# Algebra/Solving Equations

Algebra
 ← Equations Solving Equations Word Problems →

## Introduction

We've seen that an equation is like a balance. When two quantities are on either side of an equal sign in a mathematical statement, we are saying that the statement is only true under conditions that make the quantities the same. For instance, when we see the statement ${\displaystyle x+2=3}$ we now know that we are asking what number can we substitute for ${\displaystyle x}$ in the equation to make this statement true? One way you could work this out is by trying out different values of ${\displaystyle x}$ 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 ${\displaystyle {\frac {7x}{2}}+100=170}$ 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.

Since 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. This is also how a mechanical balance works. As long as you do the same thing to whatever is in both pans they stay level. If we have a value of 3 = in one pan and two things with the value of 2 and 1 in the other pan the balance is level. If we multiply both sides by 5 we get,

• ${\displaystyle 3\times 5=(2+1)\times 5}$
• ${\displaystyle 15=(2\times 5)+(1\times 5)}$
• ${\displaystyle 15=10+5}$
• ${\displaystyle 15=15}$

Notice how the equality still holds.

## Neolithic Algebra

Neolithic Functions Imagine you are the neolithic shepherd. What changes can you make to your bag of rocks?

• Addition - When new sheep are added to your flock you would use addition. For instance in the spring you would have a "lamb born" function in which you add a rock to your bag. You might also have a "multiple birth" function in which you invoke the lamb born function once for each lamb born.
• Subtraction - When you decide that a sheep is no longer part of your flock you would use a "remove sheep function" to keep your bag of rocks in balance with your herd. You might do different things with the rock for a missing sheep, a sheep you traded for something you needed, or because it became dinner one night. You might anticipate using these functions to manage your herd. If you know that you need to remove five rocks from your bag in the fall to buy firewood you might examine your bag carefully to know how many times you can invoke the "sheep dinner" function over the summer.
• Multiplication - Imagine you are a shepherd just starting out. If your neighbors had a big enough flock they might promise you a number of sheep per month as payment for watching their flock. For instance if they promised you two sheep a month for tending their flock then you would know that after 1 month you would have two sheep, after two months you would have four sheep, and after an entire year you might have enough sheep to start your own operation.
• Division - It is said that two things are inevitable: death and taxes. Division is the operation we use to make the inevitable fair. When a shepherd died his heirs could use division to divide up his flock fairly. They might use a function that allocated one sheep per heir until the flock was distributed. In what ways could the heirs distribute extra sheep if the flock was not evenly divisible? Similarly, when collecting taxes it is more fair to take a portion of the flock rather than a single amount. The clan leader would be better served taking 1/12th of a flock from each shepherd every year. If the clan leader asked for a specific number of sheep every year they might find that their clan became smaller every year just before tax time as new shepherds took their sheep to a new clan.

Thought Problem

Video Games have an attribute called game play. Look at a video game you enjoy playing and try to define how you interact with the environment using addition, subtraction, multiplication, and division.

## Using A Variable

With algebra we use variables to represent things we haven't been able to count. Since we can't count the number the variable represents the number is called an "unknown". The most common variable is denoted ${\displaystyle x}$. To use a variable we write an expression which we know is true and has ${\displaystyle x}$ on its own on one side of an equals sign and a number on the other side of the equals sign.

Sometimes a variable can represent a set of numbers. In this case the number can be represented with set notation. We generally use a letter that reminds us what the variable represents.

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

Example 1: How many dollars do I need to see a movie and buy popcorn? If we let ${\displaystyle x}$ equal the dollars that I need to get into the theater, then we could use the variable set ${\displaystyle A={3,5,7,10}}$ to represent a discount admission, a matinee admission, a regular admission, or an Imax admission. If we assume that popcorn costs 3 dollars at all the theaters we will go to then we can write the equation ${\displaystyle x-A=3}$ to represent the money we need. We could add an A to both sides to get the equation ${\displaystyle x=3+A}$. Plugging the values for A into this equation we find that ${\displaystyle x={6,8,10,13}}$.

Example 2: What is the value of ${\displaystyle x}$ in ${\displaystyle 2x=4}$?

Solution: We can do the same in this case by dividing it by 2 (as ${\displaystyle {\frac {2x}{2}}}$ is ${\displaystyle x}$ (remember that ${\displaystyle x}$ is the same as 1${\displaystyle x}$ 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,

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

Example 3: What is the value of ${\displaystyle x}$ in ${\displaystyle 3x+1=4}$

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

• ${\displaystyle 3x+1-1=4-1}$
• ${\displaystyle 3x=3}$

Then we divide both sides by 3 to get,

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

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,

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

When working problems it often helps to do additions/subtractions first and then multiplication/divisions, as this lets us avoid having to add or subtract fractions. However, both ways are equally valid.

## Simplifying Equations

Sometimes you'll come across equations which have a variable on both sides, for instance ${\displaystyle 5x-1=2x+2}$, 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 ${\displaystyle x}$ in the equation, ${\displaystyle 5x-1=2x+2}$?

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 ${\displaystyle x}$ on the left hand side.

To do this we first have to look at where ${\displaystyle x}$ occurs on the right hand side; in this case it only appears in the term ${\displaystyle 2x}$. As we don't want ${\displaystyle x}$ on the right hand side we need to get rid of it, and we can do this by subtracting ${\displaystyle 2x}$ 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.

• ${\displaystyle 5x-1-2x=2x+2-2x}$ (subtracting 2x from both sides)
• ${\displaystyle 3x-1=2}$ (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 ${\displaystyle x=1}$.

Let's say that we have a number ${\displaystyle x}$. The square root of ${\displaystyle x}$ is the number that, if multiplied by itself, equals ${\displaystyle x}$. 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 ${\displaystyle 2\times 2=4}$) or it could be -2 (because ${\displaystyle -2\times -2=4}$). We use the symbol ${\displaystyle {\sqrt {x}}}$ to indicate the positive square root of ${\displaystyle x}$.

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

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

### Rules

1. ${\displaystyle {\sqrt {x}}\cdot {\sqrt {x}}=({\sqrt {x}})^{2}={x}}$
2. ${\displaystyle {\sqrt {\frac {x}{y}}}={{\sqrt {x}} \over {\sqrt {y}}}}$
3. ${\displaystyle x^{{m} \over {n}}={({\sqrt[{n}]{x}})^{m}}}$
4. ${\displaystyle {\sqrt {x}}{\sqrt {y}}={{\sqrt {x}}{y}}}$
5. ${\displaystyle {\sqrt[{m}]{\sqrt[{n}]{x}}}={\sqrt[{m\cdot n}]{x}}=x^{1 \over {m\cdot n}}}$

### Solving for (Variable)

When solving an equation, you usually solve for a specific variable. To do so, you have to get all instances of that variable on one side of the equals sign, and everything else on the other.

### Properties of Equality

The equal sign that depicts the fact that both sides of it are equal is a very strange symbol with many properties. It tells you various traits of each side, and it allows you to manipulate each side in specific ways. Here are the different properties of that sign:

 Property Name Definition Example Reflexive a = a 8=8 Symmetric If a = b, then b = a If (3)(2) = 6, then 6 = (3)(2) Transitive If a = b & b = c, then a = c If 8 = (4)(2) and (4)(2) = (2)(4), then 8 = (2)(4) Substitution If a = b, then one can replace a with b or vice versa If a = b and 1 + a = 3, then 1 + b = 3 Addition You can add one number to both sides of the equation. ${\displaystyle x-6=14\,}$ ${\displaystyle x-6+6=14+6\,}$ ${\displaystyle x=20\,}$ Subtraction You can subtract one number from both sides of the equation. ${\displaystyle x+6=14\,}$ ${\displaystyle x+6-6=14-6\,}$ ${\displaystyle x=8\,}$ Multiplication You can multiply both sides of the equation by a number. ${\displaystyle {\frac {x}{6}}=18}$ ${\displaystyle 6({\frac {x}{6}})=(18)(6)}$ ${\displaystyle x=108\,}$ Division You can divide both sides of the equation by a number. ${\displaystyle 6x=18\,}$ ${\displaystyle 6x({\frac {1}{6}})=({\frac {18}{6}})}$ ${\displaystyle x=3\,}$

### Practice Problems

Decide whether these following problems are expressions or equations.

1. ${\displaystyle 2x+6}$

2. ${\displaystyle 3(x-14)^{2}=12}$

3. ${\displaystyle 4+6(2x+16)}$

Identify which properties are being used in the following problems.

1. ${\displaystyle a=b}$ and ${\displaystyle 3-a=4}$, so ${\displaystyle 3-b=4}$.

2. ${\displaystyle x+9=12}$, then ${\displaystyle x=3}$.

3. ${\displaystyle x-9y=4y}$, then ${\displaystyle x=13y}$.

1. Expression

2. Equation

3. Expression

1. Substitution Property of Equality

2. Subtraction Property of Equality

3. Addition Property of Equality (combine like-terms!)

## Basic Laws In Algebra

In algebra we are working with the set of real numbers. We talked about the properties of real numbers with respect to mathematical operations in the section Real Numbers. If you don't remember the Commutative Property, the Associative Property, the Distributive Property, and the Identity Property go back to the real numbers section to review them.

There are several basic laws in algebra. Understanding these will help you to manipulate and solve equations, and to understand algebraic relationships.

## Proportions or Ratios

Ratios or proportions can be expressed as an equation of fractions

(for example ${\displaystyle {\frac {Q}{R}}={\frac {S}{T}}}$ ),

or they can be expressed as a relationship Q : R = S : T , (expressed in words “ ‘Q’ is to ‘R’ as ‘S’ is to ‘T’ ”).

Using the words "is to" help us understand the physical relationship between the values, while the fractional representation can help us deal with the mathematical relationship.

For instance in America we know that 3 feet is to 1 yard as 6 feet is to 2 yards, but expressing this as a math equation: ${\displaystyle {\frac {3}{1}}={\frac {6}{2}}={\frac {2*3}{2*1}}}$ helps us see this is true because all we have done is double the original ratio.

Consider the relation 3*4 = 2*6 = 1*12. Does this mean something more than the obvious fact that 12=12=12? How many ways could you package 12 items?

In general, these relationships can be described by a general equation like ${\displaystyle (Q)(T)=(S)(R)}$

Dividing each side of the equation by ${\displaystyle (T)(R)}$ gives

${\displaystyle {\frac {(Q)(T)}{(T)(R)}}={\frac {(S)(R)}{(T)(R)}}}$ , which simplifies to ${\displaystyle {\frac {(Q)}{(R)}}={\frac {(S)}{(T)}}}$

If we divide by ${\displaystyle (Q)(S)}$ instead the results simplify to

${\displaystyle {\frac {(T)}{(S)}}={\frac {(R)}{(Q)}}}$

One needs to be careful since each term also has a proportional relation with the two other variables.

 In our examples all of the following are also valid
Q : S = R : T ,“ ‘Q’ is to ‘S’ as ‘R’ is to ‘T’ ”.
R : Q = T : S ,“ ‘R’ is to ‘Q’ as ‘T’ is to ‘S’ ”..
S : Q = T : R ,“ ‘S’ is to ‘Q’ as ‘T’ is to ‘R’ ”.

 Since 2*6=3*4
2:3 = 4:6, 2 is to 3 as 4 is to 6, and 2/3 = 4/6.
4:2 = 6:3, 4 is to 2 as 6 is to 3, and 4/2 = 6/3.
3:2 = 6:4, 3 is to 2 as 6 is to 4, and 3/2 = 6/4.
2:4 = 3:6, 2 is to 4 as 3 is to 6, and 2/4 = 3/6.


## Solving Equations

Although we have already solved a few equations, we will now discuss the formal idea of solving equations. To solve an equation, you are finding the value of any variables within the equation. To find the value of a variable, you have to manipulate the equation to state ${\displaystyle *insertvariablehere*=*somenumber*}$. Then you know the value of the variable! You will use the Properties of Equality to manipulate the equation into the desired form.

### Practice Problems

Solve for x in the following equations.

1. ${\displaystyle x+7=12\,}$

2. ${\displaystyle -{\frac {2}{3}}x+9=15}$

3. ${\displaystyle 3x-15=8x\,}$

4. ${\displaystyle c(x-4m)=d(n-9)\,}$

5. ${\displaystyle 20x+12=4(5x+3)\,}$

1. ${\displaystyle {\begin{matrix}x+7&=&12\\\ x&=&5\end{matrix}}}$ Subtract 7

2. ${\displaystyle {\begin{matrix}-{\frac {2}{3}}x+9&=&15\\\ -{\frac {2}{3}}x&=&6\\\ x&=&-9\end{matrix}}}$ Subtract 9, then multiply by -3/2 (Inverse Property of Multiplication)

3. ${\displaystyle {\begin{matrix}3x-15&=&8x\\\ -15&=&5x\\\ -3&=&x\end{matrix}}}$ Subtract 3x, then divide by 5

4. No Solution, because there are too many variables to find a single number for x.

5. ${\displaystyle {\begin{matrix}20x+12&=&4(5x+3)\\\ 20x+12&=&20x+12\end{matrix}}}$ All Real Numbers. Perform Distributive Property, and you'll get the same equation on both sides. Thus, any number would work!

## Lesson Review

Equations are two expressions that are equal to each other, and they are expressed by putting each of them on one side of the equal sign. You can add, subtract, multiply, or divide both sides of an equation while keeping it equal (For example, we know that 7 = 7, correct? What if we subtracted 2 from each side? We'd still have a true statement: 5 = 5). There are other properties of equality, such as the Reflexive, Symmetric, Transitive, and Substitution. You will be using all of these properties to solve (find the value of) variables in equations.

## Lesson Quiz

1. What property is expressed here? ${\displaystyle a=b}$ and ${\displaystyle b=c}$, then ${\displaystyle a=c}$.

2. If I divided both sides of an equation by 4, would it still be equal on both sides? If so, why?

3. Solve for y. ${\displaystyle 5\left({\frac {y}{6}}-2\right)=2y+4}$