Algebra/Equations

Balancing variables

A variable is a letter that stands for a number. For example, if I said that $x = 5$ , then any time I use $x$ , you know it's 5 for any context I use it in ( $x$ is the variable most often used). $x$ and 5 have the same value, but different appearance. Sometimes, we don't know what the variable is and we need to find out.

For instance, what number can we put in for $x$ in the equation $x + 2 = 3$ that will make it true? One way you could work this out is by trying out different values of $x$ until you get one that works. This is called guess-and-check. Alternatively you might know the answer intuitively (by thinking What do I need to add to 2 to get 3?).

However, if you have a more complicated problem such as $\frac{7 x}{2} + 100 = 170$ you are likely to have trouble solving this problem intuitively or by guess-and-check. Because of this, mathematicians worked out a technique to solve this type of problem easily. This technique is the fundamental basis of algebra.

To understand this technique, you first have to fully understand that the equal sign means that both sides of the equation are the same (Same value, different appearance), and that if you manipulate (using addition, multiplication, etc) the values on both sides of the equal sign in the same way, then they will still be equal.

For instance if we have $3 = (2 + 1)$ , and multiply all of it by 5 we get,

$3 \times 5 = (2 + 1) \times 5$
$15 = 3 \times 5$
$15 = 15$

Notice how the equality still holds.

Now when we have an unknown variable called $x$ , and we want to know what the value of $x$ is, the easiest way is to find an equation which has $x$ on its own on one side of an equals sign with a number on the other side. (For instance, if we have an equation $x = 5$ , we know that the value of $x$ must be 5).

Here are some examples of manipulating equations to get the $x$ on its own,

Example 1: What is the value of $x$ in $x - 5 = 3$ ?

Solution: We need to change the left hand side to get the $x$ on its own, we can do it in this case by adding 5 to it (as $x - 5 + 5$ is $x$ ), but to keep both sides of the equation equal we'll need to add a 5 to the other side as well to get,

$(x - 5) + 5 = (3) + 5$
$x = 3 + 5$
$x = 8$

Problem 2: What is the value of $x$ in $2 x = 4$ ?

Solution 2: We can do the same in this case by dividing it by 2 (as $\frac{2 x}{2}$ is $x$ (remember that $x$ is the same as 1 $x$ because it has implied coefficient of 1)), but again to keep both sides of the equation equal we'll need to divide the other side by 2 as well to get,

$\frac{2 x}{2} = \frac{4}{2}$
$x = \frac{4}{2}$
$x = 2$

Problem 3: What is the value of $x$ in $3 x + 1 = 4$

Solution 3: Here we first need to subtract 1 from both sides,

$3 x + 1 - 1 = 4 - 1$
$3 x = 3$

Then we divide both sides by 3 to get,

$\frac{3 x}{3} = \frac{3}{3}$
$x = 1$

Although in this case we chose to do the subtraction first and then the division, we could have done it the other way around, doing the division first followed by the subtraction, as follows,

$\frac{3 x}{3} + \frac{1}{3} = \frac{4}{3}$
$x + \frac{1}{3} = \frac{4}{3}$
$x = \frac{4}{3} - \frac{1}{3}$
$x = \frac{3}{3}$
$x = 1$

Conventionally most people do additions/subtractions first and then multiplication/divisions, as this normally makes the numbers easier to handle (for example in the last case doing it the other way resulted in us having to deal with fractions). However, both ways are equally valid.

Solving simple equations

Sometimes you'll come across equation which have a variable on both sides, for instance $5 x - 1 = 2 x + 2$ , where x can be found on both sides of the equation.

We solve this type of equation in much the same way as we've solved the previous problems, but only this time you have to first make sure all of the variables are on the same side. The easiest way to see how to do this is by example:

Example 1: How do you find the value of $x$ in the equation, $5 x - 1 = 2 x + 2$ ?

First of all you need to choose which side you want the variable to be on, the left or the right of the equals sign, in this case we'll choose to have the $x$ on the left hand side.

To do this we first have to look at where $x$ occurs on the right hand side; in this case it only appears in the term $2 x$ . As we don't want $x$ on the right hand side we need to get rid of it, and we can do this by subtracting $2 x$ from the right side. Remember that for the equality to still be accurate we need to do the same on the left side as well.

$5 x - 1 - 2 x = 2 x + 2 - 2 x$ (subtracting 2x from both sides)
$3 x - 1 = 2$ (simplifying)

Now the equation is in a form which you are familiar with from the last chapter so hopefully you should now be able to solve this problem and get the answer $x = 1$ .

Solving linear inequalities

A number on the number line is always greater than any number on its left and less than any number on its right. The symbol " $<$ " is used to represent "is less than", and " $>$ " to represent "is greater than".

For example:

 <-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|----->
      -5    -4    -3    -2    -1     0     1     2     3     4     5

From the number line, we can easily tell that 3 is greater than -2, because 3 is on the right side of -2 (or -2 is on the left of 3). We write it as $3 > - 2$ . We can also see that any positive number is always greater than negative number.

Consider any two numbers, $a$ and $b$ . One and only one of the following statements can be true:

$a > b$
$a = b$ , or
$a < b$

This is the Law of Trichotomy. Another way of describing it is that any number or variable can be to the left of, at the same point as, or to the right of another number or variable on the number line. When we look at that, it makes sense -- if something is to the left of another thing, it can't also be to the right of it!

For an inequality with one unknown, there may be many (sometimes infinitely) possible solutions.

Properties

1. Transitive property: For any three numbers $x$ , $y$ and $z$ , if $x > y$ and $y > z$ , then $x > z$ .

2. In an inequality, we can add or subtract the same value from both sides, without changing the sign (i.e. " $>$ " or " $<$ "). That is to say, for any three numbers $x$ , $y$ and $p$ :

if $x > y$ , then $x + p > y + p$ and $x - p > y - p$ .

3. We can multiply or divide both sides by a positive number without changing the sign. For example, if we have any two numbers $x$ and $y$ , and another positive number $p$ :

if $x > y$ , then $x \times p > y \times p$ and $\frac{x}{p} > \frac{y}{p}$ .

4. When we multiply or divide both sides by a negative number, we have to change the sign of the inequality (i.e, " $>$ " changes to " $<$ " and vice versa). So if we are given two numbers $x$ and $y$ , and another negative number $p$

if

x > y

,

x \times p < y \times p

and

\frac{x}{p} < \frac{y}{p}

.

Now we can go on to solve any linear inequalities.

Solving Inequalities

Solving inequalities is almost the same as solving linear equations. Let's consider an example: $x + 4 < 13$ . All we have to do is subtract 4 on both sides. We will then get $x < 9$ , and that is the answer! Note, however, that what you get is not a single answer, but a set of solutions. Any number that satisifies the condition $x < 9$ (any number that is less than 9) is a solution to the inequality. It is very convenient to represent the solution using the number line:

 <-------------------o

 <-+-----+-----+-----+-----+-----+-->
   6     7     8     9     10    11

Note: the circle(o) shows that the value 9 is not included. Later on, when we deal with less than or equal to and greater than or equal to (≤ and ≥), we use "*" to show that the value is included in the solution set.

Let's try another more complicated question: $3 x - 2 > 2 (x - 3)$ . First, you may want to expand the right hand side: $3 x - 2 > 2 x - 6$ . Then we can simply rearrange so that all the unknowns are on one side (usually the left): $3 x - 2 x > - 6 + 2$ . Hence, we can easily get the answer: $x > - 4$ .

Here's an example where the direction of the inequality changes when finding the solution: solve $4 - 6 x > 22$ .

First subtract 4 from both sizes: $- 6 x > 18$ .
Now divide through by -6, changing the direction of the inequality: $\frac{- 6 x}{- 6} < \frac{18}{- 6}$ .

So the solution to the inequality is $x < - 3$ .

Previous: Arithmetic
Next: Further Arithmetic

Manipulation of Radicals

Let's say that we have a number $x$ . The square root of $x$ is the number that, if multiplied by itself, equals $x$ . Since there are two numbers which satisfy that condition, we usually specify the positive value. For example, the square root of 4 could be 2 (because $2 \times 2 = 4$ ) or it could be -2 (because $- 2 \times - 2 = 4$ ). We use the symbol $\sqrt{x}$ to indicate the positive square root of $x$ .

The cube root of $x$ is the number that, if multiplied by itself three times, equals $x$ . We use the symbol $\sqrt[3]{x}$ to indicate the cube root of $x$ .

We use the symbol $\sqrt[n]{x}$ to indicate the number, which when multiplied $n$ times is equal to $x$ . Or in symbols: if $y = \sqrt[n]{x}$ then $y^{n} = x$ .

Rules

1.

\sqrt{x} \cdot \sqrt{x} = (\sqrt{x})^{2} = x

2.

\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}

3.

x^{\frac{m}{n}} = (\sqrt[n]{x})^{m}

4.

\sqrt{x} \sqrt{y} = \sqrt{x} y

5.

\sqrt[m]{\sqrt[n]{x}} = \sqrt[m \cdot n]{x}

Algebra/Equations

Contents

Balancing variables

Solving simple equations

Solving linear inequalities

Properties

Solving Inequalities

Manipulation of Radicals

Rules

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Algebra/Equations

Balancing variables

Solving simple equations

Solving linear inequalities

Properties

Solving Inequalities

Manipulation of Radicals

Rules

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