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Class 9 Mathematics Notes – Chapter 5: Linear Equations and Inequalities

Punjab Board | 2026

Exercise 5.1

Q1. Solve and represent the solution on a real line.

(i) \(12x+30=-6\)

Subtract 30 from both sides:

\[ 12x=-6-30=-36 \]

Divide both sides by 12:

\[ x=\frac{-36}{12}=-3 \]

Solution Set: \(\{-3\}\)

Real Line Representation: Mark a solid dot at \(-3\).

(ii) \(\frac{x}{3}+6=-12\)

Subtract 6 from both sides:

\[ \frac{x}{3}=-12-6=-18 \]

Multiply both sides by 3:

\[ x=-18(3)=-54 \]

Solution Set: \(\{-54\}\)

(iii) \(\frac{x}{2}-\frac{3x}{4}=\frac{1}{12}\)

The LCM of 2, 4, and 12 is 12. Multiply the entire equation by 12:

\[ 6x-9x=1 \]

\[ -3x=1 \]

\[ x=-\frac13 \]

Solution Set: \(\left\{-\frac13\right\}\)

(iv) \(2=7(2x+4)+12x\)

Expand the brackets:

\[ 2=14x+28+12x \]

Combine like terms:

\[ 2=26x+28 \]

Subtract 28 from both sides:

\[ -26=26x \]

\[ x=-1 \]

Solution Set: \(\{-1\}\)

(v) \(\frac{2x-1}{3}-\frac{3x}{4}=\frac56\)

Multiply the entire equation by 12:

\[ 4(2x-1)-3(3x)=10 \]

Expand:

\[ 8x-4-9x=10 \]

\[ -x-4=10 \]

\[ -x=14 \]

\[ x=-14 \]

Solution Set: \(\{-14\}\)

(vi) \(\frac{-5x}{10}=9-\frac{10}{5}x\)

Simplify the fractions:

\[ -\frac{x}{2}=9-2x \]

Add \(2x\) to both sides:

\[ -\frac{x}{2}+2x=9 \]

\[ \frac{3x}{2}=9 \]

Multiply both sides by \(\frac23\):

\[ x=6 \]

Solution Set: \(\{6\}\)

Q2. Solve each inequality and represent the solution on a real line.

(i) \(x-6\leq-2\)

Add 6 to both sides:

\[ x\leq-2+6 \]

\[ \boxed{x\leq4} \]

Interval: \((-\infty,4]\)

(ii) \(-9>-16+x\)

Add 16 to both sides:

\[ 7>x \]

\[ \boxed{x<7} \]

Interval: \((-\infty,7)\)

(iii) \(3+2x\geq3\)

Subtract 3 from both sides:

\[ 2x\geq0 \]

\[ \boxed{x\geq0} \]

Interval: \([0,\infty)\)

(iv) \(6(x+10)\leq0\)

Divide both sides by 6:

\[ x+10\leq0 \]

\[ \boxed{x\leq-10} \]

Interval: \((-\infty,-10]\)

(v) \(\frac53x-\frac34<-\frac1{12}\)

Multiply both sides by 12:

\[ 20x-9<-1 \]

\[ 20x<8 \]

\[ \boxed{x<\frac25} \]

Interval: \(\left(-\infty,\frac25\right)\)

(vi) \(\frac14x-\frac12\leq-1+\frac12x\)

Multiply both sides by 4:

\[ x-2\leq-4+2x \]

Subtract \(x\) from both sides:

\[ -2\leq-4+x \]

Add 4 to both sides:

\[ \boxed{x\geq2} \]

Interval: \([2,\infty)\)

Q3. Shade the solution region for the following linear inequalities in the \(xy\)-plane.

(i) \(2x+y\leq6\)

Associated equation:

\[ 2x+y=6 \]

Intercepts:

\(x\)-intercept: \((3,0)\)
\(y\)-intercept: \((0,6)\)

Test point \((0,0)\):

\[ 2(0)+0\leq6 \]

This statement is true.

Solution: Draw a solid boundary line through \((3,0)\) and \((0,6)\). Shade the half-plane containing the origin.

(ii) \(3x+7y\geq21\)

Associated equation:

\[ 3x+7y=21 \]

Intercepts:

\(x\)-intercept: \((7,0)\)
\(y\)-intercept: \((0,3)\)

Test point \((0,0)\):

\[ 0\geq21 \]

This statement is false.

Solution: Draw a solid boundary line and shade the half-plane not containing the origin.

(iii) \(3x-2y\geq6\)

Associated equation:

\[ 3x-2y=6 \]

Intercepts:

\(x\)-intercept: \((2,0)\)
\(y\)-intercept: \((0,-3)\)

Test point \((0,0)\):

\[ 0\geq6 \]

This statement is false.

Solution: Draw a solid boundary line and shade the half-plane away from the origin.

(iv) \(5x-4y\leq20\)

Associated equation:

\[ 5x-4y=20 \]

Intercepts:

\(x\)-intercept: \((4,0)\)
\(y\)-intercept: \((0,-5)\)

Test point \((0,0)\):

\[ 0\leq20 \]

This statement is true.

Solution: Draw a solid boundary line and shade the half-plane containing the origin.

(v) \(2x+1\geq0\)

\[ 2x\geq-1 \]

\[ \boxed{x\geq-\frac12} \]

Solution: Draw a solid vertical line at \(x=-\frac12\) and shade the region to its right.

(vi) \(3y-4\leq0\)

\[ 3y\leq4 \]

\[ \boxed{y\leq\frac43} \]

Solution: Draw a solid horizontal line at \(y=\frac43\) and shade the region below it.

Q4. Indicate the solution region of the following linear inequalities by shading.

For each pair, draw the boundary lines and shade the common region satisfying both inequalities.

(i) \(2x-3y\leq6\) and \(2x+3y\leq12\)

Draw the boundary lines:

\[ 2x-3y=6 \]

\[ 2x+3y=12 \]

Shade the intersection of the half-planes satisfying both inequalities.

(ii) \(x+y\geq5\) and \(-y+x\leq1\)

Boundary lines:

\[ x+y=5 \]

\[ x-y=1 \]

Shade the region satisfying both inequalities: above \(y=5-x\) and on the side of \(y=x-1\) where \(x-y\leq1\).

(iii) \(3x+7y\geq21\) and \(x-y\leq2\)

Boundary lines:

\[ 3x+7y=21 \]

\[ x-y=2 \]

Shade the common region satisfying both inequalities.

(iv) \(4x-3y\leq12\) and \(x\geq-\frac32\)

Boundary lines:

\[ 4x-3y=12 \]

\[ x=-\frac32 \]

Shade the region satisfying both inequalities, including the region to the right of \(x=-\frac32\).

(v) \(3x+7y\geq21\) and \(y\leq4\)

Boundary lines:

\[ 3x+7y=21 \]

\[ y=4 \]

Shade the common region above the first line and below the horizontal line \(y=4\).

(vi) \(5x+7y\leq35\) and \(x-2y\leq2\)

Boundary lines:

\[ 5x+7y=35 \]

\[ x-2y=2 \]

Shade the common region satisfying both inequalities.

Exercise 5.2

Q1. Maximize \(f(x,y)=2x+5y\) subject to the constraints

\[ 2y-x\leq8,\quad x-y\leq4,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ 2y-x=8,\qquad x-y=4 \]

Find the feasible corner points by graphing the constraints and considering the non-negativity conditions.

The objective function is:

\[ f(x,y)=2x+5y \]

The intersection of the two boundary lines is found from:

\[ 2y-x=8,\qquad x-y=4 \]

Substitute \(x=y+4\):

\[ 2y-(y+4)=8 \]

\[ y-4=8 \]

\[ y=12,\qquad x=16 \]

However, \((16,12)\) does not satisfy the first constraint because \(2(12)-16=8\), and it does satisfy the second constraint because \(16-12=4\). Thus, it is the intersection of both boundary lines and is feasible.

Evaluate the objective function at the relevant feasible corner points:

\[ f(0,0)=0 \]

\[ f(4,0)=8 \]

\[ f(0,4)=20 \]

At \((16,12)\):

\[ f(16,12)=2(16)+5(12)=32+60=92 \]

Maximum value: \(\boxed{92}\) at \(\boxed{(16,12)}\).

Q2. Maximize \(f(x,y)=x+3y\) subject to

\[ 2x+5y\leq30,\quad 5x+4y\leq20,\quad x\geq0,\quad y\geq0 \]

Find the intercepts.

For \(2x+5y=30\):

\[ x=15\text{ when }y=0,\qquad y=6\text{ when }x=0 \]

For \(5x+4y=20\):

\[ x=4\text{ when }y=0,\qquad y=5\text{ when }x=0 \]

The feasible axis corner points are \((0,0)\), \((4,0)\), and \((0,5)\).

Evaluate the objective function:

\[ f(0,0)=0 \]

\[ f(4,0)=4 \]

\[ f(0,5)=15 \]

Maximum value: \(\boxed{15}\) at \(\boxed{(0,5)}\).

Q3. Maximize \(z=2x+3y\) subject to

\[ 2x+y\leq4,\quad 4x-y\leq4,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ 2x+y=4,\qquad 4x-y=4 \]

At the intersection:

\[ y=4-2x \]

Substitute into \(4x-y=4\):

\[ 4x-(4-2x)=4 \]

\[ 6x=8 \]

\[ x=\frac43,\qquad y=\frac43 \]

The corner points are:

\[ (0,0),\quad(2,0),\quad(0,4),\quad \left(\frac43,\frac43\right) \]

Evaluate \(z=2x+3y\):

\[ z(0,0)=0 \]

\[ z(2,0)=4 \]

\[ z(0,4)=12 \]

\[ z\left(\frac43,\frac43\right)=\frac{20}{3} \]

Maximum value: \(\boxed{12}\) at \(\boxed{(0,4)}\).

Q4. Minimize \(z=2x+y\) subject to

\[ x+y\geq3,\quad 7x+5y\leq35,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ x+y=3,\qquad 7x+5y=35 \]

The feasible corner points include \((3,0)\), \((0,3)\), \((5,0)\), and \((0,7)\), subject to the constraints.

Evaluate the objective function at the relevant feasible points:

\[ z(3,0)=2(3)+0=6 \]

\[ z(0,3)=2(0)+3=3 \]

\[ z(5,0)=2(5)+0=10 \]

Minimum value: \(\boxed{3}\) at \(\boxed{(0,3)}\).

Q5. Maximize \(f(x,y)=2x+3y\) subject to

\[ 2x+y\leq8,\quad x+2y\leq14,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ 2x+y=8,\qquad x+2y=14 \]

At their intersection:

\[ y=8-2x \]

Substitute into the second equation:

\[ x+2(8-2x)=14 \]

\[ x+16-4x=14 \]

\[ -3x=-2 \]

\[ x=\frac23,\qquad y=\frac{20}{3} \]

Evaluate the objective function at the feasible corner points:

\[ f(0,0)=0 \]

\[ f(4,0)=8 \]

\[ f(0,7)=21 \]

\[ f\left(\frac23,\frac{20}{3}\right) =\frac43+20=\frac{64}{3} \]

Maximum value: \(\boxed{\frac{64}{3}}\) at \(\boxed{\left(\frac23,\frac{20}{3}\right)}\).

Q6. Find the minimum and maximum values of \(z=3x+y\) subject to

\[ 3x+5y\geq15,\quad x+6y\geq9,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ 3x+5y=15,\qquad x+6y=9 \]

Their intercepts are:

For \(3x+5y=15\):

\[ (5,0),\quad(0,3) \]

For \(x+6y=9\):

\[ (9,0),\quad\left(0,\frac32\right) \]

At the intersection:

\[ 3x+5y=15 \]

\[ x+6y=9 \]

From the second equation:

\[ x=9-6y \]

Substitute:

\[ 3(9-6y)+5y=15 \]

\[ 27-18y+5y=15 \]

\[ -13y=-12 \]

\[ y=\frac{12}{13},\qquad x=\frac{45}{13} \]

Evaluate the objective function at the relevant feasible corner points:

\[ z(5,0)=15 \]

\[ z(0,3)=3 \]

\[ z(9,0)=27 \]

\[ z\left(\frac{45}{13},\frac{12}{13}\right) =\frac{135}{13}+\frac{12}{13} =\frac{147}{13} \]

Minimum value: \(\boxed{3}\) at \(\boxed{(0,3)}\).

Maximum value: \(\boxed{27}\) at \(\boxed{(9,0)}\).

Review Exercise 5

Q1. Multiple Choice Questions (MCQs)

(i) Which of the following is a linear equation?

Correct Answer: Option (c) \(2x+1=1\)

(ii) The solution of \(5x-10=10\) is:

\[ 5x=20\implies x=4 \]

Correct Answer: Option (c) \(4\)

(iii) If \(7x+4<6x+6\), then \(x\) belongs to the interval:

\[ 7x+4<6x+6 \]

\[ x<2 \]

Correct Answer: Option (c) \((-\infty,2)\)

(iv) A vertical line divides the plane into:

Correct Answer: Option (d) Two half-planes.

(v) The linear equation formed from a linear inequality is called:

Correct Answer: Option (b) Associated equation.

(vi) \(3x+4<0\) is a:

Correct Answer: Option (b) Inequality.

(vii) A corner point is also called a:

Correct Answer: Option (b) Vertex.

(viii) The point \((0,0)\) is a solution of which inequality?

Correct Answer: Option (c) \(-2x+3y<0\)

(ix) The solution region restricted to the first quadrant is called the:

Correct Answer: Option (b) Feasible region.

(x) A function that is to be maximized or minimized is called the:

Correct Answer: Option (b) Objective function.

Q2. Solve and represent the solutions on a real line.

(i) \(\frac{x+5}{3}=1-x\)

Multiply both sides by 3:

\[ x+5=3(1-x) \]

\[ x+5=3-3x \]

\[ 4x=-2 \]

\[ \boxed{x=-\frac12} \]

(ii) \(\frac{2x+1}{3}+\frac12=1-\frac{x-1}{3}\)

Multiply both sides by 6:

\[ 2(2x+1)+3=6-2(x-1) \]

\[ 4x+2+3=6-2x+2 \]

\[ 4x+5=8-2x \]

\[ 6x=3 \]

\[ \boxed{x=\frac12} \]

(iii) \(3x+7<16\)

\[ 3x<9 \]

\[ \boxed{x<3} \]

Interval: \((-\infty,3)\)

(iv) \(5(x-3)\geq26x-(10x+4)\)

Expand the brackets:

\[ 5x-15\geq26x-10x-4 \]

\[ 5x-15\geq16x-4 \]

Subtract \(5x\) from both sides:

\[ -15\geq11x-4 \]

Add 4:

\[ -11\geq11x \]

Divide by 11:

\[ \boxed{x\leq-1} \]

Interval: \((-\infty,-1]\)

Q3. Find the solution region of the following linear inequalities.

(i) \(3x-4y\leq12\) and \(3x+2y\geq3\)

Draw the boundary lines:

\[ 3x-4y=12 \]

\[ 3x+2y=3 \]

Shade the common region satisfying both inequalities.

(ii) \(2x+y\leq4\) and \(x+2y\leq6\)

Draw the boundary lines:

\[ 2x+y=4 \]

\[ x+2y=6 \]

Shade the common feasible region satisfying both inequalities, including the relevant axes when restricted to the first quadrant.

Q4. Find the maximum value of \(g(x,y)=x+4y\) subject to

\[ x+y\leq4,\quad x\geq0,\quad y\geq0 \]

The corner points are:

\[ (0,0),\quad(4,0),\quad(0,4) \]

Evaluate the objective function:

\[ g(0,0)=0 \]

\[ g(4,0)=4 \]

\[ g(0,4)=16 \]

Maximum value: \(\boxed{16}\) at \(\boxed{(0,4)}\).

Q5. Find the minimum value of \(f(x,y)=3x+5y\) subject to

\[ x+3y\geq3,\quad x+y\geq2,\quad x\geq0,\quad y\geq0 \]

The boundary lines are:

\[ x+3y=3,\qquad x+y=2 \]

At their intersection:

\[ x=2-y \]

Substitute into the first equation:

\[ 2-y+3y=3 \]

\[ 2y=1 \]

\[ y=\frac12,\qquad x=\frac32 \]

Evaluate the objective function at the relevant feasible corner points:

\[ f(3,0)=3(3)+5(0)=9 \]

\[ f(0,2)=3(0)+5(2)=10 \]

\[ f\left(\frac32,\frac12\right) =3\left(\frac32\right)+5\left(\frac12\right) =\frac92+\frac52=7 \]

Minimum value: \(\boxed{7}\) at \(\boxed{\left(\frac32,\frac12\right)}\).

Notes by Sir Dilawar

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