# Algebra I, Article 58.

In article 58, we will learn how to subtract one algebraic quantity from another.  Note that Algebra is big boy work and you need to read every line carefully and move forward one step at a time.

The lesson begins by showing us two simple examples from Arithmetic to show how a number of different operations affects the difference of quantities subtracted.  We then move from these simple Arithmetic examples to several general Algebraic problems.

Let’s begin by looking at the two Arithmetic examples:

1. Let it be required to subtract 5+3 from 9.

If we subtract 5 from 9, the remainder will be 9−5; but we wish to subtract, not only 5, but also 3. Hence, after we have subtracted 5, we must also subtract 3.  This gives for the remainder, 9−5−3, which is equal to 1.

2. Let it be required to subtract 5−3 from 9.

If we subtract 5 from 9, the remainder is 9−5; but the quantity to be subtracted is 3 less than 5. Hence, we have subtracted 3 too much.  We must, therefore, add 3 to 9−5, which gives for the true remainder, 9−5+3, or 7.

Now, let us move from simple Arithmetic example to think of this algebraically:

3. Let it now be required to subtract b−c from a.

If we take 6 from a, the remainder is a−b; but, in doing this, we have subtracted c too much; hence, to obtain the true result, we must add c. Therefore, to subtract b-c from a is equal to a−b+c.

If a=9,  b=5, and c=3, then, from 9 we must take 5−3.  This is equal to 9-5+3.

Note that the 3 is added because we are taking 3 less than 5 from 9.  We find that 9-5+3 = 7

For further illustration, take the following:

4. Here we will look at several examples, to practice working with these algebraic expressions.

First, let us take c−a from a.

Look at this problem.  What are we being asked to do?

We are being asked to take a less than c away from a.  This can be written algebraically as a−(c−a), which is equal to a−c+a, because taking away a less than c from a is the same as adding a to a−c.  Moreover, since we are adding a to a, we can re-write this expression as 2a−c.

All of this can be expressed, algebraically as:  a−(c−a) = a−c+a = 2a−c

Note how we have simplified this problem by the use of algebra.  Taking c−a from a is the same as taking c from 2 times a.

Second, let us take a−c from a.

Look at this problem.  What are we being asked to do?

We are being asked to take c less than a away from a.  This can be written algebraically as a−(a−c), which is equal to a−a+c, because taking away c less than a from a is the same as adding c to a−a.  Moreover, since we are subtracting a from a, we are left with c.

Note how we have simplified this problem by the use of algebra.  Taking a−c from a leaves c. All of this can be expressed, algebraically as:  a−(a−c) = a−a+c = c

Third, let us take a−b from a+b.

Look at this problem.  What are we being asked to do?

We are being asked to take b less than a away from a+b.  This can be written algebraically as a+b−(a−b), which is equal to a+b−a+b, because taking away b less than a from a+b is the same as adding b to a+b−a.  Moreover, since we are subtracting a from a, and adding b to b, we are left with 2b.

All of this can be expressed, algebraically as:  a+b−(a−b) = a+b−a+b = 2b

Note how we have simplified this problem by the use of algebra.  Taking a−b from a +b leaves 2b.

Observe that in each of the preceding examples, the signs of the subtrahend are changed from plus to minus, and from minus to plus. Hence, we can learn the following rule:

Rule to Find the Difference between Two Algebraic Quantities
1. Write the quantity to be subtracted under that from which it is to be taken, placing similar terms under each other.
2. Conceive the signs of all the terms of the subtrahend to be changed, and then reduce by the rule for Addition.

As for the examples that follow in the lesson, I have provided some explanation on the forum post here:  https://classicalliberalarts.com/study/mod/forum/discuss.php?d=368

Study through these explanations carefully and let me know if you need help–and where.

God bless,
Mr. William C. Michael