in W \)
Answer:
Given the inequality:
\( 9x - 7 \leq 28 + 4x \)
Subtract \( 4x \) from both sides:
\( 9x - 4x - 7 \leq 28 + 4x - 4x \)
\( \implies 5x - 7 \leq 28 \)
Add \( 7 \) to both sides:
\( 5x - 7 + 7 \leq 28 + 7 \)
\( \implies 5x \leq 35 \)
Divide both sides by \( 5 \):
\( \frac{5x}{5} \leq \frac{35}{5} \)
\( \implies x \leq 7 \)
Since \( x \) belongs to the set of whole numbers (\( W = \{0, 1, 2, 3, \dots\} \)), the solution set is:
\( \{0, 1, 2, 3, 4, 5, 6, 7\} \)
In simple words: Bring the x terms to one side and numbers to the other. Since x is a whole number (which starts at 0), list all whole numbers up to 7.

Exam Tip: Don't forget that whole numbers (\( W \)) start from \( 0 \), unlike natural numbers (\( N \)) which start from \( 1 \). Including \( 0 \) is essential to get full marks.

Question 6. Solve: \( \frac{2}{3}x + 8 < 12 \); \( x \in W \)
Answer:
Given the inequality:
\( \frac{2}{3}x + 8 < 12 \)
Subtract \( 8 \) from both sides:
\( \frac{2}{3}x + 8 - 8 < 12 - 8 \)
\( \implies \frac{2}{3}x < 4 \)
Multiply both sides by \( \frac{3}{2} \):
\( \frac{2}{3}x \times \frac{3}{2} < 4 \times \frac{3}{2} \)
\( \implies x < 6 \)
Since \( x \) is a whole number, the solution set contains all whole numbers strictly less than \( 6 \):
\( \{0, 1, 2, 3, 4, 5\} \)
In simple words: Subtract 8 first, then multiply by 3 and divide by 2 to isolate x. Since x must be a whole number, the answer is any whole number from 0 to 5.

Exam Tip: Be careful with fractional coefficients. Multiplying by the reciprocal is a clean way to isolate the variable.

Question 7. Solve \( -5(x + 4) > 30 \); \( x \in Z \)
Answer:
Given:
\( -5(x + 4) > 30 \)
Divide both sides by \( -5 \) and reverse the inequality sign:
\( \frac{-5(x + 4)}{-5} < \frac{30}{-5} \)
\( \implies x + 4 < -6 \)
Subtract \( 4 \) from both sides:
\( x + 4 - 4 < -6 - 4 \)
\( \implies x < -10 \)
Since \( x \) is an integer (\( Z \)), the set of values strictly less than \( -10 \) is:
\( \{-11, -12, -13, \dots\} \)
In simple words: Dividing by a negative number flips the greater-than sign to a less-than sign. After solving, we get x is less than -10, which means any integer smaller than -10.

Exam Tip: For infinite sets of negative integers, write the elements in decreasing order or use ellipsis correctly: \( \{-11, -12, -13, \dots\} \) is standard.

Question 8. Solve the inequation \( 8 - 2x \geq x - 5 \); \( x \in N \).
Answer:
Given inequality:
\( 8 - 2x \geq x - 5 \)
Add \( 2x \) and \( 5 \) to both sides:
\( 8 + 5 \geq x + 2x \)
\( \implies 13 \geq 3x \)
This can be written as:
\( 3x \leq 13 \)
Divide both sides by \( 3 \):
\( x \leq \frac{13}{3} \)
\( \implies x \leq 4\frac{1}{3} \)
Since \( x \) must be a natural number (\( N \)), the values that are less than or equal to \( 4\frac{1}{3} \) are:
\( \{1, 2, 3, 4\} \)
In simple words: Rearrange the inequality to get all x terms on one side. Once simplified, x is less than or equal to 4 and 1/3, so the only natural numbers that work are 1, 2, 3, and 4.

Exam Tip: When converting mixed fractions to decide the limits, always find the nearest integer that satisfies the inequality context.

Question 9. Solve the inequality \( 18 - 3(2x - 5) > 12 \); \( x \in W \).
Answer:
Given:
\( 18 - 3(2x - 5) > 12 \)
Expand the brackets:
\( 18 - 6x + 15 > 12 \)
Combine the numerical terms:
\( 33 - 6x > 12 \)
Subtract \( 12 \) from both sides and add \( 6x \):
\( 33 - 12 > 6x \)
\( \implies 21 > 6x \)
This is equivalent to:
\( 6x < 21 \)
Divide both sides by \( 6 \):
\( x < \frac{21}{6} \)
Simplify the fraction:
\( x < \frac{7}{2} \)
\( \implies x < 3\frac{1}{2} \)
Since \( x \) belongs to the whole numbers (\( W \)), the solutions are:
\( \{0, 1, 2, 3\} \)
In simple words: First distribute the -3 across the parentheses. Simplify the terms and isolate x. Since x must be a whole number smaller than 3.5, the allowed values are 0, 1, 2, and 3.

Exam Tip: Be careful with signs when distributing negative coefficients over parentheses. Here, \( -3 \times -5 \) becomes \( +15 \).

Question 10. Solve: \( \frac{2x + 1}{3} + 15 \leq 17 \); \( x \in W \).
Answer:
Given:
\( \frac{2x + 1}{3} + 15 \leq 17 \)
Subtract \( 15 \) from both sides:
\( \frac{2x + 1}{3} \leq 17 - 15 \)
\( \implies \frac{2x + 1}{3} \leq 2 \)
Multiply both sides by \( 3 \):
\( 2x + 1 \leq 6 \)
Subtract \( 1 \) from both sides:
\( 2x \leq 5 \)
Divide both sides by \( 2 \):
\( x \leq \frac{5}{2} \)
\( \implies x \leq 2\frac{1}{2} \)
Since \( x \) is a whole number, the possible values are:
\( \{0, 1, 2\} \)
In simple words: Subtract 15 first to simplify the right side, then multiply by 3 to remove the fraction. Isolate x to find that it must be 2.5 or less, meaning whole numbers 0, 1, and 2 are the answer.

Exam Tip: When working with fractions, always isolate the fractional term first before multiplying both sides by the denominator.

Question 11. Solve: \( -3 + x < 2 \); \( x \in N \).
Answer:
Given:
\( -3 + x < 2 \)
Add \( 3 \) to both sides:
\( x < 2 + 3 \)
\( \implies x < 5 \)
Since \( x \) is a natural number, the values of \( x \) that satisfy the inequality are:
\( \{1, 2, 3, 4\} \)
In simple words: Add 3 to both sides to find that x must be less than 5. Since x has to be a natural number, the answers are 1, 2, 3, and 4.

Exam Tip: Ensure that you do not include \( 0 \) or negative numbers when the variable is defined under the set of natural numbers (\( N \)).

Question 12. Solve: \( 4x - 5 > 10 - x \); \( x \in \{0, 1, 2, 3, 4, 5, 6, 7\} \).
Answer:
Given the inequality:
\( 4x - 5 > 10 - x \)
Add \( x \) to both sides and add \( 5 \) to both sides:
\( 4x + x > 10 + 5 \)
\( \implies 5x > 15 \)
Divide both sides by \( 5 \):
\( x > \frac{15}{5} \)
\( \implies x > 3 \)
Selecting values strictly greater than \( 3 \) from the given replacement set \( \{0, 1, 2, 3, 4, 5, 6, 7\} \), we get the truth set:
\( \{4, 5, 6, 7\} \)
In simple words: Group the x terms on one side and numbers on the other side. You will find that x must be greater than 3. Choose all the numbers larger than 3 from the given set.

Exam Tip: Carefully list the elements from the specified replacement set that satisfy the final inequality, omitting the boundary value if the inequality is strict (\( > \.)).

Question 13. Solve: \( 15 - 2(2x - 1) < 15 \); \( x \in Z \).
Answer:
Given:
\( 15 - 2(2x - 1) < 15 \)
Expand the brackets:
\( 15 - 4x + 2 < 15 \)
Simplify the left side:
\( 17 - 4x < 15 \)
Subtract \( 17 \) from both sides:
\( -4x < 15 - 17 \)
\( \implies -4x < -2 \)
Divide both sides by \( -4 \), reversing the inequality sign:
\( \frac{-4x}{-4} > \frac{-2}{-4} \)
\( \implies x > \frac{1}{2} \)
Since \( x \) belongs to the set of integers (\( Z \)), the integers greater than \( \frac{1}{2} \) are:
\( \{1, 2, 3, 4, 5, \dots\} \)
In simple words: Expand the equation, then isolate x. Since dividing by -4 reverses the inequality, we get x is greater than 0.5. The integers starting from 1 onwards are the correct answer.

Exam Tip: Be cautious when dealing with fractions as boundaries. Integers larger than \( 0.5 \) start at \( 1 \), not \( 0 \).

Question 14. Solve: \( \frac{2x + 3}{5} > \frac{4x - 1}{2} \); \( x \in W \).
Answer:
Given inequality:
\( \frac{2x + 3}{5} > \frac{4x - 1}{2} \)
Cross-multiply to clear the denominators:
\( 2(2x + 3) > 5(4x - 1) \)
Expand both sides:
\( 4x + 6 > 20x - 5 \)
Rearrange terms by grouping \( x \) on one side:
\( 4x - 20x > -5 - 6 \)
\( \implies -16x > -11 \)
Divide both sides by \( -16 \) and reverse the inequality sign:
\( x < \frac{-11}{-16} \)
\( \implies x < \frac{11}{16} \)
Since \( x \) belongs to the set of whole numbers (\( W \)) and \( \frac{11}{16} \approx 0.6875 \), the only whole number less than \( \frac{11}{16} \) is \( 0 \):
\( \{0\} \)
In simple words: Cross-multiply to get rid of fractions, then solve for x. You get x is less than 11/16 (which is less than 1). The only whole number smaller than this is 0.

Exam Tip: Whole numbers include \( 0 \). When the boundary is a positive fraction less than \( 1 \), do not forget that \( 0 \) is a valid and often the only solution.

Exercise 15(B)

Question 1. Solve and graph the solution set on a number line: \( x - 5 < -2 \); \( x \in N \)
Answer:
Given the inequality:
\( x - 5 < -2 \)
Add \( 5 \) to both sides:
\( x - 5 + 5 < -2 + 5 \)
\( \implies x < 3 \)
Since \( x \) is a natural number (\( x \in N \)), the solution set is:
\( \{1, 2\} \)
The required number line graph is shown below: In simple words: Add 5 to both sides to get x < 3. Since x is a natural number, the only options are 1 and 2, which we represent with solid dots on the number line.

Exam Tip: When plotting discrete sets (like natural or whole numbers), only put solid dots on the specific numbers in the solution set. Do not draw a continuous thick line.

Question 2. Solve and graph the solution set on a number line: \( 3x - 1 > 5 \); \( x \in W \)
Answer:
Given the inequality:
\( 3x - 1 > 5 \)
Add \( 1 \) to both sides:
\( 3x - 1 + 1 > 5 + 1 \)
\( \implies 3x > 6 \)
Divide both sides by \( 3 \):
\( \frac{3x}{3} > \frac{6}{3} \)
\( \implies x > 2 \)
Since \( x \) belongs to the set of whole numbers (\( W \)), the solution set is:
\( \{3, 4, 5, \dots\} \)
The required number line graph is shown below: In simple words: Simplify the inequality to find that x must be larger than 2. Since x is a whole number, we put dots on 3, 4, and so on.

Exam Tip: For infinite discrete sets, place dots on the visible whole numbers starting from the smallest value (here \( 3, 4 \)) to indicate that the pattern continues indefinitely in that direction.

Question 3. Solve and graph the solution set on a number line: \( -3x + 12 < -15 \); \( x \in R \)
Answer:
Given the inequality:
\( -3x + 12 < -15 \)
Subtract \( 12 \) from both sides:
\( -3x < -15 - 12 \)
\( \implies -3x < -27 \)
Divide both sides by \( -3 \), and reverse the inequality direction:
\( x > \frac{-27}{-3} \)
\( \implies x > 9 \)
Since \( x \) belongs to the set of real numbers (\( R \)), the solution set is:
\( \{x : x \in R, x > 9\} \)
The required number line graph is shown below: In simple words: When we divide by -3, the inequality flips and gives x > 9. Since x is a real number, we draw an open circle at 9 and shade the entire line to the right.

Exam Tip: For real numbers (\( R \)), a strict inequality (\( > \) or \( < \)) is shown with a hollow/open circle at the boundary, and a thick continuous line for the shaded region.

Question 4. Solve and graph the solution set on a number line: \( 7 \geq 3x - 8 \); \( x \in W \)
Answer:
Given the inequality:
\( 7 \geq 3x - 8 \)
Add \( 8 \) to both sides:
\( 7 + 8 \geq 3x \)
\( \implies 15 \geq 3x \)
Divide both sides by \( 3 \):
\( 5 \geq x \)
This can be written as:
\( x \leq 5 \)
Since \( x \) is a whole number (\( x \in W \)), the solution set is:
\( \{0, 1, 2, 3, 4, 5\} \)
The required number line graph is shown below: In simple words: Solving the equation gives x less than or equal to 5. Since x is a whole number, we put solid dots on all integers from 0 to 5.

Exam Tip: Ensure that whole numbers start at \( 0 \), so do not place a dot on negative numbers like \( -1 \), but make sure to include \( 0 \).

Question 5. Solve and graph the solution set on a number line: \( 8x - 8 \leq -24 \); \( x \in Z \)
Answer:
Given inequality:
\( 8x - 8 \leq -24 \)
Add \( 8 \) to both sides:
\( 8x \leq -24 + 8 \)
\( \implies 8x \leq -16 \)
Divide both sides by \( 8 \):
\( x \leq \frac{-16}{8} \)
\( \implies x \leq -2 \)
Since \( x \) is an integer (\( Z \)), the solution set is:
\( \{\dots, -4, -3, -2\} \)
The required number line graph is shown below: In simple words: Solving gives x is less than or equal to -2. Since x must be an integer, we mark -2, -3, -4 and so on with solid dots.

Exam Tip: Since integers extend infinitely to the left, make sure the dots are plotted on the left-most visible integers in your diagram.

Question 6. Solve and graph the solution set on a number line: \( 8x - 9 \geq 35 - 3x \); \( x \in N \)
Answer:
Given inequality:
\( 8x - 9 \geq 35 - 3x \)
Add \( 3x \) to both sides:
\( 8x + 3x - 9 \geq 35 \)
\( \implies 11x - 9 \geq 35 \)
Add \( 9 \) to both sides:
\( 11x \geq 35 + 9 \)
\( \implies 11x \geq 44 \)
Divide both sides by \( 11 \):
\( x \geq 4 \)
Since \( x \) is a natural number (\( x \in N \)), the solution set is:
\( \{4, 5, 6, 7, \dots\} \)
The required number line graph is shown below: In simple words: Move variables to one side and numbers to the other to get x is greater than or equal to 4. Since x is a natural number, we place dots on 4, 5, 6, and 7.

Exam Tip: Be sure to start placing dots at \( 4 \) (the boundary) since the inequality has an "equal to" component (\( \geq \)).

Question 7. Solve and graph the solution set on a number line: \( 5x + 4 > 8x - 11 \); \( x \in Z \)
Answer:
Given inequality:
\( 5x + 4 > 8x - 11 \)
Subtract \( 5x \) from both sides:
\( 4 > 3x - 11 \)
Add \( 11 \) to both sides:
\( 4 + 11 > 3x \)
\( \implies 15 > 3x \)
Divide both sides by \( 3 \):
\( 5 > x \)
This is the same as:
\( x < 5 \)
Since \( x \) is an integer (\( Z \)), the solution set is:
\( \{\dots, 1, 2, 3, 4\} \)
The required number line graph is shown below: In simple words: Rearrange the inequality so that x is on the smaller side. We find x < 5. Since x is an integer, we mark all integers smaller than 5 on the number line.

Exam Tip: Since \( x < 5 \) is a strict inequality, \( 5 \) is not included in the solution set. Ensure there is no dot on \( 5 \).

Question 8. Solve and graph the solution set on a number line: \( \frac{2x}{5} + 1 < -3 \); \( x \in R \)
Answer:
Given inequality:
\( \frac{2x}{5} + 1 < -3 \)
Subtract \( 1 \) from both sides:
\( \frac{2x}{5} < -3 - 1 \)
\( \implies \frac{2x}{5} < -4 \)
Multiply both sides by \( 5 \):
\( 2x < -20 \)
Divide both sides by \( 2 \):
\( x < -10 \)
Since \( x \) belongs to the set of real numbers (\( R \)), the solution set is:
\( \{x : x \in R, x < -10\} \)
The required number line graph is shown below: In simple words: First subtract 1, then multiply by 5 and divide by 2 to find x < -10. Since x is a real number, we draw an open circle at -10 and shade the line to the left.

Exam Tip: Be careful when drawing real number ranges on negative scales. A value of \( x < -10 \) goes to the left (towards \( -11 \), \( -12 \)), not to the right.

Question 9. Solve and graph the solution set on a number line: \( \frac{x}{2} > -1 + \frac{3x}{4} \); \( x \in N \)
Answer:
Given inequality:
\( \frac{x}{2} > -1 + \frac{3x}{4} \)
Multiply all terms by the lowest common multiple of the denominators, which is \( 4 \):
\( \frac{x}{2} \times 4 > \left(-1 \times 4\right) + \frac{3x}{4} \times 4 \)
\( \implies 2x > -4 + 3x \)
Subtract \( 2x \) from both sides:
\( 0 > -4 + x \)
Add \( 4 \) to both sides:
\( 4 > x \)
This is equivalent to:
\( x < 4 \)
Since \( x \) is a natural number (\( x \in N \)), the solution set is:
\( \{1, 2, 3\} \)
The required number line graph is shown below: In simple words: Multiply the entire inequality by 4 to clear the fractions. You will find that x must be less than 4. Since x is a natural number, the solution is 1, 2, and 3.

Exam Tip: Natural numbers do not include \( 0 \) or negative integers. Keep your dots strictly on \( 1, 2, 3 \) for full marks.

Question 10. Solve and graph the solution set on a number line: \( \frac{2}{3}x + 5 \leq \frac{1}{2}x + 6 \); \( x \in W \)
Answer:
Given inequality:
\( \frac{2}{3}x + 5 \leq \frac{1}{2}x + 6 \)
Multiply all terms by \( 6 \) (the LCM of \( 3 \) and \( 2 \)):
\( \left(\frac{2}{3}x \times 6\right) + \left(5 \times 6\right) \leq \left(\frac{1}{2}x \times 6\right) + \left(6 \times 6\right) \)
\( \implies 4x + 30 \leq 3x + 36 \)
Subtract \( 3x \) from both sides:
\( 4x - 3x + 30 \leq 36 \)
\( \implies x + 30 \leq 36 \)
Subtract \( 30 \) from both sides:
\( x \leq 36 - 30 \)
\( \implies x \leq 6 \)
Since \( x \) is a whole number (\( W \)), the solution set is:
\( \{0, 1, 2, 3, 4, 5, 6\} \)
The required number line graph is shown below: In simple words: Clear the fractions by multiplying by 6. After isolating x, you get x is less than or equal to 6. Since x is a whole number, place solid dots starting from 0 up to 6.

Exam Tip: Be methodical when multiplying the whole equation by the LCM. Ensure that terms without fractions (like \( 5 \) and \( 6 \)) are also multiplied.

Question 11. Solve the inequation \( 5(x - 2) > 4(x + 3) - 24 \) and represent its solution on a number line. Given the replacement set is \( \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \).
Answer:
Given inequation:
\( 5(x - 2) > 4(x + 3) - 24 \)
Expand the terms:
\( 5x - 10 > 4x + 12 - 24 \)
Simplify the right-hand side:
\( 5x - 10 > 4x - 12 \)
Subtract \( 4x \) and add \( 10 \) to both sides:
\( 5x - 4x > -12 + 10 \)
\( \implies x > -2 \)
We now select values from the replacement set \( \{-4, -3, -2, -1, 0, 1, 2, 3, 4\} \) that are strictly greater than \( -2 \):
\( \{-1, 0, 1, 2, 3, 4\} \)e required number line graph is shown below: