This article explores processes used to solve algebraic expressions. Algebra involves the use of mathematical statements to describe relationships between things that vary over time. An equation is a mathematical statement which assigns two expressions to one another using the ‘equal to’ (=) sign. Therefore algebraic equations are mathematical statements that use a finite number of mathematical symbols (called variables) and which equal to zero. An example is

a + b – c = 0

Another example is

y = x -3

One may assume that the second equation does not equal to zero as mentioned in the definition for an algebraic equation. However, re-arranging this equation will show that it does equal to zero and as such, is an algebraic equation.

y – x + 3 = 0

The terms in the equations are a,b,c, and x,y.

This article will present the different types of equations and how they work, techniques for formulating and solving algebraic equations from real life problems, learning how to solve linear equations with one and two variables. The use of variables to denote quantities allows general relationships between quantities to be formally and concisely expressed, and thus enables solving a broader scope of problems. Most quantitative results in science and mathematics are expressed as algebraic equations.

Quick Summary

Variables

What are variables? Variables are unknown quantities represented by symbols (usually letters of the English alphabet) in algebraic equations. In the equation below, x, y and z.

x + y = z

These variables can be used to represent number which are not yet known in an equation. For example, if the price of biscuits is the same as that of a bunch of bananas and an apple together, we can represent this statement using an algebraic equation. If the cost of biscuits is £b, that of a bunch of bananas is £a and cost of an apple is £c, then the statement will read

b = a + c

Furthermore, variables are also used to show relationship between two or more physical properties. For example, the distance moved by a body is its speed multiplied by time taken. If the speed is represented with s, the time with t, and the distance with d, then the equation relating to these three quantities is:

d = s * t

Evaluating expressions

Algebraic expressions may be evaluated using basic mathematics operations. Some algebraic expressions are complex and may need to be simplified before they are evaluated. Rules for evaluating algebraic expressions include:

Addition/Subtraction: similar variables can be summed together or subtracted from each other. The coefficients are used to decide the quantity to be added or subtracted. For example,

x + x = 2x

4y – y = 3y

Correspondingly, note that when adding, only similar terms must be added or subtracted. When variables are not similar, they must be left the way they are. For example

2x + 2y – x + 4y = x + 6y

The similar terms are added together and the final equation left the way it is as the remaining terms are not the same.

Multiplication/Division: Similar variables can also be multiplied or divided. In this case, the quotients are used for evaluation. For example:

a * a = a²

z² * z³ = z²⁺³ = z⁵

x⁷ / x⁵ = x^7-5 = x²

In mathematics, the parenthesis (or brackets) imply multiplication. When variables in brackets are combined using addition or subtraction, the variable outside the brackets must be used to multiply out the variables in the brackets. For example

a(6a – 7) = (a * 6a) + (a * -7) = 6a² – 7a

On the other hand, algebraic equations can also be factored. Given the equation

6a² – 7a

Dividing both terms by a, the equation can be re-written as

a(6a-7)

 

Inequality

Another type of equation is known as an inequality. Instead of the equal to sign in equations, an inequality equation replaces this with a less than sign, a greater than sign, a less than or equal to sign or a greater than or equal to sign (<; >; ≤; ≥ respectively).

The names given to the signs explain exactly what they mean. For example:

1. a < b implies a is less than b
2. x > y implies x is greater than y
3. w ≤ z reads w is less than or equal to x
4. h ≥ k reads h is greater than or equal to k

Properties of Inequality

The following are properties of inequality expressions

1. a < b and b < c, then a < c.
2. If x < y and w < u, then x + w < y + u
3. For a positive variable c, if a < b, then ac < bc
4. For a negative variable c, if x < y, the cx < cy
5. If x < y, then y > x

Solving Algebraic Equations

The complexity of algebraic equations increases as the number of variables in the equation increases. Therefore for example, an equation with two variables is more complex than one with a single variable. Correspondingly, as the complexity increases, so does the ease of evaluating the equation. The following sections will discuss methods of solving various types of algebraic equations. The examples given will be possible real life problems.

Linear equations with one variable

An algebraic equation is linear if it gives a straight line graph when plotted. A linear equation with one variable is the easiest form of algebraic equation.

Example

If the number of eggs in a basket is tripled and added to 1, the answer is 13. How many eggs are in the basket?

Solution

Suppose the number of eggs in the basket is x, then our equation is

3x + 1 = 13

To evaluate this equation, first subtract 1 from both sides to give

3x + 1 – 1 = 13 – 1

3x = 12

Next divide both sides by 3

3x/3 = 12/3

X = 4

The number of eggs in the basket is 4

Linear equations with two variables

When a linear equation has two variables, the number of possible solutions are infinite, i.e. they are many. An example of a linear equation with two variables is

a + b = 4

There are infinitely many possible values for a and b. The following are possible solutions

a = 1; b = 3

a = 4; b = 0

a = 7; b = -3

The possible solutions are infinite. One needs additional information to narrow down the answer to one set of values for a and b. At this point, we should point out that for equations with two variables, there must be two equations to get the exact values of the variables. Similarly, when three variables are involved, there needs to be three equations to get the values of the variables, and so on.

A set of two equations each with two variables are known as simultaneous equations. There are a few methods of solving simultaneous equations including substitution and elimination methods.

These types of equations are known as system of linear equations.

Elimination method

The aim of elimination method is to first eliminate one of the variables leaving us with a linear equation with one variable. Once this is achieved, the equation can be solved quite easily.

For example:

4x + 2y = 14

2x – y = 1

There are two variables involved. Anyone of them can be eliminated. For this example, x will be eliminated. To do this, multiply the second equation by 2 to give

2x – y = 1     x 2

4x – 2y = 2

The two equations now become

4x + 2y = 14

4x – 2y = 2

Now subtract the second equation from the first one to give:

4x + 2y – 4x + 2y = 14 – 2

4y = 12

This is now an equation with one variable. Divide both sides by 4 to get the value of y

4y/4 = 12/4

y = 3

With the known value of y, any of the two original equations can be used to find x. Using the first one,

4x + 2y = 14

4x + 2(3) = 14

4x + 6 = 14

Subtract 6 from both sides

4x = 14- 6 = 8

x = 8/4 = 2

Substitution method

Alternatively, the substitution method can also be used to solve a simultaneous equation. Using the same set of equations as example,

4x + 2y = 14

2x – y = 1

First step is to evaluate a value for one of the variables with respect to the other variable using any of the two equations. Using the second equation in this instance and evaluating a value for y with respect to x,

2x – y = 1

y = 2x – 1

The next step will be to substitute this new ‘value’ for y into the other equation. Note that you cannot substitute into the same equation as the values will nullify each other and give 0. Substituting into first equation,

4x + 2y = 14

4x + 2(2x – 1) = 14

4x + 4x – 2 = 14

8x = 14 + 2

8x = 16

x = 16/8 = 2

This results in x = 2. Next, substitute this value of x into any of the two equations

4(2) + 2y = 14

8 + 2y = 14

2y = 14 – 8

2y = 6

y = 6/2 = 3

These are the two methods for solving a system of linear equations with two variables. When There more than two variables and hence two equations, other techniques are applied which will not be covered in this article.

 

Quadratic equations

A quadratic equation is an algebraic equation with one variable and where that variable has a power of 2 as its highest quotient. An example of a quadratic equation is

x² + 2x + 1

There are various techniques of solving a quadratic equation problem. One of such techniques makes use of the coefficients of the terms to find the variable using the quadratic formula.

In the equation, the symbols a, b and c represent the coefficients of x² and x and the constant value respectively.

Example

Solve for x in the quadratic equation x² + 2x + 1, using the quadratic formula.

From the equation, the values of a, b and c are 1, 2 and 1 respectively. Substituting these values into the formula

This implies that x = -1 or x = -1

Alternatively, this can be written as

(x + 1)(x + 1) = 0

The solution to the question is thus x = -1 twice.

 

logarithmic and exponential equations

A logarithmic equation is an equation which takes the form where a is a positive whole number.

The solution to a logarithmic equation in this format is

x = b^a

An example will suffice.

Example

Find the solution to the logarithmic equation below

First we must add 1 to both sides

Next divide both sides by 4

The solution is now

 

An exponential equation is one which follows the format aⁿ = b, where a is a positive number. Exponential equations are applied in various mathematical problems including,

Rate of change problems

Population problems

Interest rate problems

Radioactive decay problems

An example is given to explain how to solve an exponential equation problem.

Example

If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 1,000 cells? (NB: Decay constant is -0.6)

This is a radioactive decay problem. The cells are reducing at a certain rate, starting from an initial value and ending at a final value. If the ‘cells’ are changed to currency or number of people, this could also serve as an interest rate or population problem respectively.

The equation used in solving this problem is

B = Ae^nt

In this equation, B is the final value, A is the number of cells we started with, t is the time taken to arrive at the final value of cells and n is the decay constant which is given from the question

Substituting the values into the equation, we have

1000 = 5000000 e ^(-0.6t)

Now divide both sides by 5000000

1000/5000000 = e^(-0.6t)

1/5000 = e^(-0.6t)

This equation is now in the format of the general exponential equation aⁿ = b.

Take natural logarithms of both sides

ln (1/5000) = ln e^(-0.6t)

The reason for taking natural logarithm is to eliminate the exponential factor. The natural logarithm of an exponent is its quotient!

The natural logarithm of 1/5000 can be found using a scientific calculator.

-8.52 = -0.6t

t = -8.52/-0.6

t = 14.2 seconds

Radical equations

Radical signs are signs like square roots, cube roots and so on. A radical equation is one that includes a radical sign. In general, note that

For example,

With this in mind, one can now solve problems with radical equations.

Example

Solve the equation (y – 3)^2/3 = 4