More on Exponents

So now we know that \[
{x}^{3}
\] is shorthand math notation for x × x × x. In exponents, the x is called the base and the 3 is called the exponent or the power of x. Anything to the power of 2 is referred to as squared and anything to the power of 3 is referred to as cubed. These terms come from the formulas for finding the areas of a square and the volume of a cube.

There are several rules involving exponents that allow us to simplify expressions. So let’s look at

\[
\begin{array}{c}
{{x}^{2}\hspace{0.33em}\times\hspace{0.33em}{x}^{3}}\\
{\Longrightarrow\hspace{0.33em}{x}^{2}\hspace{0.33em}\times\hspace{0.33em}{x}^{3}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}\times\hspace{0.33em}{x}{)(}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}}
\end{array}
\]

 

So it appears that you can get the final result simply by adding the exponents together, this is correct as long as the exponents apply to the same base. For any two numbers a and b:

\[
{x}^{a}\hspace{0.33em}\times\hspace{0.33em}{x}^{b}\hspace{0.33em}{=}\hspace{0.33em}{x}^{a}{x}^{b}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{a}{+}{b}}
\]

Now you remember that when you see x, there is an implied “1” in front of it. Well, when dealing with exponents, if there is no visible exponent, there is an implied “1” there as well since \[
{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]. So what about a rule for division?

\[
\frac{{x}^{3}}{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}{x}}{\rlap{/}{x}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{x}}\hspace{0.33em}{=}\hspace{0.33em}\frac{x}{1}\hspace{0.33em}{=}\hspace{0.33em}{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

 

Because of the common factors of x in the numerator and denominator, I can cancel these away, leaving “1” in their place. So it appears that to get the final result, you just subtract the exponent in the denominator from the one in the numerator. Again, this is correct as long as the base is the same:

\[
\frac{{x}^{a}}{{x}^{b}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{a}{-}{b}}
\]

 

There are a few more things about exponents I want to cover before we use these in equations. This will be covered in my next post.

Exponents

Now let’s add another maths symbol: exponents. Now multiplication is really successive adding: 2 × 3 = 2 + 2 + 2 or 3 + 3. In other words, 2 × 3  can be thought of as adding 2 three times or adding 3 two times. Well exponents are indicating successive multiplication. Example:

\[
\begin{array}{l}
{{3}^{2}\hspace{0.33em}{=}\hspace{0.33em}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}\\
{{2}^{3}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{2}}\\
{{x}^{3}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}}
\end{array}
\]

I will expand on this tomorrow.

 

 

Algebra and Fractions

In yesterday’s post, we solved \[\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\]. We saw that when we first multiplied both side of the equation by 2, we had to multiply the entire left side. Let’s now look at this equation: \[\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{12}\].

Again, it appears that the first thing to do in our quest to get by itself is to multiply both sides by 2. But this time, the whole left side of the equation is divided by 2, not just x like it was in yesterday’s equation. So,

\[{2}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{2}\hspace{0.33em}\times\hspace{0.33em}{12}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{2}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{24}\]

 

Now let me stop here to explain a bit more of what’s happening. I put the implied “1” below the “2” so you can see what the next step should be. Remember that when multiplying two fractions together, you multiply the numerators together and the denominators together. But realise that when multiplying the numerators, the “2” is multiplying the whole expression “3 + 3x” so I will need to use brackets to indicate that:

\[\frac{2}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}}{2}\hspace{0.33em}{=}\hspace{0.33em}{24}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{{1}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}\]

 

Now we can see that we can cancel the “2”s, leaving the stuff in the brackets:

\[\begin{array}{l}
{\frac{{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{{1}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{2}{(}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}}{\rlap{/}{2}}\hspace{0.33em}{=}\hspace{0.33em}{24}}\\
{\Longrightarrow\hspace{0.33em}{3}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{24}}
\end{array}\]

 

Now we can subtract the 3 and the other methods we have covered to solve the equation:

3 + 3x -3 = 24 – 3 ⇒ 3x = 21

\[\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{21}{3}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{7}\]

Algebra continued

Let’s solve \[\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\] two different ways.

The first way is to remove the “2” from the denominator and get the “x” by itself. We can do this by multiplying by “2” so the “2”s will cancel. But when you multiply a side of an equation by a number, you must multiply the whole side. So, multiplying both sides of our equation by 2 gives:

\[\begin{array}{l}
{\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{2}{(}\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{2}{(}{42}{)}}\\
{\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{2}x}{\rlap{/}{2}}\hspace{0.33em}{+}\hspace{0.33em}{2}{(}{3}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{84}\hspace{0.33em}\Longrightarrow{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{x}\hspace{0.33em}{=}\hspace{0.33em}{84}}
\end{array}\]

 

Now notice I used the distributive property and properties of fractions to do the above work. We are now left with two like terms on the left, which you remember, can be added together (remember there is an implied “1” in front of x. So now we have

\[{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{x}\hspace{0.33em}{=}\hspace{0.33em}{84}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{7}{x}\hspace{0.33em}{=}\hspace{0.33em}{84}
\]

 

So what do we do here? To get x by itself, let’s divide both sides by 7:

\[\frac{\rlap{–}{7}x}{\rlap{–}{7}}\hspace{0.33em}{=}\hspace{0.33em}\frac{84}{7}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{12}\]

 

Let’s try a different way that’s a bit quicker. Remember that \[\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}\frac{x}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}{x}\]. So,

\[\frac{x}{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{0}{.}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{42}\]

 

Now the two terms on the left are like terms and we can add them together to get 3.5x = 42. Just as before, we can remove the “3.5” in front of the x be dividing both side by 3.5. It doesn’t matter if that is a decimal number – the same rules of cancelling apply:

\[\frac{3.5x}{3.5}\hspace{0.33em}{=}\hspace{0.33em}\frac{42}{3.5}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{12}\]

 

The same answer but a bit quicker since we didn’t have to work with the fraction first.

Fractions and Decimals

I want to do more interesting algebra problems but there is a bit more  to talk about regarding fractions.

Now remember that multiplying fractions is relatively simple:

\[\frac{2}{3}\hspace{0.33em}\times\hspace{0.33em}\frac{4}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{4}}{{3}\hspace{0.33em}\times\hspace{0.33em}{5}}\hspace{0.33em}{=}\hspace{0.33em}\frac{8}{15}\]

 

This and the fact that a “1” can be assumed anywhere as long as it is multiplying or dividing a number can make the same fraction look different. For example, dividing a number say x by 2, is the same as multiplying x by 1/2:

\[{x}\hspace{0.33em}\div\hspace{0.33em}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{x}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{x}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}{x}\]

 

If some of this is confusing, please review my previous post on fractions.

Now decimals are really fractions in disguise where the denominators are multiples of 10. So,

\[{0}{.}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{5}{10}{,}\hspace{0.33em}{0}{.}{67}\hspace{0.33em}{=}\hspace{0.33em}\frac{67}{100}{,}\hspace{0.33em}{0}{.}{108}\hspace{0.33em}{=}\hspace{0.33em}\frac{108}{1000}\]

 

Notice a couple of things here. The number of decimal places to the right of the “.” is the same as the number of “0”s in the denominator. Also, for pure decimals where there is no number to the left of “.”, it is customary to put a “0” because the “.” by itself is easy to miss.

Now, if you were to divide 1 by 2, long hand or on your calculator, you would get 0.5 as the answer.  Now 1 divided by 2 is indicated by the fraction \[\frac{1}{2}\]. All fractions have a decimal equivalent which you can find by doing the indicated division. Another way to show that \[\frac{1}{2}\] = 0.5 is

\[\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{5}{5}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}\times\hspace{0.33em}{5}}{{2}\hspace{0.33em}\times\hspace{0.33em}{5}}{=}\hspace{0.33em}\frac{5}{10}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\]

 

So another way to show x ÷ 2 is

\[\frac{x}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{2}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}{x}\]

 

So in the future, when we solve equations where fractions or divisions are present, we can change their form to suit us as we need to in order to solve the equation or express the answer in a desired form. Let’s see this tomorrow.

Algebra Continued

OK. I think we are ready for more complex equations to solve. We have covered the distributive property, fractions, and signs. If you do not understand any of the steps below, please see previous blogs in the “Algebra” category or use the tags to go to any specific area.

Let me introduce another symbol frequently used when doing any maths development. “⇒” means “it follows that” or “after doing some math, you get”. I will use it below.

Let’s solve for x in the equation 3x + 18 = 117. When I say “solve for x” I mean find the value of x that makes the equation a true equation. Using algebra’s prime directive: you can do any legal math to any side of an equation as long as you do the same math on both sides. So we will use this to get the equation to look like x = some number. So what guides you in this process is to do the math on both sides that gets x by itself.

Well the first thing you may want to do to our equation is to get rid of the 18 on the left side. I can do this by subtracting 18 from the left side, but I must do the same on the right:

3x + 18 = 117 ⇒ 3x + 18 -18 = 117 -18 ⇒ 3x + 0 = 99 ⇒ 3x = 99

Did you follow all that? Subtracting 18 from 18 is zero, and zero added to anything is the same anything so the zero can be removed. So what do we do with 3x = 99 to get x by itself? This is where the post on fractions comes in. If I divide the left side by 3 by using fraction notation, I will have a common factor of 3 in the numerator and denominator, and these can be cancelled, that is replaced with 1’s which can be assumed to be there:

\[\begin{array}{c}
{{3}{x}\hspace{0.33em}{=}\hspace{0.33em}{99}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{3x}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{99}{3}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}{33}}\\
{{x}\hspace{0.33em}{=}\hspace{0.33em}{33}}
\end{array}\]

So we have come up with the answer that x = 33. Did we make any mistakes? Well let’s check our answer by substituting 33 in place of x in the original equation and see if we get a true equation:

3(33) + 18 = 117 ⇒ 99 + 18 = 117 ⇒ 117 = 117

Well that’s as true as it gets. Good job!

The Distributive Property and Like Terms

Now that you know what a factor is, we can use that knowledge to solve more interesting equations than x + 7 = 10. But first, another maths shortcut and a property that is used all the time in algebra.

Remember that multiplication is implied with two sets of brackets next to each other:

(2)(-7) = -14

Well the same thing applies when there are unknowns:

3 × x = 3x, a × b = ab

Also, notice that when you see just x, you can think of that as 1x where the “1” is left off as anything multiplied by “1” is the same anything. Now let’s look at one of the most used properties in maths: the distributive property.

If you had an expression like 3(2 + 4), which means 3 × (2 + 4), you could first add the numbers in the brackets then multiply: 3(2 + 4) = 3(6) = 18. But another way is to first “distribute” the 3 among the terms in the brackets:

3(2 + 4) = 3(2) + 3(4) = 6 + 12 = 18

This holds true for any numbers:

4(6 – 5) = 4(6) – 4(5) = 24 – 20 = 4

-3(5 – 3) = (-3)(5) – (-3)(3) = -15 + 9 = -6 (Remember the sign rules?)

This is called the distributive property. But it’s not that useful to use with known numbers as I would normally just do the math in the brackets first. Now let’s use this property with unknowns:

x(5 – 3) = 5x – 3x. Now you might want to write x5 – x3, but it is customary to write the known number first, then the unknown. You can do this because when you multiply numbers together, it doesn’t matter what order you do the multiplication (this is another property called the commutative property).

Now notice that if you first did the calculation in the brackets first, you would get the answer 2x. Well because of the distributive property, you can add things like 3x and 2x together to get 5x because you can un-distribute the x to get

3x + 2xx(3 + 2) = 5x

This will work for any combination of a number multiplying an unknown. When the unknown is the same letter, or combination of letters, these are called like terms and you can add them together because of the distributive property.

14a – 11a = 3a, because 14a and -11a are like terms, however,

3x + 4a cannot be added together because the unknowns are different.

More examples:

-2y + 3y = 1yy (remember the “1” doesn’t need to be shown)

5m – 2m = 3m

102x + 33x = 135x

20x – 5x + 2y = 15x + 2y (only the like terms can be added)

Fractions and Division

One more post before I return to core algebra. We need to look a bit more at division and fractions.

Now, as you’ve seen, something like 8 ÷ 2 indicates division, but another way to show exactly the same thing is \[\frac{8}{2}\]. In other words, fractions are just another way to show division. Now before I expand on this, let’s review how fractions multiply together.

When two fractions are to be multiplied, the process is very simple. You just multiply the numerators (the numbers above the line) together and the denominators (the numbers below the line) together.

\[\frac{1}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{3}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}\times\hspace{0.33em}{3}}{{2}\hspace{0.33em}\times\hspace{0.33em}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{8}\]

 

Now this can be used to our advantage to simplify fractions. Each of the numbers in the above example are called “factors”. Factors are things that are multiplied together. So if we can show the factors of the parts of a fraction, we can effectively cancel factors that are common between the numerator and the denominator because we can split off the common factors as \[\frac{\mathrm{number}}{\mathrm{number}}\] and any number divided by itself is 1 and anything multiplied by 1 is the same anything.

\[\frac{8}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}\hspace{0.33em}\times\hspace{0.33em}{4}}{{2}\hspace{0.33em}\times\hspace{0.33em}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{2}\hspace{0.33em}\times\hspace{0.33em}\frac{4}{1}\hspace{0.33em}{=}\hspace{0.33em}{1}\hspace{0.33em}\times\hspace{0.33em}{4}\hspace{0.33em}{=}\hspace{0.33em}{4}
\]

 

A shortcut version of this is

\[\frac{8}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{4}}{\rlap{/}{2}\hspace{0.33em}\times\hspace{0.33em}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{1}\hspace{0.33em}{=}\hspace{0.33em}{4}\]

 

So note that when you cross out the only factor in the numerator or denominator, a “1” is left, and this “1” can be left out of the result since it does not change the value of the remaining numbers. Also note that this works for known and unknown numbers such as  x which we will see in my next post.

A couple more examples:

\[\begin{array}{c}
{\frac{16}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{4}}{\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{1}}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\
{\frac{6}{9}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{2}}{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{3}}
\end{array}\]

 

Signs and Multiplication and Division

Before I start this topic, let me demonstrate a shortcut in maths. I can show a multiplication by using the “×” symbol. For example, 8 × 2 = 16. But another way, which is especially useful when dealing with negative numbers is to use brackets with no “×” symbols: (8)(2) = 16, that is (8)(2) means 8 × 2.

So, if two positive numbers are multiplied or divided by one another, you already know that the result is positive.

(8)(2) = 16 (Remember the “+” symbol is assumed to be there for positive numbers)

8 ÷ 2 = 4

Now if the signs of the numbers are different, the result of the arithmetic is negative:

(-8)(2) = (8)(-2) = -16

(8) ÷ (-2) = (-8) ÷ (2) = -4

The reason for this is because multiplication is really just successive adding. For example, (-8)(2) is a shortcut for adding (-8) twice, That is (-8)(2) means (-8) + (-8) which is -16.

Now if both numbers are negative, the result is positive. It’s that double negative effect again. So

(-8)(-2) = 16

(-8) ÷ (-2) = 4

These rules are easy to remember: Like signs are positive, unlike signs are negative.

Tomorrow, let’s talk about fractions and more maths shortcuts.

Negative Numbers and Signs

There are several topics we need to cover before we can continue our discussion on algebra so that we can solve more interesting equations. The first topic is Negative Numbers and introducing another interpretation of the “+” and “-” symbols. First, negative numbers.

All real numbers can be located on the number line below. By “real” I mean numbers you are familiar with as opposed to numbers called “complex” which we will cover some time in the far future.

The numbers to the right of 0 are called positive numbers. They can technically have the “+” symbol in front of them, but when a number is positive, the “+” sign can be removed and the number is assumed to be positive if there is no “+” in front. The numbers to the left of 0 are called negative numbers. The “+” and “-” symbols are called signs and this introduces another way to view these: one way that you are familiar with is as the arithmetic operation (+ means add, – means subtract), but the other interpretation is as the sign of a number. As you will see, these interpretations can be interchanged quite freely as you work with equations.

Numbers at the tick marks are the integers. Numbers between the tick marks are fractions or other kind of number called irrational. But this post is mainly about negative numbers.

Negative numbers can represent many things you are familiar with: money you owe instead of have, distance to the west of a city instead of distance to the right, deceleration rate instead of acceleration, distance down instead of up, etc. So how do you work with negative numbers?

Now I will use brackets (parenthesis if you’re from the States) to separate numbers with their sign from the arithmetic symbols. Let’s look at the following examples:

(+7) + (+3) = 7 + 3 = 10 because the + sign in front of the numbers can be assumed and removed leaving only the + (plus) arithmetic symbol. So on the number line, you can represent this by starting at 7, then moving to the right 3 tick marks where you arrive at 10.

(+7) – (+3) = 7 – 3 = 4 for the same reason but this time the -(subtract) arithmetic operation remains.So on the number line, you can represent this by starting at 7, then moving to the left 3 tick marks where you arrive at 4. Now this is where it gets interesting.

(+7) + (-3) = 7 – 3 = 4. Adding a negative number is the same as subtracting a positive number. This is where I said the sign of a number and the arithmetic operation it represents can be freely interchanged and swapped. So you see from this and the last example, if you see “+” followed by “-” or vice versa, you can replace both with just the “-” symbol. What about

(+7) – (-3) = 7 + 3 = 10. Think of subtracting a negative number as having the same effect as a double negative in English (I will not spend no money technically is saying that you will spend money). So from this example and the first one, if you see “+” followed by “+” or “-” followed by “-“, you can replace them both by a “+” symbol.

(-7) + (+3) = -7 + 3 = -4. Start at -7 on the number line and move 3 ticks to the right, landing on -4.

(-7) – (-3) = -7 + 3 = -4. Same as above.

(-7) – (+3) = -7 – 3 = -10. Start at -7 on the number line and move to the left 3 ticks to land on -10.

Tomorrow, I will cover multiplying numbers of varying signs.