algebra x
semester finals review #1
name:
evaluate each expression. (denominators are the same)
1) $ \frac{4}{8} - \frac{3}{8} $
2) $ \frac{1}{2} + \frac{3}{2} $
3) $ \frac{11}{8} - \frac{13}{8} $
4) $ -\frac{6}{7} + \frac{11}{7} $
5) $ \frac{1}{2} - \frac{1}{2} $
6) $ \frac{3}{4} + \frac{3}{4} $
7) $ \frac{1}{3} + \frac{4}{3} - \frac{1}{3} $
8) $ -\frac{5}{3} + \frac{4}{3} + \frac{5}{3} $
9) $ -\frac{7}{4} + \frac{1}{4} + \frac{3}{4} $
10) $ \frac{3}{8} - \frac{9}{8} + \frac{3}{8} $
11) $ -\frac{1}{2} + \frac{1}{2} - \frac{3}{2} + \frac{3}{2} $
12) $ \frac{4}{3} + \frac{4}{3} + \frac{2}{3} - \frac{5}{3} $
13) $ -\frac{2}{5} + \frac{6}{5} - \frac{3}{5} + \frac{6}{5} $
14) $ -\frac{2}{3} - \frac{4}{3} + \frac{5}{3} - \frac{1}{3} $
evaluate each expression. (need to make a common denominator)
15) $ -1 + \frac{1}{5} $
16) $ -\frac{3}{8} - \frac{1}{2} $
17) $ -\frac{3}{2} - \frac{2}{3} $
18) $ -1 - \frac{7}{4} $
19) $ -2 - \frac{5}{3} $
20) $ 2 - \frac{7}{4} $
21) $ -\frac{3}{2} - \frac{3}{2} $
22) $ \frac{4}{3} - \frac{3}{2} $

Question

Accepted Answer

# Explanation:
## Step1: Solve $ -1 + \frac{1}{3} $
Convert $-1$ to thirds: $-1 = -\frac{3}{3}$. Then add: $ -\frac{3}{3} + \frac{1}{3} = \frac{-3 + 1}{3} = -\frac{2}{3} $.
## Step2: Solve $ -\frac{3}{8} - \frac{1}{2} $
Common denominator for 8 and 2 is 8. Convert $-\frac{1}{2}$ to eighths: $-\frac{1}{2} = -\frac{4}{8}$. Then subtract: $ -\frac{3}{8} - \frac{4}{8} = \frac{-3 - 4}{8} = -\frac{7}{8} $.
## Step3: Solve $ -\frac{3}{2} - \frac{2}{3} $
Common denominator for 2 and 3 is 6. Convert: $ -\frac{3}{2} = -\frac{9}{6} $, $ -\frac{2}{3} = -\frac{4}{6} $. Subtract: $ -\frac{9}{6} - \frac{4}{6} = \frac{-9 - 4}{6} = -\frac{13}{6} $.
## Step4: Solve $ -1 - (-\frac{7}{4}) $
Subtracting a negative is adding: $ -1 + \frac{7}{4} $. Convert $-1$ to fourths: $-1 = -\frac{4}{4}$. Add: $ -\frac{4}{4} + \frac{7}{4} = \frac{3}{4} $.
## Step5: Solve $ -2 - \frac{5}{3} $
Convert $-2$ to thirds: $-2 = -\frac{6}{3}$. Subtract: $ -\frac{6}{3} - \frac{5}{3} = \frac{-6 - 5}{3} = -\frac{11}{3} $.
## Step6: Solve $ 2 - (-\frac{7}{4}) $
Subtracting a negative is adding: $ 2 + \frac{7}{4} $. Convert 2 to fourths: $ 2 = \frac{8}{4} $. Add: $ \frac{8}{4} + \frac{7}{4} = \frac{15}{4} $.
## Step7: Solve $ -\frac{3}{2} - \frac{3}{2} $
Subtract numerators: $ \frac{-3 - 3}{2} = \frac{-6}{2} = -3 $.
## Step8: Solve $ \frac{4}{3} - (-\frac{3}{2}) $
Subtracting a negative is adding. Common denominator 6: $ \frac{8}{6} + \frac{9}{6} = \frac{17}{6} $.

# Answers (selected examples):
15) $ -\frac{2}{3} $
16) $ -\frac{7}{8} $
17) $ -\frac{13}{6} $
18) $ \frac{3}{4} $
19) $ -\frac{11}{3} $
20) $ \frac{15}{4} $
21) $ -3 $
22) $ \frac{17}{6} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Solve \( -1 + \frac{1}{3} \)

Step2: Solve \( -\frac{3}{8} - \frac{1}{2} \)

Step3: Solve \( -\frac{3}{2} - \frac{2}{3} \)

Step4: Solve \( -1 - (-\frac{7}{4}) \)

Step5: Solve \( -2 - \frac{5}{3} \)

Step6: Solve \( 2 - (-\frac{7}{4}) \)

Step7: Solve \( -\frac{3}{2} - \frac{3}{2} \)

Step8: Solve \( \frac{4}{3} - (-\frac{3}{2}) \)

Answer: