Chapter 5: Linear Equations and Inequalities

Mathematics

Class 92025–26 SNC28 Questions SolvedEnglish Medium

Exercise 5.1 — Class 9 Mathematics Solutions

All 28 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 5.1

4 Questions (24 Sub-parts)

Quick Solution Step by Step

Just the working. Switch to Step by Step for the full explanation.

Question 1

EasyApproved by Miss Aniya • Sep 11, 2026

Solve and represent the solution on a real line.

Hint

To solve a linear equation in one variable $a x + b = 0$ , isolate $x$ by applying inverse operations. Check your solution by substituting it back into the original equation, then plot the point on the number line.

Solution

This question consists of subparts requiring solving linear equations/inequalities and representing their solutions on the real number line with number line diagrams.

Step by Step

Q1Part (i)

EasyApproved by Miss Aniya • Sep 11, 2026

$12 x + 30 = - 6$

Quick Solution

Under Review

\begin{matrix} 12 x + 30 & = - 6 \\ 12 x & = - 36 \\ x & = - 3 \end{matrix}

Solution: x = -3 on Real Linescale: 1 unit

Step by Step

Q1Part (ii)

EasyApproved by Miss Aniya • Sep 11, 2026

$\frac{x}{3} + 6 = - 12$

Quick Solution

Under Review

\begin{matrix} \frac{x}{3} + 6 & = - 12 \\ \frac{x}{3} & = - 18 \\ x & = - 18 \times 3 \\ x & = - 54 \end{matrix}

Solution: x = -54 on Real Linescale: 10 unit

Step by Step

Q1Part (iii)

MediumApproved by Miss Aniya • Sep 11, 2026

$\frac{x}{2} - \frac{3 x}{4} = \frac{1}{12}$

Quick Solution

Under Review

\begin{matrix} (×12) 6 x - 9 x & = 1 \\ - 3 x & = 1 \\ x & = - \frac{1}{3} \end{matrix}

Solution: x = -1/3 on Real Linescale: 1 unit

Step by Step

Q1Part (iv)

EasyApproved by Miss Aniya • Sep 11, 2026

$2 = 7 (2 x + 4) + 12 x$

Quick Solution

Under Review

\begin{matrix} 2 & = 14 x + 28 + 12 x \\ 2 & = 26 x + 28 \\ 26 x & = - 26 \\ x & = - 1 \end{matrix}

Solution: x = -1 on Real Linescale: 1 unit

Step by Step

Q1Part (v)

MediumApproved by Miss Aniya • Sep 11, 2026

$\frac{2 x - 1}{3} - \frac{3 x}{4} = \frac{5}{6}$

Quick Solution

Under Review

\begin{matrix} (×12) 4 (2 x - 1) - 9 x & = 10 \\ 8 x - 4 - 9 x & = 10 \\ - x & = 14 \\ x & = - 14 \end{matrix}

Solution: x = -14 on Real Linescale: 5 unit

Step by Step

Q1Part (vi)

MediumApproved by Miss Aniya • Sep 11, 2026

$\frac{- 5 x}{10} = 9 - \frac{10}{5} x$

Quick Solution

Under Review

\begin{matrix} - \frac{1}{2} x & = 9 - 2 x \\ 2 x - \frac{1}{2} x & = 9 \\ \frac{3 x}{2} & = 9 \\ x & = 6 \end{matrix}

Solution: x = 6 on Real Linescale: 1 unit

Step by Step

Question 2

MediumApproved by Miss Aniya • Sep 11, 2026

Solve each inequality and represent the solution on a real line.

Hint

When solving linear inequalities, remember that multiplying or dividing by a negative number reverses the inequality symbol. On the number line, a strict inequality $(<, >)$ uses an open circle, whereas $(\leq, \geq)$ uses a solid closed circle.

Solution

This question consists of subparts requiring solving linear equations/inequalities and representing their solutions on the real number line with number line diagrams.

Step by Step

Q2Part (i)

Easy

$x - 6 \leq - 2$

Quick Solution

Under Review

\begin{matrix} x - 6 & \leq - 2 \\ x & \leq - 2 + 6 \\ x & \leq 4 \\ Solution set & = (- \infty, 4] \end{matrix}

Inequality: x <= 4 on Real Linescale: 1 unit

Step by Step

Q2Part (ii)

Easy

$- 9 > - 16 + x$

Quick Solution

Under Review

\begin{matrix} - 9 & > - 16 + x \\ - 9 + 16 & > x \\ x & < 7 \\ Solution set & = (- \infty, 7) \end{matrix}

Inequality: x < 7 on Real Linescale: 1 unit

Step by Step

Q2Part (iii)

Easy

$3 + 2 x \geq 3$

Quick Solution

Under Review

\begin{matrix} 3 + 2 x & \geq 3 \\ 2 x & \geq 0 \\ x & \geq 0 \\ Solution set & = [0, \infty) \end{matrix}

Inequality: x >= 0 on Real Linescale: 1 unit

Step by Step

Q2Part (iv)

Easy

$6 (x + 10) \leq 0$

Quick Solution

Under Review

\begin{matrix} 6 (x + 10) & \leq 0 \\ x + 10 & \leq 0 \\ x & \leq - 10 \\ Solution set & = (- \infty, - 10] \end{matrix}

Inequality: x <= -10 on Real Linescale: 1 unit

Step by Step

Q2Part (v)

Medium

$\frac{5}{3} x - \frac{3}{4} < \frac{- 1}{12}$

Quick Solution

Under Review

\begin{matrix} (×12) 20 x - 9 & < - 1 \\ 20 x & < 8 \\ x & < \frac{2}{5} \\ Solution set & = (- \infty, \frac{2}{5}) \end{matrix}

Inequality: x < 2/5 on Real Linescale: 1 unit

Step by Step

Q2Part (vi)

Medium

$\frac{1}{4} x - \frac{1}{2} \leq - 1 + \frac{1}{2} x$

Quick Solution

Under Review

\begin{matrix} (×4) x - 2 & \leq - 4 + 2 x \\ x - 2 x & \leq - 4 + 2 \\ - x & \leq - 2 \\ x & \geq 2 \\ Solution set & = [2, \infty) \end{matrix}

Inequality: x >= 2 on Real Linescale: 1 unit

Step by Step

Solution Region of 2x + 1 >= 0Closed right half-plane containing origin O(0, 0)

Diagram Key & Elementsmobile-optimized

L1:2x + 1 = 0

O(0, 0)

A(-0.5, 0)

Closed right half-plane containing origin O(0, 0)

Step by Step

Q3Part (vi)

Easy

$3 y - 4 \leq 0$

Quick Solution

Under Review

\begin{matrix} 3 y - 4 & = 0 \\ y & = \frac{4}{3} (horizontal line) \\ Test O (0, 0) : - 4 & \leq 0 (True) \\ ⟹ & shade origin side, solid line \end{matrix}

Solution Region of 3y - 4 <= 0Closed lower half-plane containing origin O(0, 0)

Diagram Key & Elementsmobile-optimized

L1:3y - 4 = 0

O(0, 0)

B(0, 1.3333)

Closed lower half-plane containing origin O(0, 0)

Step by Step

Question 4

HardUnder Review

Indicate the solution region of the following linear inequalities by shading:

Hint

For a system of linear inequalities: (1) Graph the boundary line of each inequality and test the origin for each half-plane, (2) find the intersection point of the two boundary lines by solving them simultaneously, and (3) shade the common intersection region satisfying both inequalities.

Solution

This question consists of subparts requiring graphing a system of two linear inequalities in the Cartesian $x y$ -plane and shading their common solution region.

Step by Step

Q4Part (i)

Hard

$2 x - 3 y \leq 6, 2 x + 3 y \leq 12$

Quick Solution

Under Review

\begin{matrix} L_{1} : 2 x - 3 y & = 6 A (3, 0), B (0, - 2) \\ L_{2} : 2 x + 3 y & = 12 C (6, 0), D (0, 4) \\ Test O : 0 & \leq 6 (True) \\ 0 & \leq 12 (True) \\ Add: 4 x & = 18 \\ x & = \frac{9}{2} \\ 9 + 3 y & = 12 \\ y & = 1 \\ ⟹ & E (4.5, 1) \end{matrix}

System: 2x - 3y <= 6 and 2x + 3y <= 12Intersection of two closed half-planes

Figure: The overlapping shaded region containing origin O(0,0) represents the solution set, with intersection vertex E(4.5, 1).

Diagram Key & Elementsmobile-optimized

L1:2x - 3y = 6

L2:2x + 3y = 12

A(3, 0)

B(0, -2)

C(6, 0)

D(0, 4)

EApex(4.5, 1)

O(0, 0)

Intersection of two closed half-planes

Step by Step

Q4Part (ii)

Hard

$x + y \geq 5, - y + x \leq 1$

Quick Solution

Under Review

\begin{matrix} L_{1} : x + y & = 5 A (5, 0), B (0, 5) \\ L_{2} : x - y & = 1 C (1, 0), D (0, - 1) \\ Test O : 0 & \geq 5 (False) \\ 0 & \leq 1 (True) \\ Add: 2 x & = 6 \\ x & = 3 \\ 3 + y & = 5 \\ y & = 2 \\ ⟹ & E (3, 2) \end{matrix}

System: x + y >= 5 and x - y <= 1Unbounded upper solution region with apex E(3, 2)

Figure: Unbounded common region lying above both boundary lines x + y = 5 and x - y = 1 with apex vertex E(3, 2).

Diagram Key & Elementsmobile-optimized

L1:x + y = 5

L2:x - y = 1

A(5, 0)

B(0, 5)

C(1, 0)

D(0, -1)

EApex(3, 2)

O(0, 0)

Unbounded upper solution region with apex E(3, 2)

Step by Step

Q4Part (iii)

Hard

$3 x + 7 y \geq 21, x - y \leq 2$

Quick Solution

Under Review

\begin{matrix} L_{1} : 3 x + 7 y & = 21 A (7, 0), B (0, 3) \\ L_{2} : x - y & = 2 C (2, 0), D (0, - 2) \\ Test O : 0 & \geq 21 (False) \\ 0 & \leq 2 (True) \\ x & = y + 2 \\ 3 (y + 2) + 7 y & = 21 \\ 10 y & = 15 \\ y & = \frac{3}{2} \\ x & = \frac{7}{2} \\ ⟹ & E (3.5, 1.5) \end{matrix}

System: 3x + 7y >= 21 and x - y <= 2Common solution region above both boundary lines

Figure: The solution region is shaded above both boundary lines with intersection point E(3.5, 1.5).

Diagram Key & Elementsmobile-optimized

L1:3x + 7y = 21

L2:x - y = 2

A(7, 0)

B(0, 3)

C(2, 0)

D(0, -2)

EApex(3.5, 1.5)

O(0, 0)

Common solution region above both boundary lines

Step by Step

Q4Part (iv)

Hard

$4 x - 3 y \leq 12, x \geq - \frac{3}{2}$

Quick Solution

Under Review

\begin{matrix} L_{1} : 4 x - 3 y & = 12 A (3, 0), B (0, - 4) \\ L_{2} : x & = - \frac{3}{2} C (- 1.5, 0) \\ Test O : 0 & \leq 12 (True) \\ 0 & \geq - \frac{3}{2} (True) \\ 4 (- \frac{3}{2}) - 3 y & = 12 \\ - 3 y & = 18 \\ y & = - 6 \\ ⟹ & E (- 1.5, - 6) \end{matrix}

System: 4x - 3y <= 12 and x >= -3/2Region to the right of x = -1.5 and above 4x - 3y = 12

Figure: The solution region lies to the right of vertical line x = -1.5 and contains origin O(0,0), meeting at vertex E(-1.5, -6).

Diagram Key & Elementsmobile-optimized

L1:4x - 3y = 12

L2:x = -1.5

A(3, 0)

B(0, -4)

C(-1.5, 0)

EApex(-1.5, -6)

O(0, 0)

Region to the right of x = -1.5 and above 4x - 3y = 12

Step by Step

Q4Part (v)

Hard

$3 x + 7 y \geq 21, y \leq 4$

Quick Solution

Under Review

\begin{matrix} L_{1} : 3 x + 7 y & = 21 A (7, 0), B (0, 3) \\ L_{2} : y & = 4 C (0, 4) \\ Test O : 0 & \geq 21 (False) \\ 0 & \leq 4 (True) \\ 3 x + 28 & = 21 \\ 3 x & = - 7 \\ x & = - \frac{7}{3} \\ ⟹ & E (- \frac{7}{3}, 4) \end{matrix}

System: 3x + 7y >= 21 and y <= 4Region below y = 4 and above 3x + 7y = 21

Figure: Infinite wedge region bounded above by horizontal line y = 4 and below by 3x + 7y = 21, meeting at apex E(-7/3, 4).

Diagram Key & Elementsmobile-optimized

L1:3x + 7y = 21

L2:y = 4

A(7, 0)

B(0, 3)

C(0, 4)

EApex(-2.33, 4)

O(0, 0)

Region below y = 4 and above 3x + 7y = 21

Step by Step

Q4Part (vi)

Hard

$5 x + 7 y \leq 35, x - 2 y \leq 2$

Quick Solution

Under Review

\begin{matrix} L_{1} : 5 x + 7 y & = 35 A (7, 0), B (0, 5) \\ L_{2} : x - 2 y & = 2 C (2, 0), D (0, - 1) \\ Test O : 0 & \leq 35 (True) \\ 0 & \leq 2 (True) \\ x & = 2 y + 2 \\ 5 (2 y + 2) + 7 y & = 35 \\ 17 y & = 25 \\ y & = \frac{25}{17} \\ x & = \frac{84}{17} \\ ⟹ & E (\frac{84}{17}, \frac{25}{17}) \end{matrix}

System: 5x + 7y <= 35 and x - 2y <= 2Common solution region containing origin O(0, 0)

Figure: The common region containing origin O(0,0) bounded by lines 5x + 7y = 35 and x - 2y = 2 meeting at E(4.94, 1.47).

Diagram Key & Elementsmobile-optimized

L1:5x + 7y = 35

L2:x - 2y = 2

A(7, 0)

B(0, 5)

C(2, 0)

D(0, -1)

EApex(4.94, 1.47)

O(0, 0)

Common solution region containing origin O(0, 0)

Step by Step

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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