Calculate Differences: Solve 8 Tricky Algebra Problems

Hey guys! Today, we are diving into some algebra problems that involve calculating differences. This might sound straightforward, but we've got a mix of decimals, fractions, and negative numbers to keep things interesting. So, let's put on our math hats and get started! We will break down each problem step by step, so you can follow along and sharpen your skills. This is going to be a fantastic journey to improve your understanding of algebra.

a. $5.5 - 8.1$

When we're tackling decimal subtraction, it's crucial to keep track of our signs. In this case, we're subtracting a larger number (8.1) from a smaller number (5.5), so we know our answer will be negative. Think of it like this: you have $5.5, but you owe $8.1. How much are you still in debt? To find the difference, we subtract the smaller number from the larger number: $8.1 - 5.5$.

Let's perform the subtraction:

8.  1
- 5.  5
------
2.  6

So, the difference between 8.1 and 5.5 is 2.6. But remember, since we were subtracting a larger number from a smaller one, our answer is negative. Therefore, $5.5 - 8.1 = -2.6$. It's super important to keep the sign in mind; otherwise, we might end up with the wrong answer. Always double-check whether your result should be positive or negative!

b. $-7.34 - (-5.51)$

Here, we encounter a double negative, which can be a bit tricky. Remember, subtracting a negative number is the same as adding a positive number. So, $-7.34 - (-5.51)$ becomes $-7.34 + 5.51$. Now, we're adding a positive number to a negative number. To solve this, we again look at the absolute values of the numbers and subtract the smaller absolute value from the larger one. The absolute value of -7.34 is 7.34, and the absolute value of 5.51 is 5.51. We subtract 5.51 from 7.34:

7.  34
- 5.  51
------
1.  83

The difference is 1.83. Since the original number with the larger absolute value was -7.34, our answer will be negative. Thus, $-7.34 - (-5.51) = -1.83$. This type of problem emphasizes the importance of understanding how negative numbers interact with subtraction. Make sure you're comfortable with these rules!

c. $-1 - 2.5$

This one is relatively straightforward but still important to understand. We are subtracting 2.5 from -1. This means we are moving further into the negative side of the number line. Think of it as starting at -1 and then moving 2.5 units to the left. To find the result, we simply add the absolute values of the numbers and keep the negative sign. So, we add 1 and 2.5, which gives us 3.5. Since we're moving further into the negatives, the answer is -3.5. Therefore, $-1 - 2.5 = -3.5$. This kind of problem helps reinforce the basic concept of adding and subtracting on the number line.

d. $-5 - \frac{5}{3}$

Now, we're dealing with a fraction and a whole number. To subtract these, we first need to convert the whole number into a fraction with the same denominator as the other fraction. In this case, we have -5 and $\frac{5}{3}$. We can convert -5 into a fraction with a denominator of 3 by multiplying it by $\frac{3}{3}$: $-5 = -5 \times \frac{3}{3} = -\frac{15}{3}$. Now our problem looks like this: $-\frac{15}{3} - \frac{5}{3}$. Since we have a common denominator, we can subtract the numerators: -rac{15}{3} - \frac{5}{3} = -rac{15 + 5}{3} = -rac{20}{3}. We can leave our answer as an improper fraction, $-\frac{20}{3}$, or convert it to a mixed number. To do this, we divide 20 by 3, which gives us 6 with a remainder of 2. So, $-\frac{20}{3} = -6\frac{2}{3}$. This problem highlights the importance of working with fractions and converting between different forms.

e. $-8\frac{3}{8} - 10\frac{1}{6}$

This problem involves subtracting mixed numbers, which adds another layer of complexity. First, we need to convert the mixed numbers into improper fractions. To do this, we multiply the whole number by the denominator and add the numerator, then put the result over the original denominator. So, -8\frac{3}{8} = -rac{(8 \times 8) + 3}{8} = -rac{64 + 3}{8} = -rac{67}{8} and 10\frac{1}{6} = rac{(10 \times 6) + 1}{6} = rac{60 + 1}{6} = rac{61}{6}. Now our problem is -rac{67}{8} - rac{61}{6}. To subtract these fractions, we need a common denominator. The least common multiple of 8 and 6 is 24. We convert both fractions to have a denominator of 24: -rac{67}{8} = -rac{67 \times 3}{8 \times 3} = -rac{201}{24} and rac{61}{6} = rac{61 \times 4}{6 \times 4} = rac{244}{24}. Now we can subtract: -rac{201}{24} - rac{244}{24} = -rac{201 + 244}{24} = -rac{445}{24}. We can leave our answer as an improper fraction, or we can convert it to a mixed number. Dividing 445 by 24 gives us 18 with a remainder of 13. So, -rac{445}{24} = -18\frac{13}{24}. This problem is a great way to practice working with mixed numbers and finding common denominators.

f. $-\frac{1}{2} - \left(-\frac{5}{9}\right)$

Here, we have another double negative situation with fractions. Remember, subtracting a negative is the same as adding a positive. So, $-\frac{1}{2} - \left(-\frac{5}{9}\right)$ becomes $-\frac{1}{2} + \frac{5}{9}$. To add these fractions, we need a common denominator. The least common multiple of 2 and 9 is 18. We convert both fractions to have a denominator of 18: -\frac{1}{2} = -rac{1 \times 9}{2 \times 9} = -rac{9}{18} and $\frac{5}{9} = \frac{5 \times 2}{9 \times 2} = \frac{10}{18}$. Now we can add: $-\frac{9}{18} + \frac{10}{18} = \frac{-9 + 10}{18} = \frac{1}{18}$. This problem is a good review of fraction addition with negatives and the importance of finding common denominators.

g. $\frac{5}{8} - \left(-\frac{7}{8}\right)$

This problem is interesting because we are subtracting a negative fraction from a positive fraction. Again, subtracting a negative is equivalent to adding a positive. So, the expression $\frac{5}{8} - \left(-\frac{7}{8}\right)$ becomes $\frac{5}{8} + \frac{7}{8}$. Since both fractions already have the same denominator, we can simply add the numerators: $\frac{5}{8} + \frac{7}{8} = \frac{5 + 7}{8} = \frac{12}{8}$. Now, we can simplify this fraction. Both 12 and 8 are divisible by 4, so we divide both the numerator and the denominator by 4: $\frac{12}{8} = \frac{12 \div 4}{8 \div 4} = \frac{3}{2}$. We can also convert this improper fraction to a mixed number: $\frac{3}{2} = 1\frac{1}{2}$. This problem reinforces the concept of adding fractions with the same denominator and simplifying the result.

h. $-1\frac{1}{3} - 1\frac{2}{3}$

Lastly, we have another subtraction problem involving mixed numbers. As we did before, we'll first convert these mixed numbers to improper fractions. -1\frac{1}{3} = -rac{(1 \times 3) + 1}{3} = -rac{3 + 1}{3} = -rac{4}{3} and 1\frac{2}{3} = rac{(1 \times 3) + 2}{3} = rac{3 + 2}{3} = rac{5}{3}. Now we subtract the fractions: -rac{4}{3} - rac{5}{3}. Since they have the same denominator, we can subtract the numerators: -rac{4}{3} - rac{5}{3} = -rac{4 + 5}{3} = -rac{9}{3}. Finally, we simplify the fraction: -rac{9}{3} = -3. This final problem wraps up our practice with mixed numbers and improper fractions, demonstrating how to convert and simplify to arrive at the answer.

Wrapping up our algebraic adventure, we've tackled a variety of subtraction problems involving decimals, fractions, and mixed numbers. Remember, the key to mastering these problems is to understand the rules for working with negative numbers and fractions, and to practice, practice, practice! Keep up the great work, and you'll become an algebra ace in no time!