The solution of the new math book for 9th class step by step is given below.
Question 1: Solve and represent the solution on a real line
(i) \( 12x + 30 = -6 \)
\[12x + 30 = -6\]
\[12x = -6 – 30\]
\[12x = -36\]
\[x = -3\]
Solution: \( x = -3 \)
(ii) \( \frac{x}{3} + 6 = -12 \)
\[\frac{x}{3} + 6 = -12\]
\[\frac{x}{3} = -12 – 6\]
\[\frac{x}{3} = -18\]
\[x = -18 \times 3\]
\[x = -54\]
Solution: \( x = -54 \)
(iii) \( \frac{x}{2} – \frac{3x}{4} = \frac{1}{12} \)
\[\frac{x}{2} – \frac{3x}{4} = \frac{1}{12}\]
\[\frac{2x}{4} – \frac{3x}{4} = \frac{1}{12}\]
\[\frac{-x}{4} = \frac{1}{12}\]
\[-12x = 4\]
\[x = -\frac{4}{12}\]
\[x = -\frac{1}{3}\]
Solution: \( x = -\frac{1}{3} \)
(iv) \( 2 = 7(2x + 4) + 12x \)
\[2 = 7(2x + 4) + 12x\]
\[2 = 14x + 28 + 12x\]
\[2 = 26x + 28\]
\[26x = 2 – 28\]
\[26x = -26\]
\[x = -1\]
Solution: \( x = -1 \)
(v) \( \frac{2x – 1}{3} – \frac{3x}{4} = \frac{5}{6} \)
\[\frac{2x – 1}{3} – \frac{3x}{4} = \frac{5}{6}\]
\[4(2x – 1) – 3(3x) = 10\]
\[8x – 4 – 9x = 10\]
\[-x – 4 = 10\]
\[-x = 14\]
\[x = -14\]
Solution: \( x = -14 \)
(vi) \( \frac{-5x}{10} = 9 – \frac{10}{5}x \)
\[\frac{-5x}{10} = 9 – \frac{10}{5}x\]
\[\frac{-x}{2} = 9 – 2x\]
\[-x = 18 – 4x\]
\[3x = 18\]
\[x = 6\]
Solution: \( x = 6 \)
Question 2 Solve each inequality and represent the solution on a real line
(i) \(x – 6 \leq -2\)
\[x – 6 \leq -2\]
\[x – 6 + 6 \leq -2 + 6\]
\[x \leq 4\]
(ii) \(-9 > -16 + x\)
\[-9 > -16 + x\]
\[-9 + 16 > -16 + x + 16\]
\[7 > x\]
\[x < 7\]
(iii) \(3 + 2x \geq 3\)
\[3 + 2x \geq 3\]
\[3 + 2x – 3 \geq 3 – 3\]
\[2x \geq 0\]
\[x \geq 0\]
(iv) \(6(x + 10) \leq 0\)
\[6(x + 10) \leq 0\]
\[x + 10 \leq 0\]
\[x + 10 – 10 \leq 0 – 10\]
\[x \leq -10\]
(v) \(\frac{5}{3}x – \frac{3}{4} < -\frac{1}{12}\)
\[\frac{5}{3}x – \frac{3}{4} < -\frac{1}{12}\]
\[\frac{5}{3}x < -\frac{1}{12} + \frac{3}{4}\]
\[\frac{5}{3}x < \frac{-1}{12} + \frac{9}{12}\]
\[\frac{5}{3}x < \frac{8}{12}\]
\[\frac{5}{3}x < \frac{2}{3}\]
\[x < \frac{2}{3} \div \frac{5}{3}\]
\[x < \frac{2}{3} \cdot \frac{3}{5}\]
\[x < \frac{2}{5}\]
(vi) \(\frac{1}{4}x – \frac{1}{2} \leq -1 + \frac{1}{2}x\)
\[\frac{1}{4}x – \frac{1}{2} \leq -1 + \frac{1}{2}x\]
\[\frac{1}{4}x – \frac{1}{2} – \frac{1}{2}x \leq -1\]
\[-\frac{1}{4}x – \frac{1}{2} \leq -1\]
\[-\frac{1}{4}x – \frac{1}{2} + \frac{1}{2} \leq -1 + \frac{1}{2}\]
\[-\frac{1}{4}x \leq -\frac{1}{2}\]
\[x \geq 2\]
Question 3 Shade the solution region for the following linear inequalities in the xy-plane.
(i) \(2x + y \leq 6\)
Equation \(2x + y = 6\)
Make Points:
Put \(x = 0\) into the equation:
\[2(0) + y = 6 \]
\[y = 6\]
\[ (0, 6)\]
Put \(y = 0\) into the equation:
\[2x + 0 = 6 \]
\[ x = 3\]
\[ (3, 0)\]
Draw a solid line passing through \((0, 6)\) and \((3, 0)\).
Take a test point:
Put \((0, 0)\) into the inequality:
\[2(0) + 0 \leq 6\]
\[ 0 \leq 6\]
\[ \text{(True)}\]
As the test point is true then Shade the region towards (0,0).
(ii) \(3x + 7y \geq 21\)
Equation \(3x + 7y = 21\)
Make Points:
Put \(x = 0\) into the equation :
\[3(0) + 7y = 21\]
\[ y = 3 \]
\[(0, 3)\]
Put \(y = 0\) into the equation:
\[3x + 7(0) = 21 \]
\[ x = 7\]
\[(7, 0)\]
Draw a solid line passing through \((0, 3)\) and \((7, 0)\).
Take Test point:
Put \((0, 0)\) into the inequality:
\[3(0) + 7(0) \geq 21 \]
\[0 \geq 21\]
\[ \text{(False)}\]
As the test point is false then Shade the region away from (0,0).
(iii) \(3x – 2y \geq 6\)
Equation \(3x – 2y = 6\)
Make Points:
Put \(x = 0\) into the equation:
\[3(0) – 2y = 6\]
\[ y = -3 \]
\[ (0, -3)\]
Put \(y = 0\) into the equation:
\[3x – 2(0) = 6 \]
\[ x = 2 \]
\[(2, 0)\]
Draw a solid line passing through \((0, -3)\) and \((2, 0)\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[3(0) – 2(0) \geq 6 \]
\[ 0 \geq 6 \]
\[ \text{(False)}\]
As the test point is false then Shade the region away from (0,0).
(iv) \(5x – 4y \leq 20\)
Equation \(5x – 4y = 20\)
Make Points:
Put \(x = 0\) into the equation:
\[5(0) – 4y = 20 \]
\[ y = -5 \]
\[ (0, -5)\\]
Put \(y = 0\) into the equation:
\[5x – 4(0) = 20 \]
\[ x = 4 \]
\[ (4, 0)\]
Draw a solid line passing through \((0, -5)\) and \((4, 0)\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[5(0) – 4(0) \leq 20 \]
\[ 0 \leq 20 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
(v) \(2x + 1 \geq 0\)
Equation \(2x + 1 = 0\)
Solve for \(x\):
\[2x = -1 \]
\[ x = -\frac{1}{2}\]
Vertical line at \(x = -\frac{1}{2}\).
Draw a solid vertical line at \(x = -\frac{1}{2}\).
Take Test point:
Put \((0, 0)\) into the inequality:
\[2(0) + 1 \geq 0 \]
\[ 1 \geq 0\]
\[ \text{(True)}\]
Shade the region towards (0,0).
(vi) \(3y – 4 \leq 0\)
Equation \(3y – 4 = 0\)
Solve for \(y\):
\[3y = 4 \]
\[ y = \frac{4}{3}\]
Horizontal line at \(y = \frac{4}{3}\).
Draw a solid horizontal line at \(y = \frac{4}{3}\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[3(0) – 4 \leq 0 \]
\[ -4 \leq 0 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
Question 4 Indicate the solution region of the following linear inequality by shading
(i) \(2x – 3y \leq 6\) and \(2x + 3y \leq 12\)
For \(2x – 3y \leq 6\) take Equation \(2x – 3y = 6\)
Make Points:
Put \(x = 0\) into the equation:
\[2(0) – 3y = 6 \]
\[ y = -2 \]
\[(0, -2)\]
Put \(y = 0\) into the equation:
\[2x – 3(0) = 6 \]
\[ x = 3 \]
\[(3, 0)\]
Draw a solid line passing through \((0, -2)\) and \((3, 0)\).
Take Test point:
Put \((0, 0)\) into the inequality:
\[2(0) – 3(0) \leq 6 \]
\[ 0 \leq 6 \]
\[ \text{(True)}\\]
Shade the region towards (0,0).
For \(2x + 3y \leq 12\) take Equation \(2x + 3y = 12\)
Make Points:
Put \(x = 0\) into the equation:
\[2(0) + 3y = 12 \]
\[ y = 4 \]
\[(0, 4)\]
Put \(y = 0\) into the equation:
\[2x + 3(0) = 12 \]
\[ x = 6 \]
\[ (6, 0)\]
Draw a solid line passing through \((0, 4)\) and \((6, 0)\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[2(0) + 3(0) \leq 12 \]
\[ 0 \leq 12 \]
\[\text{(True)}\]
Shade the region towards (0,0).
Label the feasible region on the graph of the both inequality
(ii) \(x + y \geq 5\) and \(-y + x \leq 1\)
For \(x + y \geq 5\) the Equation take \(x + y = 5\)
Make Points:
Put \(x = 0\) into the equation:
\[0 + y = 5 \]
\[ y = 5 \]
\[(0, 5)\]
Put \(y = 0\) into the equation:
\[x + 0 = 5 \]
\[ x = 5 \]
\[(5, 0)\]
Draw a solid line passing through \((0, 5)\) and \((5, 0)\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[0 + 0 \geq 5 \]
\[ 0 \geq 5 \]
\[ \text{(False)}\]
Shade the region away from (0,0).
For \(-y + x \leq 1\) take Equation \(-y + x = 1\)
Make Points:
Put \(x = 0\) into the equation:
\[0 – y = 1 \]
\[ y = -1 \]
\[(0, -1)\]
Put \(y = 0\) into the equation:
\[-0 + x = 1 \]
\[ x = 1 \]
\[(1, 0)\]
Draw a solid line passing through \((0, -1)\) and \((1, 0)\).
Take a Test point:
Put \((0, 0)\) into the inequality:
\[-0 + 0 \leq 1 \]
\[ 0 \leq 1 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
Label the feasible region on the graph of the both inequality.
(iii) \(3x + 7y \geq 21\) and \(x – y \leq 2\)
For \(3x + 7y \geq 21\) take Equation \(3x + 7y = 21\)
Make the Points:
Put \(x = 0\) into the equaton:
\[3(0) + 7y = 21 \]
\[y = 3\]
\[ (0, 3)\]
Put \(y = 0\) into the equation:
\[3x + 7(0) = 21 \]
\[ x = 7 \]
\[ (7, 0)\]
Draw a solid line passing through \((0, 3)\) and \((7, 0)\).
Test point:
Put \((0, 0)\) into the inequality:
\[3(0) + 7(0) \geq 21 \]
\[ 0 \geq 21 \]
\[ \text{(False)}\]
Shade the region away from (0,0).
For \(x – y \leq 2\) the Equation \(x – y = 2\)
Make Points:
Put \(x = 0\) into the equation:
\[0 – y = 2 \]
\[ y = -2 \]
\[ (0, -2)\]
Put \(y = 0\) into the equation:
\[x – 0 = 2 \]
\[x = 2 \]
\[ (2, 0)\]
Draw a solid line passing through \((0, -2)\) and \((2, 0)\).
Test point:
Put \((0, 0)\) into the inequality:
\[0 – 0 \leq 2 \]
\[ 0 \leq 2 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
Label the feasible region on the graph of the both inequality.
(iv) \(4x – 3y \leq 12\) and \(x \geq -\frac{3}{2}\)
For \(4x – 3y \leq 12\) take Equation \(4x – 3y = 12\)
Make Points:
Put \(x = 0\):
\[4(0) – 3y = 12 \]
\[ y = -4 \]
\[ (0, -4)\]
Put \(y = 0\):
\[4x – 3(0) = 12 \]
\[ x = 3 \]
\[(3, 0)\]
Draw a solid line passing through \((0, -4)\) and \((3, 0)\).
Test point:
Put \((0, 0)\) into the inequality:
\[4(0) – 3(0) \leq 12 \]
\[0 \leq 12 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
For \(x \geq -\frac{3}{2}\):
Vertical line at \(x = -\frac{3}{2}\).
Test point:
Put \((0, 0)\) into the inequality:
\[0 \geq -\frac{3}{2} \]
\[\text{(True)}\]
Shade the region to the right of the line.
Label the feasible region on the graph of the both inequality.
(v) \(3x + 7y \geq 21\) and \(y \leq 4\)
For \(3x + 7y \geq 21\) take Equation \(3x + 7y = 21\)
Make Points:
Put \(x = 0\):
\[3(0) + 7y = 21 \]
\[y = 3 \]
\[(0, 3)\]
Put \(y = 0\) into the equation:
\[3x + 7(0) = 21 \]
\[ x = 7 \]
\[(7, 0)\]
Draw a solid line passing through \((0, 3)\) and \((7, 0)\).
Test point:
Put \((0, 0)\) into the inequality:
\[3(0) + 7(0) \geq 21 \]
\[ 0 \geq 21 \]
\[\text{(False)}\]
Shade the region away from (0,0).
For \(y \leq 4\):
Horizontal line at \(y = 4\).
Test point:
Put \((0, 0)\) into the inequality:
\[0 \leq 4 \]
\[\text{(True)}\]
Shade the region below the line.
Label the feasible region on the graph of the both inequality.
(vi) \(5x + 7y \leq 35\) and \(x – 2y \leq 2\)
For \(5x + 7y \leq 35\) the Equation \(5x + 7y = 35\)
Make Points:
Put \(x = 0\) into the equation:
\[5(0) + 7y = 35 \]
\[ y = 5 \]
\[(0, 5)\]
Put \(y = 0\) into the equation:
\[5x + 7(0) = 35 \]
\[ x = 7 \]
\[ (7, 0)\]
Draw a solid line passing through \((0, 5)\) and \((7, 0)\).
Test point:
Put \((0, 0)\) into the inequality:
\[5(0) + 7(0) \leq 35 \]
\[ 0 \leq 35 \]
\[\text{(True)}\]
Shade the region towards (0,0).
For \(x – 2y \leq 2\) take Equation \(x – 2y = 2\)
Make Points:
Put \(x = 0\) into the equation:
\[0 – 2y = 2 \]
\[ y = -1 \]
\[ (0, -1)\]
Put \(y = 0\) into the equation:
\[x – 2(0) = 2 \]
\[x = 2 \]
\[ (2, 0)\]
Draw a solid line passing through \((0, -1)\) and \((2, 0)\).
Test a test point:
Put \((0, 0)\) into the inequality:
\[0 – 2(0) \leq 2\]
\[ 0 \leq 2 \]
\[ \text{(True)}\]
Shade the region towards (0,0).
Label the feasible region on the graph of the both inequality.
