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In my last post, you saw a technique to solve equations when one side consists of factors (things multiplied together) and the other side is zero. Generally, if you have any number of factors with unknowns in them, the only way that the equation can be solved is by setting each factor to zero and solving for the unknown.

That is, if you have say four unknown numbers which multiplied together equal zero: abcd = 0, the only way that this can be true is if any of the unknown numbers are zero.

Let’s have a more complex example. Consider

(x – 4)(x + 7)(x-3)(x+5) = 0

Here are four expressions multiplied together, each with an unknown number x. This can be solved by setting each factor to zero as you can only get zero through multiplication if any of the things multiplied are zero themselves:

x – 4 = 0  ⇒  x = 4

x + 7 = 0  ⇒  x = -7

x – 3 = 0  ⇒  x = 3

x + 5 = 0  ⇒  x = -5

So there are four solutions to this equation.

This is in fact a technique frequently used to solve equations with powers of x

But there are many ways to factor and those skills will be covered in future posts.

I think we are ready for more interesting equations to solve. Given the knowledge from the past posts, you have the tools to solve:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}
\]

 

This equation is an example of things called quadratic equations. A quadratic equation is one that only has three kinds of terms: \[
{\mathrm{a}}{x}^{2}{,}\hspace{0.33em}{\mathrm{b}}{x}{,}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}{\mathrm{c}}
\] where the a, b, and c things are known numbers. I will refine this definition a bit later. Now let’s get back to our equation.

Remember, our goal is to do legal things to both sides of the equation to eventually get x = a known number. So you may notice that there is a +5 on both side of the equation. Well a good first step would be to subtract 5 from both sides and this would eliminate them:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{-}\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{-}\hspace{0.33em}{5}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}
\]

 

Now I’ve decided on the next step because I know what the answer is. Please bear with me and all will be revealed. So let’s now subtract 18x from both sides. Now remember yesterday I said there is a mental shortcut to doing this. The result can be obtained by just moving the 18x to the left side and change its sign. But showing the full detail:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{-}\hspace{0.33em}{18}{x}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{-}\hspace{0.33em}{18}{x}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

 

Notice that on the left side, I combined the like terms 7x and -18x and that the right side is now zero. In fact, this equation is now in the form:

\[
{\mathrm{a}}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{\mathrm{b}}{x}\hspace{0.33em}{+}\hspace{0.33em}{\mathrm{c}}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where in this case, a = 1, b = -11 and c = 0. This is the formal definition of a quadratic equation. The only restriction on a, b, or c is that “a” cannot be zero.

Now most quadratic equations can only be solved with something called the quadratic formula. We’ll eventually cover that, but this one can be solved using what we know now.

Now we will not be using square roots for this one. The next step will be to “un-distribute” the x from the terms on the left side:

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{11}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

 

So now we have two things multiplied together that equal zero. How can two numbers multiplied together equal zero? Well only if one of them is zero. So this sets up two possible solutions: either x = 0 or x – 11 = 0. If you go back to the original equation, you will see that substituting 0 for x will give the true equation 5 = 5. I will leave solving x – 11 = 0 to you, but the answer is x = 11. This is another solution to our equation.

A quadratic equation will have up to two solutions, and in this case, it does have two solutions, 0 and 11. Again, if the equation comes from a real world problem, only one of the answer may make sense so you would discard the other solution.

So let see how to solve \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}
\]. A good first step would be to remove the -7 from the left side. We do this by adding +7:

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}\hspace{0.33em}{+}\hspace{0.33em}{7}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\]

 

Now before I proceed further, after doing things like adding or subtracting a number to both sides of an equation, you start to notice mental shortcuts. For example, adding the 7 to both sides of the equation, effectively is the same as moving the 7 from the left side to the right side and changing the sign. This can be done with any term: move it to the other side of the equation and change the sign. By the way, a term is anything that is added or subtracted to or from the entire side of an equation. I’ll illustrate this further in future equations.

So now we have \[{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\]. So how do we get x by itself? From yesterday’s post, we know that:

\[
\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

So this suggests that we need to take the square root of both sides of the equation:

\[
{x}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[]{25}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{5}
\]

 

Did you remember that taking the square root results in two solutions? If you replace the x in the original equation with either +5 or -5, you will get a true equation 18 = 18.

Now many time the equation comes from a physical problem, like solving for a length. If so, then you can ignore the negative solution as that would not make sense. There are times, however, when the negative solution is the desired one.

Now eventually, I would like to be able to solve equations like \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}
\]. But you see that if I am to get x by itself, I need something to get rid of the exponent “2”. This is where square roots come in. But let’s first talk about inverse operations.

You may have already noticed that addition and subtraction are inverse operations because if you add a number to something and then subtract it or vice versa, you are left with the original something:

x + 3 – 3 = x,       x – 3 + 3 = x

The same can be said of multiplication and division:

\[
{3}{x}\hspace{0.33em}\div\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

The square root of a number is the inverse of squaring a number. Where squaring a number means “what do I get when I multiply a number by itself”, the square root of a number means “what number multiplied by itself is equal to the original number”. The math symbol which means “take the square root of this number” is called a radical and looks like\[
\sqrt{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}
\]. You will see later that there are other ways to indicate a square root. Now a few examples:

\[
\begin{array}{l}
{\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}}\\
{\sqrt{4}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\
{\sqrt{100}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{10}^{2}\hspace{0.33em}{=}\hspace{0.33em}{100}}
\end{array}
\]

 

So let’s illustrate the “inverseness” of this:

\[
\begin{array}{l}
{\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{5}}\\
{\sqrt{{2}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{2}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\
{\sqrt{{10}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{10}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{10}}
\end{array}
\]

 

Are you starting to see the potential of this to get rid of the “2” in \[
{x}^{2}
\]?

Now I’ve only been half truthful so far. For the square root examples I’ve shown you, I only showed you one of the possible solutions. Is 5 the only square root of 25? Think about it. Is there not another number that when multiplied by itself is 25? What about -5?  When you take the square root of a number, you actually are introducing two solutions:

\[
\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{5}
\]

The symbol ± means “plus or minus” and indicates that the two solutions are +5 and -5.

Let’s use this new tool to solve an equation on my next post.

In my last post, you saw another maths shortcut – exponents. But I used examples where the exponents were positive integers. It turns out that mathematically, any number, be it negative, fractional, or irrational, can be used as an exponent. Let’s expand our knowledge here by looking at exponents that are not positive integers.

In my last post, I presented the rule that

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

so that

\[
\frac{{x}^{3}}{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{3}{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{1}\hspace{0.33em}{=}\hspace{0.33em}{x}
\].

What if you have \[
\frac{{x}^{3}}{{x}^{3}}
\]?

Well, the rule says that

\[
\frac{{x}^{3}}{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{3}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{0}
\]

But we know that the same number divided by itself is 1. So it makes sense, and it makes maths consistent, if we define anything raised to the “0” power as 1. That is,

\[
{x}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
\] for any number x.

Now let’s go further. What if we have \[
\frac{{x}^{2}}{{x}^{3}}
\]? According to the rule, this is \[
{x}^{{2}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{-}{1}}
\]. What does a negative exponent mean? Well let’s do the same problem without using the rule:

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

 

So it appears that a negative exponent means that the number raised to a negative power, is the same as the number in the denominator raised to its positive power. This is true for any exponent:

\[
{x}^{{-}{a}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{a}}
\]

 

This means that factors in a fraction can be moved at will from the numerator to the denominator or vice versa, by just changing the sign of the exponents:

\[
\begin{array}{l}
{\frac{7}{{x}^{{-}{2}}}\hspace{0.33em}{=}\hspace{0.33em}{7}{x}^{2}}\\
{{3}{y}^{{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{{y}^{3}}}\\
{\frac{4{xy}^{5}}{{wz}^{{-}{6}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4{xy}^{5}{z}^{6}}{w}}
\end{array}
\]

 

However, be careful. You can only do this with factors, that is things that are multiplied together. It does not work with fractions where things are added or subtracted:

\[
\frac{x}{{y}\hspace{0.33em}{+}\hspace{0.33em}{z}^{{-}{3}}}\hspace{0.33em}\ne\hspace{0.33em}\frac{{x}\hspace{0.33em}{+}\hspace{0.33em}{z}^{3}}{y}
\]

 

By the way, the symbol ≠ means “does not equal”.

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.

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.

 

 

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 x 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

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

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.

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.