Rational Numbers (Complete Chapter) interactive worksheet | Live ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

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

#### Solution:
Let the unknown number be \( x \). According to the problem:
\[
x + (-5) = -\frac{4}{3}
\]
Simplify:
\[
x - 5 = -\frac{4}{3}
\]
Add 5 to both sides:
\[
x = -\frac{4}{3} + 5
\]
Convert 5 to a fraction with a denominator of 3:
\[
5 = \frac{15}{3}
\]
So:
\[
x = -\frac{4}{3} + \frac{15}{3} = \frac{-4 + 15}{3} = \frac{11}{3}
\]

Answer:
\[
\boxed{\frac{11}{3}}
\]

---

Problem 2:

The sum of two rational numbers is \(-8\). If one of the numbers is \(-\frac{15}{7}\), find the other.

#### Solution:
Let the unknown number be \( y \). According to the problem:
\[
y + \left(-\frac{15}{7}\right) = -8
\]
Simplify:
\[
y - \frac{15}{7} = -8
\]
Add \(\frac{15}{7}\) to both sides:
\[
y = -8 + \frac{15}{7}
\]
Convert \(-8\) to a fraction with a denominator of 7:
\[
-8 = -\frac{56}{7}
\]
So:
\[
y = -\frac{56}{7} + \frac{15}{7} = \frac{-56 + 15}{7} = \frac{-41}{7}
\]

Answer:
\[
\boxed{-\frac{41}{7}}
\]

---

Problem 3:

What should be added to \(-\frac{7}{8}\) so as to get \(\frac{5}{9}\)?

#### Solution:
Let the number to be added be \( z \). According to the problem:
\[
-\frac{7}{8} + z = \frac{5}{9}
\]
Solve for \( z \):
\[
z = \frac{5}{9} - \left(-\frac{7}{8}\right) = \frac{5}{9} + \frac{7}{8}
\]
Find a common denominator for 9 and 8, which is 72:
\[
\frac{5}{9} = \frac{5 \times 8}{9 \times 8} = \frac{40}{72}, \quad \frac{7}{8} = \frac{7 \times 9}{8 \times 9} = \frac{63}{72}
\]
So:
\[
z = \frac{40}{72} + \frac{63}{72} = \frac{40 + 63}{72} = \frac{103}{72}
\]

Answer:
\[
\boxed{\frac{103}{72}}
\]

---

Problem 4:

What number should be added to \(-\frac{5}{11}\) so as to get \(\frac{26}{33}\)?

#### Solution:
Let the number to be added be \( w \). According to the problem:
\[
-\frac{5}{11} + w = \frac{26}{33}
\]
Solve for \( w \):
\[
w = \frac{26}{33} - \left(-\frac{5}{11}\right) = \frac{26}{33} + \frac{5}{11}
\]
Find a common denominator for 33 and 11, which is 33:
\[
\frac{5}{11} = \frac{5 \times 3}{11 \times 3} = \frac{15}{33}
\]
So:
\[
w = \frac{26}{33} + \frac{15}{33} = \frac{26 + 15}{33} = \frac{41}{33}
\]

Answer:
\[
\boxed{\frac{41}{33}}
\]

---

Problem 5:

What number should be added to \(-\frac{5}{7}\) to get \(-\frac{2}{3}\)?

#### Solution:
Let the number to be added be \( v \). According to the problem:
\[
-\frac{5}{7} + v = -\frac{2}{3}
\]
Solve for \( v \):
\[
v = -\frac{2}{3} - \left(-\frac{5}{7}\right) = -\frac{2}{3} + \frac{5}{7}
\]
Find a common denominator for 3 and 7, which is 21:
\[
-\frac{2}{3} = -\frac{2 \times 7}{3 \times 7} = -\frac{14}{21}, \quad \frac{5}{7} = \frac{5 \times 3}{7 \times 3} = \frac{15}{21}
\]
So:
\[
v = -\frac{14}{21} + \frac{15}{21} = \frac{-14 + 15}{21} = \frac{1}{21}
\]

Answer:
\[
\boxed{\frac{1}{21}}
\]

---

Problem 6:

What should be subtracted from \(\left(\frac{3}{4} - \frac{2}{3}\right)\) to get \(-\frac{1}{6}\)?

#### Solution:
First, simplify \(\frac{3}{4} - \frac{2}{3}\):
Find a common denominator for 4 and 3, which is 12:
\[
\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}, \quad \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}
\]
So:
\[
\frac{3}{4} - \frac{2}{3} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12}
\]
Let the number to be subtracted be \( u \). According to the problem:
\[
\frac{1}{12} - u = -\frac{1}{6}
\]
Solve for \( u \):
\[
-u = -\frac{1}{6} - \frac{1}{12}
\]
Find a common denominator for 6 and 12, which is 12:
\[
-\frac{1}{6} = -\frac{1 \times 2}{6 \times 2} = -\frac{2}{12}
\]
So:
\[
-u = -\frac{2}{12} - \frac{1}{12} = \frac{-2 - 1}{12} = -\frac{3}{12} = -\frac{1}{4}
\]
Thus:
\[
u = \frac{1}{4}
\]

Answer:
\[
\boxed{\frac{1}{4}}
\]

---

Problem 7:

Evaluate \((-5 \times \frac{2}{15}) - (-6 \times \frac{2}{9})\).

#### Solution:
First, simplify each term:
\[
-5 \times \frac{2}{15} = \frac{-5 \times 2}{15} = \frac{-10}{15} = -\frac{2}{3}
\]
\[
-6 \times \frac{2}{9} = \frac{-6 \times 2}{9} = \frac{-12}{9} = -\frac{4}{3}
\]
So the expression becomes:
\[
-\frac{2}{3} - \left(-\frac{4}{3}\right) = -\frac{2}{3} + \frac{4}{3} = \frac{-2 + 4}{3} = \frac{2}{3}
\]

Answer:
\[
\boxed{\frac{2}{3}}
\]

---

Problem 8:

Evaluate \(\left(-\frac{9}{4} \times \frac{5}{3}\right) + \left(\frac{13}{2} \times \frac{5}{6}\right)\).

#### Solution:
First, simplify each term:
\[
-\frac{9}{4} \times \frac{5}{3} = \frac{-9 \times 5}{4 \times 3} = \frac{-45}{12} = -\frac{15}{4}
\]
\[
\frac{13}{2} \times \frac{5}{6} = \frac{13 \times 5}{2 \times 6} = \frac{65}{12}
\]
So the expression becomes:
\[
-\frac{15}{4} + \frac{65}{12}
\]
Find a common denominator for 4 and 12, which is 12:
\[
-\frac{15}{4} = -\frac{15 \times 3}{4 \times 3} = -\frac{45}{12}
\]
So:
\[
-\frac{15}{4} + \frac{65}{12} = -\frac{45}{12} + \frac{65}{12} = \frac{-45 + 65}{12} = \frac{20}{12} = \frac{5}{3}
\]

Answer:
\[
\boxed{\frac{5}{3}}
\]

---

Problem 9:

By what rational number should we multiply \(-\frac{15}{56}\) to get \(-\frac{5}{7}\)?

#### Solution:
Let the rational number be \( r \). According to the problem:
\[
r \times \left(-\frac{15}{56}\right) = -\frac{5}{7}
\]
Solve for \( r \):
\[
r = \frac{-\frac{5}{7}}{-\frac{15}{56}} = \frac{5}{7} \div \frac{15}{56} = \frac{5}{7} \times \frac{56}{15} = \frac{5 \times 56}{7 \times 15} = \frac{280}{105} = \frac{8}{3}
\]

Answer:
\[
\boxed{\frac{8}{3}}
\]

---

Problem 10:

By what number should \(-\frac{33}{8}\) be divided to get \(-\frac{11}{2}\)?

#### Solution:
Let the number be \( d \). According to the problem:
\[
-\frac{33}{8} \div d = -\frac{11}{2}
\]
This can be rewritten as:
\[
-\frac{33}{8} \times \frac{1}{d} = -\frac{11}{2}
\]
Solve for \( d \):
\[
\frac{1}{d} = \frac{-\frac{11}{2}}{-\frac{33}{8}} = \frac{11}{2} \div \frac{33}{8} = \frac{11}{2} \times \frac{8}{33} = \frac{11 \times 8}{2 \times 33} = \frac{88}{66} = \frac{4}{3}
\]
So:
\[
d = \frac{3}{4}
\]

Answer:
\[
\boxed{\frac{3}{4}}
\]

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Final Answers:

\[
\boxed{\frac{11}{3}, -\frac{41}{7}, \frac{103}{72}, \frac{41}{33}, \frac{1}{21}, \frac{1}{4}, \frac{2}{3}, \frac{5}{3}, \frac{8}{3}, \frac{3}{4}}
\]