Algebra combines numbers and variables using mathematical operations like addition, subtraction, multiplication and division to represent a specific problem. The solutions to the problems are found by using predefined rules to manipulate each mathematical expression.
An example of an algebraic expression is:
\(3x+2=5\)
In this example, x is the unknown value, 3 is the coefficient of x, 2 and 5 are constants (fixed values), and the operation being performed is an addition (+).
Remember that the coefficient is the number that is multiplied by a variable
Algebra can be classified into different sub-branches according to the level of complexity of their algebraic expressions and where they are applied. These branches range from elementary algebra to more abstract and complex equations, which require more advanced mathematics. Elementary algebra deals with solving algebraic expressions to find a solution, and it is used in most fields like science, medicine, economics and engineering.
Abu Ja'far Muhammad ibn Musa al-Khwarizmi invented algebra. He was a writer, scientist, astronomer, geographer, and mathematician, born in the 780s in Baghdad. The term algebra comes from the Arabic word al-jabr, which means "the reunion of broken parts".
Why is algebraic expression important in the real world?
Being able to understand algebra not only helps you to represent algebraic expressions and find their solutions. It also allows you to improve your problem-solving skills, helping you to think critically and logically, identify patterns, and solve more complex problems involving numbers and unknown values.
Knowledge of algebra can be applied to solve everyday problems. A business manager can use algebraic expressions to calculate costs and profits. Think about a shop manager that wants to calculate the number of chocolate milk cartons sold at the end of the day, to decide whether to continue stocking them or not. He knows that at the start of the day he had 30 cartons in stock, and at the end, there were 12 left. He can use the following algebraic expression:
\(30 - x = 12\) x is the number of chocolate milk cartons sold
We need to work out the value of x by solving the expression above:
\(30 - 12 = x\) isolating x to one side of the equation and solving the operation
x = 18
The number of chocolate milk cartons sold on that day was 18.
This is just a simple example, but the benefits of understanding algebra go a lot further than that. It helps us with daily activities like shopping, managing a budget, paying our bills, planning a holiday, among others.
Types of algebraic equations
The degree of an algebraic equation is the highest power present in the variables of the equation. Algebraic equations can be classified according to their degree as follows:
Linear equations
Linear equations are used to represent problems where the degree of the variables (i.e. x, y or z) is one. For example, \(ax+b = 0\), where x is the variable, and a and b are constants.
Quadratic equations
Quadratic equations are generically represented as \(ax^2+bx+c = 0\), where x is the variable, and a, b and c are constants. They contain variables with power 2. Quadratic equations will produce two possible solutions for x that satisfy the equation.
Cubic equations
Cubic equations are represented in a generic form as \(ax^3 + bx^2+cx +d=0\), where x is the variable, and a, b, c and d are constants. They contain variables with power 3.
What are the basic properties of algebra?
The basic properties of algebra that you need to keep in mind when solving algebraic equations are:
\(a + b = b + a\)
\(a \cdot b = b \cdot a\)
\(a + (b +c) = (a+b)+c\)
\(a \cdot (b \cdot c) = (a \cdot b) \cdot c\)
\(a \cdot (b +c)= a \cdot b + a \cdot c\)
Reciprocal of \(a = \frac{1}{a}\)
\(a + 0 = 0 + a = a\)
\(a \cdot 1 = 1 \cdot a =a\)
\(a + (-a) = 0\)
\(a \cdot \frac{1}{a} = 1\)
Solving linear algebraic equations
To solve linear algebraic equations, you should follow the following steps:
Step 1: each side of the equation must be simplified by removing parentheses and combining terms
Step 2: add or subtract to isolate the variable on one side of the equation
Step 3: multiply or divide to obtain the value of the unknown variable
Example 1: Variable on one side of the algebraic equation
\(3 (x + 1) + 4 = 16\)
- Step 1: \(\begin{align} 3x + 3 + 4 = 16 \\ 3x + 7 = 16 \end{align}\)
- Step 2: \(\begin{align} 3x = 16 - 7 \\ 3x = 9 \end{align}\)
- Step 3: \(\begin{align} x = \frac{9}{3} \\ x = 3 \end{align}\)
Example 2: Variable on both sides of the algebraic equation
\(4x + 3 = x - 6\)
- Step 1: We can skip this step as there are no parentheses in this equation
- Step 2: \(\begin{align} 4x - x = -6 - 3 \\ 3x = -9 \end{align}\)
- Step 3: \(\begin{align} x = \frac{-9}{3} \\ x = -3 \end{align}\)
Example 3: Word problem
You have a box of blue and red balls. The total of balls is 50, and the amount of red balls is twice the amount of blue balls minus 10. How many red balls are there in the box?
To solve word problems you need to follow this strategy:
Our variables are:
B = amount of blue balls
R = amount of red balls
Equations:
1) \(B + R = 50\)
2) \(R = 2B - 10\)
Now we solve the equations:
We know that \(R = 2B - 10\), so we can substitute the value of R in equation 1 with that expression
\(B + (2B - 10) = 50\)
\(B + 2B - 10 = 50\)
\(3B = 50 + 10\)
\(3B = 60\)
\(B = \frac{60}{3}\)
\(B = 20\)
Now we substitute the value of B in equation 2:
\(R = 2B - 10\)
\(R = 2 \cdot 20 - 10\)
\(R = 40 - 10\)
\(R = 30\)
There are 30 red balls in the box.
What are the different types of problems in algebra?
The different types of problems that you can find in algebra vary depending on the type of algebraic expressions involved and their complexity. The main ones are:
Algebra & functions - key takeaways
Algebra is a branch of mathematics that uses letters or variables to represent unknown values that can change.
Real-life problems can be represented using algebraic expressions.
Algebra uses predefined rules to manipulate each mathematical expression.
Understanding algebra helps to improve problem-solving skills, critical and logical thinking, identifying patterns, and skills to solve more complex problems involving numbers and unknown values.
The different types of algebraic equations according to their degree are: linear, quadratic and cubic.
To solve linear algebraic equations each side of the equation must be simplified by removing parentheses and combining terms, then add or subtract to isolate the variable on one side of the equation, and finally multiply or divide to obtain the value of the unknown variable.
To solve word problems start by assigning variables to unknown values, construct the equations, then solve the equations.
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