How to Use Algebra in Arithmetic

algebra-arithmetic

USING ALGEBRA IN ARITHMETIC

It’s interesting that algebra can be useful in arithmetic.

Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic.

However, you can apply your algebra skills to arithmetic problems.

Following is an example.

Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy.

One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together.

Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58.

You may not have realized it at the time, but your teacher was applying the distributive property of mathematics.

When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364.

Now let’s see how algebra can supply an alternative solution.

I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator.

So let’s write 58 as 60 minus 2.

Then the problem is 58 × 58 = (60 – 2) × (60 – 2).

Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.”

One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y².

Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2.

58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2).

Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2).

58 × 58 = 3600 – 240 + 4 = 3364.

Note that (–2) × (–2) = +4 because the two minus signs cancel out.

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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Tips for Finding the Slope of a Straight Line

Graphing

HOW TO FIND THE SLOPE OF A STRAIGHT LINE

Slope is a measure that indicates how steep or shallow a straight line is:

  • A line with greater slope is steeper.
  • A line with less slope is shallower.
  • A horizontal line has zero slope.

Slope Sign

Slope can be positive or negative:

  • A line with positive slope slants upward.
  • A line with negative slope slants downward.
  • A line with zero slope is horizontal.

Slope Sign B

The slope of a straight line equals rise over run.

  • The rise between two points is vertical. It’s the change in y.
  • The run between two points is horizontal. It’s the change in x.

Slope Eqn

TIPS FOR FINDING SLOPE

From the graph of a straight line, determine the slope as follows:

  • Mark two points on the line.
  • Read the x- and y-coordinates of the two points, (x1, y1) and (x2, y2).
  • Subtract y2 – y1 to get the rise.
  • Subtract x2 – x1 to get the run.
  • Divide the rise by the run.

Slope Triangle

Here are a few tips:

  • When choosing the two points, try to find points where both x and y are easy to read without interpolating. This isn’t always possible: In that case, at least one coordinate should be easy to read without interpolating.
  • Choose two points far apart. This reduces the relative error in interpolating.
  • Make sure that both points lie on the straight line. Don’t choose a point that’s close to the line, but doesn’t lie on it.

EXAMPLE OF HOW TO DETERMINE SLOPE

Example: Find the slope of the straight line in the graph below.

Slope Example

Solution: First, choose two points on the line. Ideally, these points should be far apart and easy to read.

In this case, it’s easy to read both the x- and y-coordinates for the leftmost and rightmost points shown in the graph. So let’s choose those.

Slope Example 2

  • The leftmost point has coordinates (0, 3).
  • The rightmost point has coordinates (10, 8).

Subtract the y-values to determine the rise:

y2 – y1 = 8 – 3 = 5

Subtract the x-values to determine the run:

x2 – x1 = 10 – 0 = 10

(In coordinate graphing, recall that x is horizontal and y is vertical.)

Divide the rise by the run to find the slope:

Slope Example 3

The slope of the line is 0.5.

Check: You can check your answer as follows.

Look at the graph. Starting from (0, 3), the next point that’s easy to read is (2, 1).

From (0, 3) to (2, 1), the line goes one unit up (vertically) and 2 units over (horizontally).

The ratio of the rise to the run is 1 to 2. Divide the rise (of 1) by the run (of 2). The slope is 0.5. ♦

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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Fibonacci Sequence & a Cool Pattern

Image from ShutterStock.

Image from ShutterStock.

FIBONACCI SEQUENCE

The Fibonacci sequence adds consecutive terms:

  • 1
  • 1
  • 1 + 1 = 2
  • 2 + 1 = 3
  • 3 + 2 = 5
  • 5 + 3 = 8
  • 8 + 5 = 13
  • 13 + 8 = 21
  • 21 + 13 = 34
  • 34 + 21 = 55
  • 55 + 34 = 89

Since the last two terms were 55 and 89, we would add these together to get 89 + 55 = 144.

Then you would add 144 and 89 to make 233, and so on.

I saw a cool pattern involving the Fibonacci sequence recently at the Mathemagical Site:

Fibonacci Triples via Mathemagical Site

This involves Fibonacci triples.

A Fibonacci triple consists of three consecutive numbers from the Fibonacci sequence, such as:

  • 1, 1, 2
  • 1, 2, 3
  • 2, 3, 5
  • 3, 5, 8
  • 5, 8, 13
  • 8, 13, 21
  • 13, 21, 34

As shown on the Mathemagical Site, the square of the middle number is always one less or one more than the product of the first and third numbers:

FT

Here are a few examples:

  • (2, 3, 5): 3 x 3 = 2 x 5 – 1
  • (3, 5, 8): 5 x 5 = 3 x 8 + 1
  • (5, 8, 13): 8 x 8 = 5 x 13 – 1

Curious about this, I’ve been working through the algebra, and finally came up with an algebraic proof, which follows.

My proof is algebraic and not necessarily obvious, but since I worked it out, I thought I would share it here. 🙂

We begin with two facts about the Fibonacci sequence:

FT3

These are two different ways of saying that if you add two consecutive numbers from the Fibonacci sequence, you get the next number in the sequence.

Now solve for xn in each sequence:

FT4

Multiply these together:

FT2

Foil this out:

FT5

Recall that

FT6

Plug this into the first term on the right-hand side of the previous equation:

FT7

Distribute:

FT8

Two of these terms cancel (the remaining terms are rearranged):

FT9

Believe it or not, this basically concludes the proof. The remainder is basically interpreting this result.

This is a recursion relation that relates the square of the n-th term to the square of the previous term (xn-1 times itself).

Following is the Fibonacci sequence, labeling values of n:

  • n = 1 is 1.
  • n = 2 is 1.
  • n = 3 is 2.
  • n = 4 is 3.
  • n = 5 is 5.
  • n = 6 is 8.
  • n = 7 is 13.
  • n = 8 is 21.

Let’s plug in n = 3 and see what happens:

FT10

If instead you plug in n = 4, you get:

Ft11

Now just plug in these two expressions (x3x3 – x4x2 and x4x4 – x5x3) into the previous recursion relation and you can prove that all of the Fibonacci triples satisfy one of the following relations:

FT

That is, if x3x3 – x4x2 = 1 and x4x4 – x5x3 = -1, the previous recursion relation gives similar expressions for x5x5 – x6x4, x6x6 – x7x5, and so on.

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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Algebra with Roman Numerals

Algebra Roman Numerals

ALGEBRA WITH ROMAN NUMERALS

Aren’t you glad you don’t have to do algebra with Roman numerals?

The problem is: Xx + X = CX.

This translates to 10x + 10 = 110.

Subtract X from both sides: Xx = CX – X = C.

Translation from Roman numerals: 10x = 110 – 10 = 100.

Now divide both sides by the coefficient: x = C/X = X.

Not expressed in Roman numerals, it looks like this: x = 100/10 = 10.

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks