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

Every step explained, with hints and the method used.

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.

Quick Solution

Q1Part (i)

EasyApproved by Miss Aniya • Sep 11, 2026

$12 x + 30 = - 6$

Hint

Subtract $30$ from both sides, then divide by $12$ to obtain $x = - 3$ .

Solution

Isolate the variable term.

Subtract $30$ from both sides:
$\begin{matrix} 12 x + 30 & = - 6 \\ 12 x & = - 6 - 30 \\ 12 x & = - 36 \end{matrix}$
Solve for $x$ .

Divide both sides by $12$ :
$\begin{matrix} x & = \frac{- 36}{12} \\ x & = - 3 \end{matrix}$
Check the solution.

Substitute $x = - 3$ into the original equation:
$\begin{matrix} L.H.S. & = 12 (- 3) + 30 \\ = - 36 + 30 \\ = - 6 \\ = R.H.S. (True) \end{matrix}$
Representation on the real number line.

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

Answer

$x = - 3 or Solution Set = {- 3}$

Quick Solution

Q1Part (ii)

EasyApproved by Miss Aniya • Sep 11, 2026

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

Hint

Subtract $6$ to get $\frac{x}{3} = - 18$ , then multiply both sides by $3$ .

Solution

Isolate the fractional term.

Subtract $6$ from both sides:
$\begin{matrix} \frac{x}{3} + 6 & = - 12 \\ \frac{x}{3} & = - 12 - 6 \\ \frac{x}{3} & = - 18 \end{matrix}$
Solve for $x$ .

Multiply both sides by $3$ :
$\begin{matrix} x & = - 18 \times 3 \\ x & = - 54 \end{matrix}$
Check the solution.

Substitute $x = - 54$ into the original equation:
$\begin{matrix} L.H.S. & = \frac{- 54}{3} + 6 \\ = - 18 + 6 \\ = - 12 \\ = R.H.S. (True) \end{matrix}$
Representation on the real number line.

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

Answer

$x = - 54 or Solution Set = {- 54}$

Quick Solution

Q1Part (iii)

MediumApproved by Miss Aniya • Sep 11, 2026

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

Hint

Multiply the entire equation by the L.C.M. $12$ to clear all fractional denominators.

Solution

Eliminate denominators using the L.C.M.

The L.C.M. of the denominators $2, 4, 12$ is $12$ . Multiply each term by $12$ :
$\begin{matrix} 12 (\frac{x}{2}) - 12 (\frac{3 x}{4}) & = 12 (\frac{1}{12}) \\ 6 x - 3 (3 x) & = 1 \\ 6 x - 9 x & = 1 \\ - 3 x & = 1 \end{matrix}$
Solve for $x$ .

Divide both sides by $- 3$ :
$\begin{matrix} x & = - \frac{1}{3} \end{matrix}$
Check the solution.

Substitute $x = - \frac{1}{3}$ into the original equation:
$\begin{matrix} L.H.S. & = \frac{- \frac{1}{3}}{2} - \frac{3 (- \frac{1}{3})}{4} \\ = - \frac{1}{6} - \frac{- 1}{4} \\ = - \frac{1}{6} + \frac{1}{4} \\ = \frac{- 2 + 3}{12} \\ = \frac{1}{12} \\ = R.H.S. (True) \end{matrix}$
Representation on the real number line.

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

Answer

$x = - \frac{1}{3} or Solution Set = {- \frac{1}{3}}$

Quick Solution

Q1Part (iv)

EasyApproved by Miss Aniya • Sep 11, 2026

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

Hint

Distribute $7 (2 x + 4) = 14 x + 28$ , then group like terms: $14 x + 12 x = 26 x$ .

Solution

Expand parentheses and combine like terms.
$\begin{matrix} 2 & = 7 (2 x) + 7 (4) + 12 x \\ 2 & = 14 x + 28 + 12 x \\ 2 & = 26 x + 28 \end{matrix}$
Isolate the variable term.

Subtract $28$ from both sides:
$\begin{matrix} 2 - 28 & = 26 x \\ - 26 & = 26 x \\ 26 x & = - 26 \end{matrix}$
Solve for $x$ .

Divide by $26$ :
$\begin{matrix} x & = \frac{- 26}{26} \\ x & = - 1 \end{matrix}$
Check the solution.

Substitute $x = - 1$ into the right-hand side:
$\begin{matrix} R.H.S. & = 7 [2 (- 1) + 4] + 12 (- 1) \\ = 7 [- 2 + 4] - 12 \\ = 7 (2) - 12 \\ = 14 - 12 \\ = 2 \\ = L.H.S. (True) \end{matrix}$
Representation on the real number line.

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

Answer

$x = - 1 or Solution Set = {- 1}$

Quick Solution

Q1Part (v)

MediumApproved by Miss Aniya • Sep 11, 2026

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

Hint

Multiply by L.C.M. $12$ : $4 (2 x - 1) - 9 x = 10 ⟹ 8 x - 4 - 9 x = 10$ .

Solution

Clear denominators using the L.C.M.

The L.C.M. of the denominators $3, 4, 6$ is $12$ . Multiply every term by $12$ :
$\begin{matrix} 12 (\frac{2 x - 1}{3}) - 12 (\frac{3 x}{4}) & = 12 (\frac{5}{6}) \\ 4 (2 x - 1) - 3 (3 x) & = 2 (5) \\ 8 x - 4 - 9 x & = 10 \\ - x - 4 & = 10 \end{matrix}$
Solve for $x$ .

Add $4$ to both sides and multiply by $- 1$ :
$\begin{matrix} - x & = 10 + 4 \\ - x & = 14 \\ x & = - 14 \end{matrix}$
Check the solution.

Substitute $x = - 14$ into the original equation:
$\begin{matrix} L.H.S. & = \frac{2 (- 14) - 1}{3} - \frac{3 (- 14)}{4} \\ = \frac{- 28 - 1}{3} - \frac{- 42}{4} \\ = \frac{- 29}{3} + \frac{21}{2} \\ = \frac{- 58 + 63}{6} \\ = \frac{5}{6} \\ = R.H.S. (True) \end{matrix}$
Representation on the real number line.

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

Answer

$x = - 14 or Solution Set = {- 14}$

Quick Solution

Q1Part (vi)

MediumApproved by Miss Aniya • Sep 11, 2026

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

Hint

Simplify $\frac{- 5 x}{10} = - \frac{x}{2}$ and $\frac{10 x}{5} = 2 x$ , then combine like terms: $2 x - \frac{x}{2} = \frac{3 x}{2} = 9$ .

Solution

Simplify fractions.

Reduce each fractional coefficient:
$\begin{matrix} \frac{- 5}{10} & = - \frac{1}{2} \\ \frac{10}{5} & = 2 \end{matrix}$
So the equation becomes:
$\begin{matrix} - \frac{1}{2} x & = 9 - 2 x \end{matrix}$
Collect variable terms on one side.

Add $2 x$ to both sides:
$\begin{matrix} 2 x - \frac{1}{2} x & = 9 \\ \frac{4 x - x}{2} & = 9 \\ \frac{3 x}{2} & = 9 \end{matrix}$
Solve for $x$ .

Multiply both sides by $\frac{2}{3}$ :
$\begin{matrix} x & = 9 \times \frac{2}{3} \\ x & = 3 \times 2 \\ x & = 6 \end{matrix}$
Check the solution.

Substitute $x = 6$ :
$\begin{matrix} L.H.S. & = \frac{- 5 (6)}{10} = \frac{- 30}{10} = - 3 \\ R.H.S. & = 9 - \frac{10}{5} (6) = 9 - 2 (6) = 9 - 12 = - 3 \\ L.H.S. & = R.H.S. (True) \end{matrix}$
Representation on the real number line.

Solution: x = 6 on Real Linescale: 1 unit

Answer

$x = 6 or Solution Set = {6}$

Quick Solution

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.

Quick Solution

Q2Part (i)

EasyUnder Review

$x - 6 \leq - 2$

Hint

Add $6$ to both sides to get $x \leq 4$ . Since the inequality is non-strict ( $\leq$ ), the endpoint $4$ is included (solid dot).

Solution

Isolate $x$ .

Add $6$ to both sides:
$\begin{matrix} x - 6 & \leq - 2 \\ x & \leq - 2 + 6 \\ x & \leq 4 \end{matrix}$
Express in set and interval notation.
- Set-builder notation: ${x | x \leq 4, x \in ℝ}$
- Interval notation: $(- \infty, 4]$
Representation on the real number line.

The solution consists of all real numbers less than or equal to $4$ . On the number line, a solid dot is placed at $x = 4$ with the shaded ray extending infinitely to the left.

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

Answer

$x \leq 4 or Solution Set = {x | x \leq 4} = (- \infty, 4]$

Quick Solution

Q2Part (ii)

EasyUnder Review

$- 9 > - 16 + x$

Hint

Add $16$ to both sides: $7 > x ⟺ x < 7$ . The strict inequality $<$ uses an open circle at $7$ .

Solution

Isolate $x$ .

Add $16$ to both sides:
$\begin{matrix} - 9 & > - 16 + x \\ - 9 + 16 & > x \\ 7 & > x \\ x & < 7 \end{matrix}$
Express in set and interval notation.
- Set-builder notation: ${x | x < 7, x \in ℝ}$
- Interval notation: $(- \infty, 7)$
Representation on the real number line.

A hollow/open circle is placed at $x = 7$ (since $7$ is excluded) with the ray extending indefinitely to the left.

Inequality: x < 7 on Real Linescale: 1 unit

Answer

$x < 7 or Solution Set = {x | x < 7} = (- \infty, 7)$

Quick Solution

Q2Part (iii)

EasyUnder Review

$3 + 2 x \geq 3$

Hint

Subtract $3$ from both sides: $2 x \geq 0 ⟹ x \geq 0$ .

Solution

Isolate the variable term.

Subtract $3$ from both sides:
$\begin{matrix} 3 + 2 x & \geq 3 \\ 2 x & \geq 3 - 3 \\ 2 x & \geq 0 \end{matrix}$
Solve for $x$ .

Divide both sides by $2$ :
$\begin{matrix} x & \geq 0 \end{matrix}$
Express in set and interval notation.
- Set-builder notation: ${x | x \geq 0, x \in ℝ}$
- Interval notation: $[0, \infty)$
Representation on the real number line.

A solid dot is placed at the origin $x = 0$ with the shaded ray extending to the right.

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

Answer

$x \geq 0 or Solution Set = {x | x \geq 0} = [0, \infty)$

Quick Solution

Q2Part (iv)

EasyUnder Review

$6 (x + 10) \leq 0$

Hint

Divide by $6$ to obtain $x + 10 \leq 0 ⟹ x \leq - 10$ .

Solution

Divide both sides by 6.

Since $6 > 0$ , the inequality direction does not change:
$\begin{matrix} 6 (x + 10) & \leq 0 \\ x + 10 & \leq \frac{0}{6} \\ x + 10 & \leq 0 \end{matrix}$
Solve for $x$ .

Subtract $10$ from both sides:
$\begin{matrix} x & \leq - 10 \end{matrix}$
Express in set and interval notation.
- Set-builder notation: ${x | x \leq - 10, x \in ℝ}$
- Interval notation: $(- \infty, - 10]$
Representation on the real number line.

A solid dot is placed at $x = - 10$ with the shaded ray extending indefinitely to the left.

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

Answer

$x \leq - 10 or Solution Set = {x | x \leq - 10} = (- \infty, - 10]$

Quick Solution

Q2Part (v)

MediumUnder Review

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

Hint

Multiply through by $12$ : $20 x - 9 < - 1 ⟹ 20 x < 8 ⟹ x < \frac{2}{5}$ .

Solution

Clear denominators using the L.C.M.

Multiply all terms by the L.C.M. $12$ :
$\begin{matrix} 12 (\frac{5}{3} x) - 12 (\frac{3}{4}) & < 12 (\frac{- 1}{12}) \\ 4 (5 x) - 3 (3) & < - 1 \\ 20 x - 9 & < - 1 \end{matrix}$
Isolate $x$ .

Add $9$ to both sides:
$\begin{matrix} 20 x & < - 1 + 9 \\ 20 x & < 8 \end{matrix}$
Solve for $x$ .

Divide by $20$ and simplify:
$\begin{matrix} x & < \frac{8}{20} \\ x & < \frac{2}{5} (or x < 0.4) \end{matrix}$
Express in set and interval notation.
- Set-builder notation: ${x | x < \frac{2}{5}, x \in ℝ}$
- Interval notation: $(- \infty, \frac{2}{5})$
Representation on the real number line.

An open circle at $x = \frac{2}{5} = 0.4$ with the shaded ray extending to the left.

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

Answer

$x < \frac{2}{5} or Solution Set = {x | x < \frac{2}{5}} = (- \infty, \frac{2}{5})$

Quick Solution

Q2Part (vi)

MediumUnder Review

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

Hint

Multiply both sides by the least common multiple $4$ : $4 (\frac{1}{4} x - \frac{1}{2}) \leq 4 (- 1 + \frac{1}{2} x)$ , which gives $x - 2 \leq - 4 + 2 x$ . Rearranging yields $2 \leq x ⟺ x \geq 2$ . In interval notation, the solution set is $[2, \infty)$ . On the real line, plot a solid closed circle at $2$ with a ray directed toward $+ \infty$ .

Solution

Clear denominators using the least common multiple (L.C.M.).

The denominators are $4$ and $2$ . Their least common multiple is $L.C.M. (4, 2) = 4$ . Multiply each term on both sides by $4$ (since $4 > 0$ , the inequality direction is unchanged):
$\begin{matrix} 4 (\frac{1}{4} x - \frac{1}{2}) & \leq 4 (- 1 + \frac{1}{2} x) \\ 4 \cdot (\frac{1}{4} x) - 4 \cdot (\frac{1}{2}) & \leq 4 (- 1) + 4 \cdot (\frac{1}{2} x) \\ x - 2 & \leq - 4 + 2 x \end{matrix}$
Collect the variable terms on one side.

Subtract $2 x$ from both sides and add $2$ to both sides:
$\begin{matrix} x - 2 x & \leq - 4 + 2 \\ - x & \leq - 2 \end{matrix}$
Divide both sides by $- 1$ . Dividing by a negative number reverses the inequality sign from $\leq$ to $\geq$ :
$\begin{matrix} \frac{- x}{- 1} & \geq \frac{- 2}{- 1} \\ x & \geq 2 \end{matrix}$
Alternative rearrangement (keeping coefficients positive):
$\begin{matrix} x - 2 & \leq - 4 + 2 x \\ - 2 + 4 & \leq 2 x - x \\ 2 & \leq x \\ x & \geq 2 \end{matrix}$
Express the solution set in standard notations.
- Inequality form:
x≥2or2≤x<∞
- Set-builder notation:
{x|x≥2,x∈ℝ}
- Interval notation:
[2,∞)
Verification check.
1. Boundary check at $x = 2$ :
L.H.S.=14(2)−12=12−12=0R.H.S.=−1+12(2)=−1+1=00≤0(True, boundary included)
1. Interior test point at $x = 4$ :
L.H.S.=14(4)−12=1−12=12R.H.S.=−1+12(4)=−1+2=112≤1(True)
1. Exterior test point at $x = 0$ :
L.H.S.=14(0)−12=−12R.H.S.=−1+12(0)=−1−12≤−1(False, −0.5>−1)
Thus, the solution $x \geq 2$ is fully verified.
Representation on the real number line.

Since the inequality includes the endpoint $x = 2$ ( $\geq$ ), a solid closed circle is plotted at $x = 2$ , and a shaded ray extends indefinitely to the right toward $+ \infty$ .

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

Answer

$x \geq 2 or 2 \leq x < \infty or Solution Set = {x | x \geq 2, x \in ℝ} = [2, \infty)$

Quick Solution

Question 3

MediumUnder Review

Shade the solution region for the following linear inequalities in $x y$ -plane:

Hint

To graph a 2D linear inequality: (1) Write its associated equation $a x + b y = c$ , (2) find the $x$ -intercept and $y$ -intercept, (3) draw a solid line for $\leq, \geq$ or dashed line for $<, >$ , and (4) test the origin $O (0, 0)$ to determine which half-plane to shade.

Solution

This question consists of subparts requiring graphing a linear inequality in two variables and shading its solution region (half-plane) in the Cartesian $x y$ -plane.

Quick Solution

Q3Part (i)

MediumUnder Review

$2 x + y \leq 6$

Hint

Form the associated equation $2 x + y = 6$ . Find the intercepts $A (3, 0)$ and $B (0, 6)$ . Draw a solid boundary line through these points. Test the origin $O (0, 0)$ : $2 (0) + 0 \leq 6 ⟹ 0 \leq 6$ , which is true. Therefore, shade the half-plane containing the origin.

Solution

Associated equation and coordinate intercepts.

The associated linear boundary equation is:
2x+y=6
1. Find the $x$ -intercept: Substitute $y = 0$ into the associated equation:
2x+(0)=62x=6x=62x=3
Thus, the boundary line intersects the $x$ -axis at $A (3, 0)$ .
1. Find the $y$ -intercept: Substitute $x = 0$ into the associated equation:
2(0)+y=60+y=6y=6
Thus, the boundary line intersects the $y$ -axis at $B (0, 6)$ .
Drawing the boundary line.

Plot the points $A (3, 0)$ and $B (0, 6)$ on the Cartesian coordinate plane. Since the inequality symbol is $\leq$ (non-strict), draw a solid line passing through $A$ and $B$ (the boundary points satisfy the inequality).
Test point check at the origin $O (0, 0)$ .

Since the boundary line does not pass through the origin $(0, 0)$ , we use $O (0, 0)$ as our test point: Substitute $x = 0$ and $y = 0$ into the given inequality:
$\begin{matrix} 2 (0) + (0) & \leq 6 \\ 0 + 0 & \leq 6 \\ 0 & \leq 6 (True) \end{matrix}$
Conclusion and shaded solution region.

Because the origin satisfies the inequality, the solution region is the closed lower-left half-plane containing the origin $O (0, 0)$ . For $2 x + y \leq 6$ , the solid boundary line is included in the solution set.

Solution Region of 2x + y <= 6Closed half-plane containing origin O(0, 0)
Diagram Key & Elementsmobile-optimized
L1:2x + y = 6
O(0, 0)
A(3, 0)
B(0, 6)
Closed half-plane containing origin O(0, 0)

Answer

Associated equation: $2 x + y = 6$ Intercepts: $A (3, 0) and B (0, 6)$ Boundary line: Solid straight line through $A (3, 0)$ and $B (0, 6)$ . Solution region: Closed half-plane containing the origin $O (0, 0)$ (the lower-left region towards the origin).

Quick Solution

Q3Part (ii)

MediumUnder Review

$3 x + 7 y \geq 21$

Hint

Form the associated equation $3 x + 7 y = 21$ . Find the intercepts $A (7, 0)$ and $B (0, 3)$ . Draw a solid boundary line through these points. Test the origin $O (0, 0)$ : $3 (0) + 7 (0) \geq 21 ⟹ 0 \geq 21$ , which is false. Therefore, shade the half-plane not containing the origin (the side opposite to $O (0, 0)$ ).

Solution

Associated equation and coordinate intercepts.

The associated linear boundary equation is:
3x+7y=21
1. Find the $x$ -intercept: Substitute $y = 0$ into the associated equation:
3x+7(0)=213x+0=213x=21x=213x=7
Thus, the boundary line intersects the $x$ -axis at $A (7, 0)$ .
1. Find the $y$ -intercept: Substitute $x = 0$ into the associated equation:
3(0)+7y=210+7y=217y=21y=217y=3
Thus, the boundary line intersects the $y$ -axis at $B (0, 3)$ .
Drawing the boundary line.

Plot the points $A (7, 0)$ and $B (0, 3)$ on the coordinate plane. Since the inequality symbol is $\geq$ (non-strict), join these points with a solid continuous line.
Test point check at the origin $O (0, 0)$ .

Substitute $x = 0$ and $y = 0$ into $3 x + 7 y \geq 21$ :
$\begin{matrix} 3 (0) + 7 (0) & \geq 21 \\ 0 + 0 & \geq 21 \\ 0 & \geq 21 (False) \end{matrix}$
Conclusion and shaded solution region.

Because $0 \geq 21$ is false, the origin $(0, 0)$ does not satisfy the inequality. Therefore, the solution region is the closed upper-right half-plane on the opposite side of the origin (away from $O (0, 0)$ ), including the boundary line $3 x + 7 y = 21$ .

Solution Region of 3x + 7y >= 21Closed half-plane not containing origin O(0, 0)
Diagram Key & Elementsmobile-optimized
L1:3x + 7y = 21
O(0, 0)
A(7, 0)
B(0, 3)
Closed half-plane not containing origin O(0, 0)

Answer

Associated equation: $3 x + 7 y = 21$ Intercepts: $A (7, 0) and B (0, 3)$ Boundary line: Solid straight line through $A (7, 0)$ and $B (0, 3)$ . Solution region: Closed half-plane not containing the origin $O (0, 0)$ (upper-right region opposite to the origin).

Quick Solution

Q3Part (iii)

MediumUnder Review

$3 x - 2 y \geq 6$

Hint

Form the associated equation $3 x - 2 y = 6$ . Find the intercepts $A (2, 0)$ and $B (0, - 3)$ . Draw a solid boundary line through these points. Test the origin $O (0, 0)$ : $3 (0) - 2 (0) \geq 6 ⟹ 0 \geq 6$ , which is false. Therefore, shade the half-plane on the opposite side of the origin (the lower-right half-plane away from $O (0, 0)$ ).

Solution

Associated equation and coordinate intercepts.

The associated linear boundary equation is:
3x−2y=6
1. Find the $x$ -intercept: Substitute $y = 0$ into the associated equation:
3x−2(0)=63x−0=63x=6x=63x=2
Thus, the boundary line intersects the $x$ -axis at $A (2, 0)$ .
1. Find the $y$ -intercept: Substitute $x = 0$ into the associated equation:
3(0)−2y=60−2y=6−2y=6y=6−2y=−3
Thus, the boundary line intersects the $y$ -axis at $B (0, - 3)$ .
Drawing the boundary line.

Plot $A (2, 0)$ and $B (0, - 3)$ on the coordinate plane. Because the inequality symbol is $\geq$ (non-strict), connect the points with a solid continuous line.
Test point check at the origin $O (0, 0)$ .

Substitute $x = 0$ and $y = 0$ into $3 x - 2 y \geq 6$ :
$\begin{matrix} 3 (0) - 2 (0) & \geq 6 \\ 0 - 0 & \geq 6 \\ 0 & \geq 6 (False) \end{matrix}$
Conclusion and shaded solution region.

Because $0 \geq 6$ is false, the origin $(0, 0)$ is not part of the solution set. Therefore, the solution region is the closed lower-right half-plane on the non-origin side of the boundary line $3 x - 2 y = 6$ (the region below and to the right of the line, including the boundary line itself).

(Note: In slope-intercept form, $3 x - 2 y \geq 6 ⟺ - 2 y \geq - 3 x + 6 ⟺ y \leq \frac{3}{2} x - 3$ , which defines the region below the line.)

Solution Region of 3x - 2y >= 6Closed half-plane not containing origin O(0, 0)
Diagram Key & Elementsmobile-optimized
L1:3x - 2y = 6
O(0, 0)
A(2, 0)
B(0, -3)
Closed half-plane not containing origin O(0, 0)

Answer

Associated equation: $3 x - 2 y = 6$ Intercepts: $A (2, 0) and B (0, - 3)$ Boundary line: Solid straight line through $A (2, 0)$ and $B (0, - 3)$ . Solution region: Closed half-plane not containing the origin $O (0, 0)$ (lower-right region opposite to the origin).

Quick Solution

Q3Part (iv)

MediumUnder Review

$5 x - 4 y \leq 20$

Hint

Form the associated equation $5 x - 4 y = 20$ . Find the intercepts $A (4, 0)$ and $B (0, - 5)$ . Draw a solid boundary line through these points. Test the origin $O (0, 0)$ : $5 (0) - 4 (0) \leq 20 ⟹ 0 \leq 20$ , which is true. Therefore, shade the half-plane containing the origin $O (0, 0)$ (the upper-left half-plane).

Solution

Associated equation and coordinate intercepts.

The associated linear boundary equation is:
5x−4y=20
1. Find the $x$ -intercept: Substitute $y = 0$ into the associated equation:
5x−4(0)=205x−0=205x=20x=205x=4
Thus, the boundary line intersects the $x$ -axis at $A (4, 0)$ .
1. Find the $y$ -intercept: Substitute $x = 0$ into the associated equation:
5(0)−4y=200−4y=20−4y=20y=20−4y=−5
Thus, the boundary line intersects the $y$ -axis at $B (0, - 5)$ .
Drawing the boundary line.

Plot $A (4, 0)$ and $B (0, - 5)$ on the coordinate plane. Because the inequality symbol is $\leq$ (non-strict), connect them with a solid continuous line.
Test point check at the origin $O (0, 0)$ .

Substitute $x = 0$ and $y = 0$ into $5 x - 4 y \leq 20$ :
$\begin{matrix} 5 (0) - 4 (0) & \leq 20 \\ 0 - 0 & \leq 20 \\ 0 & \leq 20 (True) \end{matrix}$
Conclusion and shaded solution region.

Because $0 \leq 20$ is true, the origin $(0, 0)$ satisfies the inequality. Therefore, the solution region is the closed upper-left half-plane containing the origin $O (0, 0)$ , including the solid boundary line $5 x - 4 y = 20$ .

(Note: In slope-intercept form, $5 x - 4 y \leq 20 ⟺ - 4 y \leq - 5 x + 20 ⟺ y \geq \frac{5}{4} x - 5$ , which defines the region above the line.)

Solution Region of 5x - 4y <= 20Closed half-plane containing origin O(0, 0)
Diagram Key & Elementsmobile-optimized
L1:5x - 4y = 20
O(0, 0)
A(4, 0)
B(0, -5)
Closed half-plane containing origin O(0, 0)

Answer

Associated equation: $5 x - 4 y = 20$ Intercepts: $A (4, 0) and B (0, - 5)$ Boundary line: Solid straight line through $A (4, 0)$ and $B (0, - 5)$ . Solution region: Closed half-plane containing the origin $O (0, 0)$ (upper-left region towards the origin).

Quick Solution

Q3Part (v)

EasyUnder Review

$2 x + 1 \geq 0$

Hint

Form the associated equation $2 x + 1 = 0$ , which gives the vertical line $x = - \frac{1}{2}$ passing through $A (- \frac{1}{2}, 0)$ . Draw a solid vertical line. Test the origin $O (0, 0)$ : $2 (0) + 1 \geq 0 ⟹ 1 \geq 0$ , which is true. Therefore, shade the half-plane containing the origin (the right half-plane where $x \geq - \frac{1}{2}$ ).

Solution

Associated equation.

Set the inequality to equality:
$\begin{matrix} 2 x + 1 & = 0 \\ 2 x & = - 1 \\ x & = - \frac{1}{2} \end{matrix}$
Since there is no $y$ -variable, $x = - \frac{1}{2}$ (or $x = - 0.5$ ) represents a vertical line parallel to the $y$ -axis intersecting the $x$ -axis at $A (- \frac{1}{2}, 0)$ .
Drawing the boundary line.

Since the inequality symbol is $\geq$ (non-strict), draw a solid vertical line passing through $x = - \frac{1}{2}$ (boundary included).
Test point check at the origin $O (0, 0)$ .

Substitute $x = 0$ into $2 x + 1 \geq 0$ :
$\begin{matrix} 2 (0) + 1 & \geq 0 \\ 0 + 1 & \geq 0 \\ 1 & \geq 0 (True) \end{matrix}$
Conclusion and shaded solution region.

Because $1 \geq 0$ is true, the origin $(0, 0)$ belongs to the solution region. Therefore, the solution region is the closed right half-plane on the origin side of the vertical line $x = - \frac{1}{2}$ (consisting of all points $(x, y)$ with $x \geq - \frac{1}{2}$ ).

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)

Answer

Associated equation: $2 x + 1 = 0 ⟹ x = - \frac{1}{2}$ Intercept: $A (- \frac{1}{2}, 0)$ Boundary line: Solid vertical line $x = - \frac{1}{2}$ parallel to the $y$ -axis. Solution region: Closed right half-plane containing the origin $O (0, 0)$ (all points where $x \geq - \frac{1}{2}$ ).

Quick Solution

Q3Part (vi)

EasyUnder Review

$3 y - 4 \leq 0$

Hint

Form the associated equation $3 y - 4 = 0$ , which gives the horizontal line $y = \frac{4}{3}$ passing through $B (0, \frac{4}{3})$ . Draw a solid horizontal line. Test the origin $O (0, 0)$ : $3 (0) - 4 \leq 0 ⟹ - 4 \leq 0$ , which is true. Therefore, shade the half-plane containing the origin (the lower half-plane where $y \leq \frac{4}{3}$ ).

Solution

Associated equation.

Set the inequality to equality:
$\begin{matrix} 3 y - 4 & = 0 \\ 3 y & = 4 \\ y & = \frac{4}{3} \end{matrix}$
Since there is no $x$ -variable, $y = \frac{4}{3} = 1 \frac{1}{3} \approx 1.33$ represents a horizontal line parallel to the $x$ -axis intersecting the $y$ -axis at $B (0, \frac{4}{3})$ .
Drawing the boundary line.

Draw a solid continuous horizontal line at $y = \frac{4}{3}$ (since the inequality symbol is $\leq$ ).
Test point check at the origin $O (0, 0)$ .

Substitute $y = 0$ into $3 y - 4 \leq 0$ :
$\begin{matrix} 3 (0) - 4 & \leq 0 \\ 0 - 4 & \leq 0 \\ - 4 & \leq 0 (True) \end{matrix}$
Conclusion and shaded solution region.

Because $- 4 \leq 0$ is true, the origin $(0, 0)$ belongs to the solution region. Therefore, the solution region is the closed lower half-plane on the origin side of the horizontal line $y = \frac{4}{3}$ (consisting of all points $(x, y)$ with $y \leq \frac{4}{3}$ ), including the solid boundary line.

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)

Answer

Associated equation: $3 y - 4 = 0 ⟹ y = \frac{4}{3}$ Intercept: $B (0, \frac{4}{3})$ Boundary line: Solid horizontal line $y = \frac{4}{3}$ parallel to the $x$ -axis. Solution region: Closed lower half-plane containing the origin $O (0, 0)$ (all points where $y \leq \frac{4}{3}$ ).

Quick Solution

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.

Quick Solution

Q4Part (i)

HardUnder Review

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

Hint

For the system $2 x - 3 y \leq 6$ and $2 x + 3 y \leq 12$ : graph boundary line $L_{1}$ through $A (3, 0)$ and $B (0, - 2)$ , and boundary line $L_{2}$ through $C (6, 0)$ and $D (0, 4)$ . Origin $O (0, 0)$ satisfies both inequalities. Solving simultaneously gives intersection point $E (\frac{9}{2}, 1) = E (4.5, 1)$ . The solution region is the common half-plane containing $O (0, 0)$ .

Solution

Associated equations and intercepts.

For the first inequality $2 x - 3 y \leq 6$ , the associated linear equation $L_{1}$ is:
2x−3y=6
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 2 x - 3 (0) & = 6 \\ 2 x & = 6 \\ x & = 3 ⟹ A (3, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 2 (0) - 3 y & = 6 \\ - 3 y & = 6 \\ y & = - 2 ⟹ B (0, - 2) \end{matrix}$
For the second inequality $2 x + 3 y \leq 12$ , the associated linear equation $L_{2}$ is:
2x+3y=12
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 2 x + 3 (0) & = 12 \\ 2 x & = 12 \\ x & = 6 ⟹ C (6, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 2 (0) + 3 y & = 12 \\ 3 y & = 12 \\ y & = 4 ⟹ D (0, 4) \end{matrix}$
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $2 x - 3 y \leq 6$ : $\begin{matrix} 2 (0) - 3 (0) & \leq 6 \\ 0 & \leq 6 (True) \end{matrix}$ The solution set of $2 x - 3 y \leq 6$ is the closed half-plane containing the origin $O (0, 0)$ (the region lying above and to the left of $L_{1}$ ).
- Substitute $(0, 0)$ into $2 x + 3 y \leq 12$ : $\begin{matrix} 2 (0) + 3 (0) & \leq 12 \\ 0 & \leq 12 (True) \end{matrix}$ The solution set of $2 x + 3 y \leq 12$ is the closed half-plane containing the origin $O (0, 0)$ (the region lying below and to the left of $L_{2}$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} 2 x - 3 y & = 6 — (1) \\ 2 x + 3 y & = 12 — (2) \end{matrix}$
Adding equation (1) and equation (2):
$\begin{matrix} (2 x - 3 y) + (2 x + 3 y) & = 6 + 12 \\ 4 x & = 18 \\ x & = \frac{18}{4} \\ x & = \frac{9}{2} = 4.5 \end{matrix}$
Substitute $x = \frac{9}{2}$ into equation (2):
$\begin{matrix} 2 (\frac{9}{2}) + 3 y & = 12 \\ 9 + 3 y & = 12 \\ 3 y & = 12 - 9 \\ 3 y & = 3 \\ y & = 1 \end{matrix}$
Thus, the boundary lines intersect at point $E (\frac{9}{2}, 1) = E (4.5, 1)$ .
Description of the common solution region.

The required solution region is the common intersection of both closed half-planes containing the origin $O (0, 0)$ . It forms an infinite wedge-like region bounded on the right by the apex vertex $E (4.5, 1)$ and the two solid boundary rays $2 x - 3 y = 6$ and $2 x + 3 y = 12$ , including all points on both boundary lines.
Graphical representation of the system.

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

Answer

Boundary Line $L_{1}$ ( $2 x - 3 y = 6$ ): Passes through $A (3, 0)$ and $B (0, - 2)$ ; solution half-plane contains $O (0, 0)$ .
Boundary Line $L_{2}$ ( $2 x + 3 y = 12$ ): Passes through $C (6, 0)$ and $D (0, 4)$ ; solution half-plane contains $O (0, 0)$ .
Point of Intersection: $E (\frac{9}{2}, 1) = E (4.5, 1)$ .
Solution Region: The common shaded region containing the origin $O (0, 0)$ , bounded on the right by vertex $E (4.5, 1)$ and the solid boundary lines $2 x - 3 y = 6$ and $2 x + 3 y = 12$ .

Quick Solution

Q4Part (ii)

HardUnder Review

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

Hint

For $x + y \geq 5$ and $x - y \leq 1$ : boundary line $L_{1}$ passes through $A (5, 0)$ and $B (0, 5)$ (shading away from $O (0, 0)$ ), and boundary line $L_{2}$ passes through $C (1, 0)$ and $D (0, - 1)$ (shading toward $O (0, 0)$ ). Solving simultaneously gives intersection apex $E (3, 2)$ . The solution region is the common region above both boundary lines.

Solution

Associated equations and intercepts.

Rewrite $- y + x \leq 1$ in standard form as $x - y \leq 1$ .

For the first inequality $x + y \geq 5$ , the associated linear equation $L_{1}$ is:
x+y=5
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} x + 0 & = 5 \\ x & = 5 ⟹ A (5, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 0 + y & = 5 \\ y & = 5 ⟹ B (0, 5) \end{matrix}$
For the second inequality $x - y \leq 1$ , the associated linear equation $L_{2}$ is:
x−y=1
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} x - 0 & = 1 \\ x & = 1 ⟹ C (1, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 0 - y & = 1 \\ - y & = 1 \\ y & = - 1 ⟹ D (0, - 1) \end{matrix}$
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $x + y \geq 5$ : $\begin{matrix} 0 + 0 & \geq 5 \\ 0 & \geq 5 (False) \end{matrix}$ The solution set of $x + y \geq 5$ is the closed half-plane not containing the origin $O (0, 0)$ (the region above and to the right of $L_{1}$ ).
- Substitute $(0, 0)$ into $x - y \leq 1$ : $\begin{matrix} 0 - 0 & \leq 1 \\ 0 & \leq 1 (True) \end{matrix}$ The solution set of $x - y \leq 1$ is the closed half-plane containing the origin $O (0, 0)$ (the region above and to the left of $L_{2}$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} x + y & = 5 — (1) \\ x - y & = 1 — (2) \end{matrix}$
Adding equation (1) and equation (2):
$\begin{matrix} (x + y) + (x - y) & = 5 + 1 \\ 2 x & = 6 \\ x & = \frac{6}{2} \\ x & = 3 \end{matrix}$
Substitute $x = 3$ into equation (1):
$\begin{matrix} 3 + y & = 5 \\ y & = 5 - 3 \\ y & = 2 \end{matrix}$
Thus, the boundary lines intersect at point $E (3, 2)$ .
Description of the common solution region.

The solution region is the unbounded region lying simultaneously above $x + y = 5$ and above $x - y = 1$ . It extends infinitely in the positive $y$ -direction with its lowest apex vertex at $E (3, 2)$ , including both solid boundary rays.
Graphical representation of the system.

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)

Answer

Boundary Line $L_{1}$ ( $x + y = 5$ ): Passes through $A (5, 0)$ and $B (0, 5)$ ; solution half-plane is on the non-origin side (above the line).
Boundary Line $L_{2}$ ( $x - y = 1$ ): Passes through $C (1, 0)$ and $D (0, - 1)$ ; solution half-plane is on the origin side (above the line).
Point of Intersection: $E (3, 2)$ .
Solution Region: The unbounded common region lying above both boundary lines $x + y = 5$ and $x - y = 1$ , extending upward with apex at $E (3, 2)$ .

Quick Solution

Q4Part (iii)

HardUnder Review

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

Hint

For $3 x + 7 y \geq 21$ and $x - y \leq 2$ : boundary line $L_{1}$ passes through $A (7, 0)$ and $B (0, 3)$ (origin test is false, shade above), and boundary line $L_{2}$ passes through $C (2, 0)$ and $D (0, - 2)$ (origin test is true, shade above). The lines intersect at $E (\frac{7}{2}, \frac{3}{2}) = E (3.5, 1.5)$ . Shade the region common to both half-planes.

Solution

Associated equations and intercepts.

For the first inequality $3 x + 7 y \geq 21$ , the associated linear equation $L_{1}$ is:
3x+7y=21
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 3 x + 7 (0) & = 21 \\ 3 x & = 21 \\ x & = 7 ⟹ A (7, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 3 (0) + 7 y & = 21 \\ 7 y & = 21 \\ y & = 3 ⟹ B (0, 3) \end{matrix}$
For the second inequality $x - y \leq 2$ , the associated linear equation $L_{2}$ is:
x−y=2
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} x - 0 & = 2 \\ x & = 2 ⟹ C (2, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 0 - y & = 2 \\ - y & = 2 \\ y & = - 2 ⟹ D (0, - 2) \end{matrix}$
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $3 x + 7 y \geq 21$ : $\begin{matrix} 3 (0) + 7 (0) & \geq 21 \\ 0 & \geq 21 (False) \end{matrix}$ The solution set of $3 x + 7 y \geq 21$ is the closed half-plane not containing the origin $O (0, 0)$ (the region above $L_{1}$ ).
- Substitute $(0, 0)$ into $x - y \leq 2$ : $\begin{matrix} 0 - 0 & \leq 2 \\ 0 & \leq 2 (True) \end{matrix}$ The solution set of $x - y \leq 2$ is the closed half-plane containing the origin $O (0, 0)$ (the region above and to the left of $L_{2}$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} 3 x + 7 y & = 21 — (1) \\ x - y & = 2 ⟹ x = y + 2 — (2) \end{matrix}$
Substitute $x = y + 2$ into equation (1):
$\begin{matrix} 3 (y + 2) + 7 y & = 21 \\ 3 y + 6 + 7 y & = 21 \\ 10 y + 6 & = 21 \\ 10 y & = 21 - 6 \\ 10 y & = 15 \\ y & = \frac{15}{10} \\ y & = \frac{3}{2} = 1.5 \end{matrix}$
Substitute $y = \frac{3}{2}$ into equation (2):
$\begin{matrix} x & = \frac{3}{2} + 2 \\ x & = \frac{3 + 4}{2} \\ x & = \frac{7}{2} = 3.5 \end{matrix}$
Thus, the boundary lines intersect at point $E (\frac{7}{2}, \frac{3}{2}) = E (3.5, 1.5)$ .
Description of the common solution region.

The solution region is the common region lying above both boundary lines $3 x + 7 y = 21$ and $x - y = 2$ . It extends indefinitely in the upper half-plane, bounded from below by the solid line segments meeting at vertex $E (3.5, 1.5)$ .
Graphical representation of the system.

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

Answer

Boundary Line $L_{1}$ ( $3 x + 7 y = 21$ ): Passes through $A (7, 0)$ and $B (0, 3)$ ; solution half-plane is on the non-origin side (above the line).
Boundary Line $L_{2}$ ( $x - y = 2$ ): Passes through $C (2, 0)$ and $D (0, - 2)$ ; solution half-plane is on the origin side (above the line).
Point of Intersection: $E (\frac{7}{2}, \frac{3}{2}) = E (3.5, 1.5)$ .
Solution Region: The common region lying above both boundary lines $3 x + 7 y = 21$ and $x - y = 2$ , extending infinitely upward with apex at $E (3.5, 1.5)$ .

Quick Solution

Q4Part (iv)

HardUnder Review

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

Hint

For $4 x - 3 y \leq 12$ and $x \geq - \frac{3}{2}$ : boundary line $L_{1}$ passes through $A (3, 0)$ and $B (0, - 4)$ (contains $O (0, 0)$ ), and $L_{2}$ is the vertical line $x = - \frac{3}{2} = - 1.5$ (shading to the right toward $O (0, 0)$ ). Substitute $x = - \frac{3}{2}$ into $4 x - 3 y = 12$ to find the intersection point $E (- \frac{3}{2}, - 6) = E (- 1.5, - 6)$ .

Solution

Associated equations and intercepts.

For the first inequality $4 x - 3 y \leq 12$ , the associated linear equation $L_{1}$ is:
4x−3y=12
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 4 x - 3 (0) & = 12 \\ 4 x & = 12 \\ x & = 3 ⟹ A (3, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 4 (0) - 3 y & = 12 \\ - 3 y & = 12 \\ y & = - 4 ⟹ B (0, - 4) \end{matrix}$
For the second inequality $x \geq - \frac{3}{2}$ , the associated equation $L_{2}$ is:
x=−32=−1.5
This represents a vertical line parallel to the $y$ -axis:
- $x$ -intercept: $C (- \frac{3}{2}, 0) = C (- 1.5, 0)$
- $y$ -intercept: None (the line is parallel to the $y$ -axis).
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $4 x - 3 y \leq 12$ : $\begin{matrix} 4 (0) - 3 (0) & \leq 12 \\ 0 & \leq 12 (True) \end{matrix}$ The solution set of $4 x - 3 y \leq 12$ is the closed half-plane containing the origin $O (0, 0)$ (the region above and to the left of $L_{1}$ ).
- Substitute $x = 0$ into $x \geq - \frac{3}{2}$ : $\begin{matrix} 0 & \geq - \frac{3}{2} (True) \end{matrix}$ The solution set of $x \geq - \frac{3}{2}$ is the closed right half-plane containing the origin $O (0, 0)$ (all points with $x \geq - 1.5$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} 4 x - 3 y & = 12 — (1) \\ x & = - \frac{3}{2} — (2) \end{matrix}$
Substitute $x = - \frac{3}{2}$ into equation (1):
$\begin{matrix} 4 (- \frac{3}{2}) - 3 y & = 12 \\ - 6 - 3 y & = 12 \\ - 3 y & = 12 + 6 \\ - 3 y & = 18 \\ y & = \frac{18}{- 3} \\ y & = - 6 \end{matrix}$
Thus, the boundary lines intersect at point $E (- \frac{3}{2}, - 6) = E (- 1.5, - 6)$ .
Description of the common solution region.

The solution region is the common region lying to the right of the vertical line $x = - \frac{3}{2}$ and above/left of the line $4 x - 3 y = 12$ . It contains the origin $O (0, 0)$ and has its lowest corner vertex at $E (- 1.5, - 6)$ , extending infinitely upward and to the right, including both solid boundary lines.
Graphical representation of the system.

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

Answer

Boundary Line $L_{1}$ ( $4 x - 3 y = 12$ ): Passes through $A (3, 0)$ and $B (0, - 4)$ ; solution half-plane contains the origin $O (0, 0)$ .
Boundary Line $L_{2}$ ( $x = - \frac{3}{2}$ ): Vertical line at $x = - 1.5$ ; solution half-plane contains the origin $O (0, 0)$ (right half-plane).
Point of Intersection: $E (- \frac{3}{2}, - 6) = E (- 1.5, - 6)$ .
Solution Region: The region bounded on the left by the vertical line $x = - \frac{3}{2}$ and above/left by the line $4 x - 3 y = 12$ , meeting at vertex $E (- 1.5, - 6)$ .

Quick Solution

Q4Part (v)

HardUnder Review

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

Hint

For $3 x + 7 y \geq 21$ and $y \leq 4$ : boundary line $L_{1}$ passes through $A (7, 0)$ and $B (0, 3)$ (shading above), and $L_{2}$ is the horizontal line $y = 4$ (shading below toward $O (0, 0)$ ). Substitute $y = 4$ into $3 x + 7 y = 21$ to obtain the intersection point $E (- \frac{7}{3}, 4) \approx E (- 2.33, 4)$ .

Solution

Associated equations and intercepts.

For the first inequality $3 x + 7 y \geq 21$ , the associated linear equation $L_{1}$ is:
3x+7y=21
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 3 x + 7 (0) & = 21 \\ 3 x & = 21 \\ x & = 7 ⟹ A (7, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 3 (0) + 7 y & = 21 \\ 7 y & = 21 \\ y & = 3 ⟹ B (0, 3) \end{matrix}$
For the second inequality $y \leq 4$ , the associated equation $L_{2}$ is:
y=4
This represents a horizontal line parallel to the $x$ -axis:
- $y$ -intercept: $C (0, 4)$
- $x$ -intercept: None (the line is parallel to the $x$ -axis).
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $3 x + 7 y \geq 21$ : $\begin{matrix} 3 (0) + 7 (0) & \geq 21 \\ 0 & \geq 21 (False) \end{matrix}$ The solution set of $3 x + 7 y \geq 21$ is the closed half-plane not containing the origin $O (0, 0)$ (the region above $L_{1}$ ).
- Substitute $y = 0$ into $y \leq 4$ : $\begin{matrix} 0 & \leq 4 (True) \end{matrix}$ The solution set of $y \leq 4$ is the closed lower half-plane containing the origin $O (0, 0)$ (all points with $y \leq 4$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} 3 x + 7 y & = 21 — (1) \\ y & = 4 — (2) \end{matrix}$
Substitute $y = 4$ into equation (1):
$\begin{matrix} 3 x + 7 (4) & = 21 \\ 3 x + 28 & = 21 \\ 3 x & = 21 - 28 \\ 3 x & = - 7 \\ x & = - \frac{7}{3} \approx - 2.33 \end{matrix}$
Thus, the boundary lines intersect at point $E (- \frac{7}{3}, 4) \approx E (- 2.33, 4)$ .
Description of the common solution region.

The solution region is the infinite wedge bounded from above by the horizontal line $y = 4$ and from below by the inclined line $3 x + 7 y = 21$ . It originates at the apex vertex $E (- \frac{7}{3}, 4)$ and extends infinitely to the right, including all points on both solid boundary rays.
Graphical representation of the system.

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

Answer

Boundary Line $L_{1}$ ( $3 x + 7 y = 21$ ): Passes through $A (7, 0)$ and $B (0, 3)$ ; solution half-plane is on the non-origin side (above the line).
Boundary Line $L_{2}$ ( $y = 4$ ): Horizontal line at $y = 4$ ; solution half-plane contains the origin $O (0, 0)$ (lower half-plane).
Point of Intersection: $E (- \frac{7}{3}, 4) \approx E (- 2.33, 4)$ .
Solution Region: The infinite wedge lying below the horizontal line $y = 4$ and above the line $3 x + 7 y = 21$ , with vertex at $E (- \frac{7}{3}, 4)$ .

Quick Solution

Q4Part (vi)

HardUnder Review

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

Hint

For $5 x + 7 y \leq 35$ and $x - 2 y \leq 2$ : boundary line $L_{1}$ passes through $A (7, 0)$ and $B (0, 5)$ , and boundary line $L_{2}$ passes through $C (2, 0)$ and $D (0, - 1)$ . Both inequalities contain origin $O (0, 0)$ . Solving simultaneously yields intersection vertex $E (\frac{84}{17}, \frac{25}{17}) \approx E (4.94, 1.47)$ .

Solution

Associated equations and intercepts.

For the first inequality $5 x + 7 y \leq 35$ , the associated linear equation $L_{1}$ is:
5x+7y=35
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} 5 x + 7 (0) & = 35 \\ 5 x & = 35 \\ x & = 7 ⟹ A (7, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 5 (0) + 7 y & = 35 \\ 7 y & = 35 \\ y & = 5 ⟹ B (0, 5) \end{matrix}$
For the second inequality $x - 2 y \leq 2$ , the associated linear equation $L_{2}$ is:
x−2y=2
- $x$ -intercept: Substitute $y = 0$ : $\begin{matrix} x - 2 (0) & = 2 \\ x & = 2 ⟹ C (2, 0) \end{matrix}$
- $y$ -intercept: Substitute $x = 0$ : $\begin{matrix} 0 - 2 y & = 2 \\ - 2 y & = 2 \\ y & = - 1 ⟹ D (0, - 1) \end{matrix}$
Test point analysis at origin $O (0, 0)$ .
- Substitute $(0, 0)$ into $5 x + 7 y \leq 35$ : $\begin{matrix} 5 (0) + 7 (0) & \leq 35 \\ 0 & \leq 35 (True) \end{matrix}$ The solution set of $5 x + 7 y \leq 35$ is the closed half-plane containing the origin $O (0, 0)$ (the region below and to the left of $L_{1}$ ).
- Substitute $(0, 0)$ into $x - 2 y \leq 2$ : $\begin{matrix} 0 - 2 (0) & \leq 2 \\ 0 & \leq 2 (True) \end{matrix}$ The solution set of $x - 2 y \leq 2$ is the closed half-plane containing the origin $O (0, 0)$ (the region above and to the left of $L_{2}$ ).
Simultaneous solution for intersection point $E$ .

Solve equations $L_{1}$ and $L_{2}$ simultaneously:
$\begin{matrix} 5 x + 7 y & = 35 — (1) \\ x - 2 y & = 2 ⟹ x = 2 y + 2 — (2) \end{matrix}$
Substitute $x = 2 y + 2$ into equation (1):
$\begin{matrix} 5 (2 y + 2) + 7 y & = 35 \\ 10 y + 10 + 7 y & = 35 \\ 17 y + 10 & = 35 \\ 17 y & = 35 - 10 \\ 17 y & = 25 \\ y & = \frac{25}{17} \approx 1.47 \end{matrix}$
Substitute $y = \frac{25}{17}$ into equation (2):
$\begin{matrix} x & = 2 (\frac{25}{17}) + 2 \\ x & = \frac{50}{17} + \frac{34}{17} \\ x & = \frac{84}{17} \approx 4.94 \end{matrix}$
Thus, the boundary lines intersect at point $E (\frac{84}{17}, \frac{25}{17}) \approx E (4.94, 1.47)$ .
Description of the common solution region.

The solution region is the common closed region containing the origin $O (0, 0)$ , bounded from above by the solid line $5 x + 7 y = 35$ and bounded from below by the solid line $x - 2 y = 2$ . These boundary lines meet at vertex $E (\frac{84}{17}, \frac{25}{17}) \approx (4.94, 1.47)$ , and all points on both boundary lines are included in the solution set.
Graphical representation of the system.

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)

Answer

Boundary Line $L_{1}$ ( $5 x + 7 y = 35$ ): Passes through $A (7, 0)$ and $B (0, 5)$ ; solution half-plane contains the origin $O (0, 0)$ (below the line).
Boundary Line $L_{2}$ ( $x - 2 y = 2$ ): Passes through $C (2, 0)$ and $D (0, - 1)$ ; solution half-plane contains the origin $O (0, 0)$ (above the line).
Point of Intersection: $E (\frac{84}{17}, \frac{25}{17}) \approx E (4.94, 1.47)$ .
Solution Region: The common closed region containing the origin $O (0, 0)$ , bounded above by $5 x + 7 y = 35$ and below by $x - 2 y = 2$ , meeting at vertex $E (\frac{84}{17}, \frac{25}{17})$ .

Quick Solution

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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