Rational Numbers

Q1: Find the additive inverse of:

i. 9

Additive Inverse: The number which when added to 9 gives 0 is: -9
Answer: -9

ii. -11

Additive Inverse: The number which when added to -11 gives 0 is: 11
Answer: 11

iii. \(\frac{-8}{13}\)

Additive Inverse: The number which when added to \(\frac{-8}{13}\) gives 0 is: \(\frac{8}{13}\)
Answer: \(\frac{8}{13}\)

iv. \(\frac{5}{-6}\)

Step 1: Rewrite with negative sign in numerator: \(\frac{-5}{6}\)
Step 2: Its additive inverse is: \(\frac{5}{6}\)
Answer: \(\frac{5}{6}\)

v. 0

Additive Inverse: The number which when added to 0 gives 0 is: \(0\)
Answer: 0

Q2: Subtract:

i. \(\frac{3}{5}\) from \(\frac{1}{2}\)

Step 1: Write as: \(\frac{1}{2} – \frac{3}{5}\)
Step 2: LCM of 2 and 5 is 10 \[ \frac{1}{2} = \frac{5}{10},\quad \frac{3}{5} = \frac{6}{10} \\ \frac{5}{10} – \frac{6}{10} = \frac{-1}{10} \] Answer: \(\frac{-1}{10}\)

ii. \(\frac{-4}{7}\) from \(\frac{2}{3}\)

Step 1: Write as: \(\frac{2}{3} – \left(\frac{-4}{7}\right)\) \[ = \frac{2}{3} + \frac{4}{7} \] Step 2: LCM of 3 and 7 is 21 \[ \frac{2}{3} = \frac{14}{21},\quad \frac{4}{7} = \frac{12}{21} \\ \frac{14}{21} + \frac{12}{21} = \frac{26}{21} \] Answer: \(\frac{26}{21}\)

iii. \(\frac{-5}{6}\) from \(\frac{-3}{4}\)

Step 1: Write as: \(\frac{-3}{4} – \left(\frac{-5}{6}\right)\) \[ = \frac{-3}{4} + \frac{5}{6} \] Step 2: LCM of 4 and 6 is 12 \[ \frac{-3}{4} = \frac{-9}{12},\quad \frac{5}{6} = \frac{10}{12} \\ \frac{-9}{12} + \frac{10}{12} = \frac{1}{12} \] Answer: \(\frac{1}{12}\)

iv. \(\frac{-7}{9}\) from \(0\)

Step 1: Write as: \(0 – \left(\frac{-7}{9}\right) = \frac{7}{9}\)
Answer: \(\frac{7}{9}\)

v. \(4\) from \(\frac{-6}{11}\)

Step 1: Write as: \(\frac{-6}{11} – 4\)
Step 2: Convert 4 to fraction with denominator 11: \[ 4 = \frac{44}{11} \\ \frac{-6}{11} – \frac{44}{11} = \frac{-50}{11} \] Answer: \(\frac{-50}{11}\)

vi. \(\frac{3}{8}\) from \(\frac{-5}{6}\)

Step 1: Write as: \(\frac{-5}{6} – \frac{3}{8}\)
Step 2: LCM of 6 and 8 is 24 \[ \frac{-5}{6} = \frac{-20}{24},\quad \frac{3}{8} = \frac{9}{24} \\ \frac{-20}{24} – \frac{9}{24} = \frac{-29}{24} \] Answer: \(\frac{-29}{24}\)

Q3: Evaluate:

i. \(\frac{5}{6} – \frac{7}{8}\)

Step 1: LCM of 6 and 8 is 24 \[ \frac{5}{6} = \frac{20}{24},\quad \frac{7}{8} = \frac{21}{24} \\ \frac{20}{24} – \frac{21}{24} = \frac{-1}{24} \] Answer: \(\frac{-1}{24}\)

ii. \(\frac{5}{12} – \frac{17}{18}\)

Step 1: LCM of 12 and 18 is 36 \[ \frac{5}{12} = \frac{15}{36},\quad \frac{17}{18} = \frac{34}{36} \\ \frac{15}{36} – \frac{34}{36} = \frac{-19}{36} \] Answer: \(\frac{-19}{36}\)

iii. \(\frac{11}{15} – \frac{13}{20}\)

Step 1: LCM of 15 and 20 is 60 \[ \frac{11}{15} = \frac{44}{60},\quad \frac{13}{20} = \frac{39}{60} \\ \frac{44}{60} – \frac{39}{60} = \frac{5}{60} = \frac{1}{12} \] Answer: \(\frac{1}{12}\)

iv. \(\frac{-5}{9} – \frac{-2}{3}\)

Step 1: \(\frac{-5}{9} + \frac{2}{3}\) \[ \frac{-5}{9} + \frac{2}{3} = \frac{-5}{9} + \frac{6}{9} = \frac{1}{9} \] Answer: \(\frac{1}{9}\)

v. \(\frac{6}{11} – \frac{-3}{4}\)

Step 1: \(\frac{6}{11} + \frac{3}{4}\)
LCM of 11 and 4 is 44 \[ \frac{6}{11} = \frac{24}{44},\quad \frac{3}{4} = \frac{33}{44} \] \[ \frac{24}{44} + \frac{33}{44} = \frac{57}{44} \] Answer: \(\frac{57}{44}\)

vi. \(\frac{-2}{3} – \frac{3}{4}\)

Step 1: LCM of 3 and 4 is 12 \[ \frac{-2}{3} = \frac{-8}{12},\quad \frac{3}{4} = \frac{9}{12} \\ \frac{-8}{12} – \frac{9}{12} = \frac{-17}{12} \] Answer: \(\frac{-17}{12}\)

Q4: The sum of two rational numbers is \(\frac{-5}{8}\). If one of them is \(\frac{7}{16}\), find the other.

Let the other rational number be \(x\).
According to the question: \[ x + \frac{7}{16} = \frac{-5}{8} \]Step 1: Transpose \(\frac{7}{16}\) to the RHS \[ x = \frac{-5}{8} – \frac{7}{16} \]Step 2: Take LCM of 8 and 16
LCM of 8 and 16 is 16 \[ \frac{-5}{8} = \frac{-10}{16} \] So, \[ x = \frac{-10}{16} – \frac{7}{16} = \frac{-17}{16} \]Answer: The other number is \(\frac{-17}{16}\)

Q5: The sum of two rational numbers is -4. If one of them is \(\frac{-3}{5}\), find the other.

Let the other rational number be \(x\).
According to the question: \[ x + \frac{-3}{5} = -4 \]Step 1: Transpose \(\frac{-3}{5}\) to the RHS \[ x = -4 – \left(-\frac{3}{5}\right) \]Step 2: Simplify the signs \[ x = -4 + \frac{3}{5} \]Step 3: Express \(-4\) with denominator 5 \[ -4 = \frac{-20}{5} \\ x = \frac{-20}{5} + \frac{3}{5} = \frac{-17}{5} \]Answer: The other number is \(\frac{-17}{5}\)

Q6: The sum of two rational numbers is \(\frac{-5}{4}\). If one of them is -3, find the other.

Let the other rational number be \(x\).
According to the question: \[ x + (-3) = \frac{-5}{4} \]Step 1: Transpose -3 to the RHS \[ x = \frac{-5}{4} + 3 \]Step 2: Convert 3 to a rational number with denominator 4 \[ 3 = \frac{12}{4} \\ x = \frac{-5}{4} + \frac{12}{4} \]Step 3: Add the rational numbers \[ x = \frac{-5 + 12}{4} = \frac{7}{4} \]Answer: The other number is \(\frac{7}{4}\)

Q7: What should be added to \(\frac{-5}{6}\) to get \(\frac{-2}{3}\)?

Let the required number be \(x\).
According to the question: \[ x + \frac{-5}{6} = \frac{-2}{3} \]Step 1: Transpose \(\frac{-5}{6}\) to the RHS \[ x = \frac{-2}{3} – \frac{-5}{6} \]Step 2: Replace minus-minus with plus \[ x = \frac{-2}{3} + \frac{5}{6} \]Step 3: Take LCM of denominators LCM of 3 and 6 is 6. \[ \frac{-2}{3} = \frac{-4}{6} \]Step 4: Add the rational numbers \[ x = \frac{-4}{6} + \frac{5}{6} = \frac{1}{6} \]Answer: \(\frac{1}{6}\) should be added to \(\frac{-5}{6}\) to get \(\frac{-2}{3}\)

Q8: What should be added to \(\frac{2}{5}\) get -1?

Let the required number be \(x\).
According to the question: \[ x + \frac{2}{5} = -1 \]Step 1: Transpose \(\frac{2}{5}\) to the RHS \[ x = -1 – \frac{2}{5} \]Step 2: Express -1 with denominator 5 \[ -1 = \frac{-5}{5} \] So, \[ x = \frac{-5}{5} – \frac{2}{5} \]Step 3: Subtract the numerators \[ x = \frac{-5 – 2}{5} = \frac{-7}{5} \]Answer: \(\frac{-7}{5}\) should be added to \(\frac{2}{5}\) to get -1

Q9: What should be subtracted from \(\frac{-3}{4}\) to get \(\frac{-5}{6}\)?

Let the required number be \(x\).
According to the question: \[ \frac{-3}{4} – x = \frac{-5}{6} \]Step 1: Transpose \(x\) to RHS and \(\frac{-5}{6}\) to LHS \[ x = \frac{-3}{4} – \left(\frac{-5}{6}\right) \\ x = \frac{-3}{4} + \frac{5}{6} \]Step 2: Find LCM of denominators 4 and 6 \[ \text{LCM of 4 and 6} = 12 \]Step 3: Convert to like denominators \[ \frac{-3}{4} = \frac{-9}{12}, \quad \frac{5}{6} = \frac{10}{12} \]Step 4: Add the fractions \[ x = \frac{-9}{12} + \frac{10}{12} = \frac{1}{12} \]Answer: \(\frac{1}{12}\) should be subtracted from \(\frac{-3}{4}\) to get \(\frac{-5}{6}\)

Q10: What should be subtracted from \(\frac{-2}{3}\) to get 1?

Let the required number be \(x\).
According to the question: \[ \frac{-2}{3} – x = 1 \]Step 1: Transpose \(x\) to RHS and 1 to LHS \[ x = \frac{-2}{3} – 1 \]Step 2: Express 1 as a fraction with denominator 3 \[ 1 = \frac{3}{3} \]Step 3: Subtract the fractions \[ x = \frac{-2}{3} – \frac{3}{3} = \frac{-5}{3} \]Answer: \(\frac{-5}{3}\) should be subtracted from \(\frac{-2}{3}\) to get 1