Introduction to algebra

To solve algebraic equations and find unknown values, which is a key part of algebra, you first need to understand the main terms and skills explained on this page.

To solve algebraic equations and find unknown values, which is a key part of algebra, you first need to understand the main terms and skills explained on this page.

In brief, you will learn:

Definitions of important terms

When a problem is written using algebra it can be called either:

  • an algebraic expression (when there is no equals sign), or
  • an algebraic equation (when there is an equals sign).

Both algebraic expressions and algebraic equations involve the use of constants and variables:

Constants are numbers wthat have a fixed value (for example \(1, 5, 2108, 37.5, -103,\) etc.).

Variables are usually small letters of the alphabet, or sometimes other symbols. They are used to represent an unknown number in an expression or equation (for example \(x, y, a, \theta,\) etc.). The value of this number can change (or vary, as the word variable suggests) depending on the situation.

In the algebraic expression \(a + 5,\) for example, \(5\) is a constant, while \(a\) is a variable. This means that different numbers can be substituted for \(a\) and the value of the expression will change accordingly. For example, if \(a = 1\) then \(a + 5\) becomes \(1 + 5,\) which we know is equal to \(6\).

Another important idea is that each ‘part’ of an algebraic expression or equation separated is called a term. A term is separated from other terms by addition, subtraction, or division. A term can be a constant, a variable, or a combination of both.

For example, the following algebraic expression contains \(5\) different terms, each separated by addition (note that we usually write \(x\) and \(y\) rather than \(1x\) and \(1y\), even though they mean the same thing):

\[2x + y + 3y + 5x + x\]

Because this expression has more than one variable, it includes both like terms and unlike terms.

Like terms are terms which have the same variable. For example \(2x,\) \(5x\) and \(x\) are all like terms in the expression above. Similarly, \(y\) and \(3y\) are also like terms.

Note that \(x\) and \(x^2\) are not like terms because one term is \(x\) while the other has \(x\) raised to the power of \(2.\) In the same way, \(x\) and \(xy\) are not like terms because one is \(x\) only while the other has \(x\) and \(y.\)

Unlike terms are terms that have different variables. For example \(2x\) and \(y\) are unlike terms. As shown above, \(x\) and \(x^2\) are also unlike terms, and so are \(x\) and \(xy\).

It is important to understand that only like terms can be added or subtracted. For example, you would not say that \(2\) apples plus \(5\) oranges equals \(7\) apples (or \(7\) oranges), In the same way, you cannot add \(2x\) and \(5y\) together.

Finally, when terms are combined in certain ways, they form polynomials.

A polynomial is an expression made up of constants and/or variables raised to positive powers (also known as exponents), combined using addition, subtraction, and/or multiplication. For example, \(x^2 + 2x - 3\) is a polynomial, and so is \(20\), but \(x^{-2} + 3\) is not.

Simplifying algebraic expressions

As with most topics, it is best to write an algebraic expression or equation in the simplest form possible. There is no need to make it more complicated than necessary. There are a few key methods you can use to simplify algebraic expressions and equations, and these are explained in this section.

Note that the examples focus on algebraic expressions, as these are easier to use when first learning the techniques. However, remember that the main goal is to apply these techniques to solve algebraic equations, as shown in later sections of this module.

Adding or subtracting like terms

You can simplify an algebraic expression by adding or subtracting like terms. To do this, follow these steps:

  1. Rewrite the expression so that all like terms are grouped together. When moving terms, make sure you move the \(+\) or \(–\) sign in front of each term with it.

  2. Simplify the expression by adding or subtracting like terms.

Example: Simplify the algebraic expression \(2a\,–\,3b\,–\,a\,+\,4ab\,–\,2b\)

Solution: Following the steps above, we have:

  1. Rewriting the expression so that all like terms are next to each other gives \(2a\,–\,a\,–\,3b\,–\,2b\,+\,4ab\) (or you can write it in other ways, as long as all the \(a\) terms are together and all the \(b\) terms are together; the \(ab\) term doesn’t have any like terms)

  2. Adding and subtracting the like terms in the rearranged algebraic expression \(2a\,–\,a\,–\,3b\,–\,2b\,+\,4ab\) gives \(2a\,–\,a = a\) and \(-3b\,–\,2b = -5b\), so our simplified expression is \(a\,-\,5b\,+\,4ab\)

Multiplying or dividing terms

Note that there are several ways to show that two or more terms need to be multiplied. For example, if \(x\) and \(y\) are to be multiplied, this might be written as \(x \times y\), as \(x.y\), as \((x)(y)\) or \(x(y)\).

No matter how the multiplication is written, you can simplify an algebraic expression by multiplying or dividing terms. Follow these steps:

  1. Rewrite the expression (or each part of the expression, when dividing) as a product of its factors.

  2. If needed, rewrite the expression (or each part of the expression, when dividing) so that like terms are grouped together.

  3. Multiply or divide the constants and multiply or divide the variables, as needed. Remember that if you multiply or divide a negative by a negative, the result is positive (for example, \(-2 \times -2 = 4\)).

Example 1: Simplify the algebraic expression \(4ab \times 3acd\)

Solution: Following the steps above, we have:

  1. Rewriting the expression as a product of its factors gives \(4 \times a \times b \times 3 \times a \times c \times d\)

  2. Rewriting the expression so that the like terms are next to each other gives \(4 \times 3 \times a \times a \times b \times c \times d\)

  3. Since \(4 \times 3 = 12\) and \(a \times a = a^2\), it follows that the simplified expression is \(12a^2bcd\)

Note that \(a^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

Example 2: Simplify the algebraic expression \(\frac{20x^3y}{4x^2y^2}\)

Solution: Following the steps above, we have:

  1. Rewriting each part of the expression as a product of its factors gives \(\frac{20 \times x \times x \times x \times y}{4 \times x \times x \times y \times y}\)

  2. The like terms are already next to each other for both parts of this expression, so rewriting is not required

  3. Since \(\frac{20}{4} = 5\) and \(\frac{x \times x \times x}{x \times x} = x\) and \(\frac{y}{y \times y} = \frac{1}{y}\), it follows that the simplified expression is \(\frac{5x}{y}\)

Note that the above division above has been performed by simplifying fractions, which works in the same way for variables as it does for constants. If you would like a refresher on how simplifying fractions works, please visit the Fractions section of the Numeracy fundamentals module.

After you have worked through the examples above, try simplifying some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

ANSWER: This expression contains all like terms, which can be added and subtracted to give \(4c\)

ANSWER: Rearranging this expression gives \(3d - d - 5c + 3c + 4 - 9\), which can then be simplified to \(2d - 2c -5\)

ANSWER: Rewriting this expression gives \(13 \times a \times 2 \times b\), which then gives \(13 \times 2 \times a \times b\) and therefore simplifies to \(26ab\)

ANSWER: Rewriting this expression gives \(\frac{10 \times a \times a \times b}{5 \times a \times c}\), and since \(\frac{10}{5} = 2\) and \(\frac{a \times a}{a} = a\), it follows that it simplifies to \(\frac{2ab}{c}\)

ANSWER: Rearranging this expression gives \(3a - a + b - 3b\), which can then be simplified to \(2a - 2b\)

ANSWER: Rearranging this expression gives \(2f + 3h + 5h - g + g\), which can then be simplified to \(2f + 8h\)

ANSWER: Rearranging this expression gives \(20c + 5c - 4d - 15d + 10\), which can then be simplified to \(25c - 19d + 10\)

ANSWER: Rearranging this expression gives \(-5f + 10f + 102g + 6 + 3\), which can then be simplified to \(5f + 102g + 9\)

ANSWER: Rewriting this expression gives \(4 \times c \times a \times b \times 9\), which then gives \(4 \times 9 \times c \times a \times b\) and therefore simplifies to \(36abc\) (while the variables can be listed in any order, the usual convention is to list them alphabetically)

ANSWER: Rewriting this expression gives \(2 \times a \times a \times 5 \times a \times a\), which then gives \(2 \times 5 \times a \times a \times a \times a\) and therefore simplifies to \(10a^4\)

ANSWER: Rewriting this expression gives \(\frac{12 \times a \times c}{a \times b}\), and since \(\frac{a}{a} = 1\), it follows that it simplifies to \(\frac{12c}{b}\)

ANSWER: Rewriting this expression gives \(\frac{20 \times d \times d \times e \times f}{5 \times d \times e}\), and since \(\frac{20}{5} = 4\) and \(\frac{d \times d}{d} = d\) and \(\frac{e}{e} = 1\), it follows that it simplifies to \(4df\)

Expanding brackets

Algebraic expressions or equations often include brackets. When this happens, it can be helpful to expand the brackets to make the expression or equation easier to work with. This section explains how to expand brackets for both one and two sets of brackets. Note again that the examples use algebraic expressions for simplicity. You will learn how to apply these methods to solve algebraic equations in later sections of this module.

Expanding one set of brackets

An algebraic expression that has one term outside a set of brackets can be expanded using the following steps:

  1. Write out each multiplication that needs to be done. Do this by multiplying the term outside the brackets by each term inside the brackets. If you place brackets around each product, it can help you keep track of the steps and deal with negative signs later.

  2. Carry out each multiplication using the method explained earlier. Keep each result in brackets so that it is easier to track any negative signs.

  3. Remove the brackets from each term. If there is a negative sign inside a set of brackets, change the sign of the terms when you remove the brackets.

Example: Expand the brackets in the algebraic expression \(-3a(2b + 4a - 9)\)

Solution: Following the steps above, we have:

  1. Writing out the multiplications we need to perform, using brackets to indicate multiplication and adding each pair of terms, gives \((-3a)(2b) + (-3a)(4a) + (-3a)(-9)\)

  2. Performing the multiplication above, keeping the results in brackets, gives \((-6ab) + (-12a^2) + (27a)\)

  3. Removing brackets and making the required terms negative gives \(-6ab - 12a^2 + 27a\)

Note that \(a^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

Expanding pairs of brackets

An algebraic expression with two sets of brackets being multiplied can be expanded using the following steps:

  1. Write out each multiplication that needs to be done. Multiply each term in the first set of brackets by each term in the second set of brackets, and then add the results together. Use brackets to show each multiplication and to help avoid mistakes with negative signs. A helpful way to remember which terms to multiply when there are two terms in each set of brackets is with the acronym FOIL:
    • First: Multiply the first terms in each bracket
    • Outside: Multiply the two outside terms
    • Inside: Multiply the two inside terms
    • Last: Multiply the last terms in each bracket
      \(\)

  2. Perform each multiplication, keeping each result in brackets so you can clearly track any negative signs.

  3. Remove the brackets from each term. If there is a negative sign in a set of brackets, change the sign of the terms when you remove the brackets.

  4. Simplify the expression by adding or subtracting like terms, as explained in the earlier section.

Example: Expand the brackets in the algebraic expression \((a - 3)(a + 4)\)

Solution: Following the steps above, we have:

  1. Writing out the multiplications we need to perform, using brackets to indicate multiplication and adding each pair of terms, gives \((a)(a) + (a)(4) + (-3)(a) + (-3)(4)\)
    Again note that FOIL can help with this:

  2. Performing the multiplication above, keeping the results in brackets, gives \((a^2) + (4a) + (-3a) + (-12)\)

  3. Removing brackets and making the required terms negative gives \(a^2 + 4a - 3a - 12\)

  4. Finally, simplifying the expression gives \(a^2 + a - 12\)

After you have worked through the examples above, try expanding the brackets in some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

\(= (b)(3) + (b)(b)\)

\(= 3b + b^2\)

\(= (3xy)(6x) + (3xy)(-y^2) + (3xy)(x^2y)\)

\(= 18x^2y) + (-3xy^3) + (3x^3y^2)\)

\(= 18x^2y - 3xy^3 + 3x^3y^2\)

\(= (e)(g) + (e)(f) + (f)(g) + (f)(f)\)

\(= eg + ef + fg + f^2\)

\(= (c)(-c) + (c)(-3d) + (-4d)(-c) + (-4d)(-3d)\)

\(= (-c^2) + (-3cd) + (4cd) + (12d^2)\)

\(= -c^2 - 3cd + 4cd + 12d^2\)

\(= -c^2 + cd + 12d^2\)

\(= (a)(-a) + (a)(1) + (a)(-ac^2)\)

\(= (-a^2) + (a) + (-a^2c^2)\)

\(= -a^2 + a - a^2c^2\)

\(= (-3d)(3) + (-3d)(-2c) + (-3d)(5d)\)

\(= (-9d) + (6cd) + (-15d^2)\)

\(= -9d + 6cd - 15d^2\)

\(= (-a^3)(-a^2) + (-a^3)(-3a) + (-a^3)(4)\)

\(= (a^5) + (3a^4) + (-4a^3)\)

\(= a^5 + 3a^4 - 4a^3\)

\(= (4dg^2)(2d) + (4dg^2)(3g^3)\)

\(= 8d^2g^2 + 12dg^5\)

\(= (7x)(8x) + (7x)(-3y) + (2y)(8x) + (2y)(-3y)\)

\(= (56x^2) + (-21xy) + (16xy) + (-6y^2)\)

\(= 56x^2 - 21xy + 16xy - 6y^2\)

\(= 56x^2 - 5xy - 6y^2\)

\(=(x^2)(x) + (x^2)(y) + (y)(x) + (y)(y)\)

\(= x^3 + x^2y + xy + y^2\)

\(= (x)(2x) + (x)(-4) + (3y)(2x) + (3y)(-4)\)

\(= (2x^2) + (-4x) + (6xy) + (-12y)\)

\(= 2x^2 - 4x + 6xy - 12y\)

\(= (2x)(2x) + (2x)(-1) + (3)(2x) + (3)(-1)\)

\(= (4x^2) + (-2x) + (6x) + (-3)\)

\(= 4x^2 - 2x + 6x - 3\)

\(= 4x^2 + 4x - 3\)

Factorising

Factorising is the opposite of expanding brackets. Instead of removing brackets, you add them into an algebraic expression or equation. Factorising is another method that can help you solve algebraic equations in some situations (however, it would not be helpful to keep switching between expanding and factorising the same equation).

This section explains how to factorise using one set of brackets, and also looks at a common example of factorising with two sets of brackets. Note that the examples use algebraic expressions for simplicity. You will learn how to apply these methods to solve algebraic equations in later sections of this module.

Factorising with one set of brackets

You can factorise an algebraic expression using one set of brackets by following these steps:

  1. Identify the highest common factor (HCF) of all the terms in the expression. The highest common factor is the largest value that divides into each term. It may include a constant, a variable (or variables), or a combination of both.

  2. Place the highest common factor outside of a set of brackets.

  3. Write the remaining part of each term inside the brackets.

Example: Factorise the algebraic expression \(3x^2 + 21x\)

Solution: Following the steps above, we have:

  1. The highest common factor is \(3x\), since \(3\) divides into each term in the expression a whole number of times and there is one \(x\) in each term.

  2. Putting this outside a set of brackets gives \(3x(…..)\)

  3. Since there is an \(x\) remaining in \(3x^2\) when it is divided through by \(3x\), the first term inside the brackets should be \(x\).
    Since there is a \(7\) remaining in \(21x\) when it is divided through by \(3x\), the second term inside the brackets should be \(7\).
    Therefore it follows that the factorised form of the expression is \(3x(x + 7)\)

Note that \(x^2\) is an example of an exponent (otherwise known as an indice or power). If you are unsure of how to work with exponents, it is recommended that you visit the Exponents page of this module before proceeding with further examples on this page.

Factorising with two sets of brackets

Sometimes you need two sets of brackets to factorise an expression. A common example is a quadratic expression, which has the form \(ax^2 + bx + c\) (where \(a\), \(b\) and \(c\) are numbers, and \(a\) does not equal 0 - as this would not form a quadratic).

For more information on quadratic equations see the Solving quadratic equations page of this module. For now, you should know that you can factorise a quadratic expression where \(a = 1\) by following these steps:

  1. Write two sets of brackets, and place the variable from the expression inside both brackets.

  2. Find pairs of factors for the last term in the expression. Make sure you consider whether the factors are positive or negative.

  3. Decide which pair of factors can be added or subtracted to give the constant in the middle term of the expression. Then place one factor in each set of brackets, using the correct signs.

Example: Factorise the algebraic expression \(x^2 + 7x + 12\)

Solution: Following the steps above, we have:

  1. Writing out two sets of brackets with the variable inside each gives \((x\textrm{…..})(x\textrm{…..})\)

  2. The pairs of factors for \(12\) are \(1\) and \(12\), \(2\) and \(6\) and \(3\) and \(4\). Note that both terms must be positive, since they need to multiply together to make positive \(12\) and add together to make positive \(7\)

  3. The pair of factors that can be added together to make \(7\) is \(3\) and \(4\), so the factorised form of \(x^2 + 7x + 12\) is \((x + 3)(x + 4)\), or \((x + 4)(x + 3)\) (it doesn’t matter which way around you write it)

After you have worked through the examples above, try factorising some or all of the following algebraic expressions (again, visit the Exponents page of this module first if needed).

Common factor is \(2a\), so factorising gives \(2a(7 + 4b)\)

Common factor is \(3x^2y\), so factorising gives \(3x^2y(4x - y + 2z)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-12\) and add together to make \(1\) (since \(x = 1x\)). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and one of each pair must be negative and one positive in order to multiply together to make \(-12\). The correct pair is \(-3\) and \(4\), since \(-3 + 4 = 1\), so the expression can be factorised as \((x - 3)(x + 4)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(12\) and add together to make \(-8\). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and both numbers must be negative in order to multiply together to make \(12\) and add together to make \(-8\). The correct pair is \(-2\) and \(-6\), since \(-2 - 6 = -8\), so the expression can be factorised as \((y - 2)(y - 6)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-12\) and add together to make \(-4\). The factors of \(12\) are \(1\) & \(12\), \(2\) & \(6\) and \(3\) & \(4\), and one of each pair must be negative and one positive in order to multiply together to make \(-12\). The correct pair is \(2\) and \(-6\), since \(2 - 6 = -4\), so the expression can be factorised as \((z + 2)(z - 6)\)

Common factor is \(5c\), so factorising gives \(5c(2 + 3c)\)

Common factor is \(4ac\), so factorising gives \(4ac(4b + 1 - 3bc)\)

Common factor is \(3m\), so factorising gives \(3m(n + 2m - 4n^2)\)

Common factor is \(-5x^2\), so factorising gives \(-5x^2(y + 2x + 4)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(6\) and add together to make \(5\). The factors of \(6\) are \(1\) & \(6\) and \(2\) & \(3\), and both numbers must be positive in order to multiply together to make \(6\) and add together to make \(5\). The correct pair is \(2\) and \(3\), since \(2 + 3 = 5\), so the expression can be factorised as \((x + 2)(x + 3)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-10\) and add together to make \(-3\). The factors of \(10\) are \(1\) & \(10\) and \(2\) & \(5\), and one of each pair must be negative and one positive in order to multiply together to make \(-10\). The correct pair is \(2\) and \(-5\), since \(2 - 5 = -3\), so the expression can be factorised as \((y + 2)(y - 5)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(24\) and add together to make \(-10\). The factors of \(24\) are \(1\) & \(24\), \(2\) & \(12\), \(3\) & \(8\) and \(4\) & \(6\), and both numbers must be negative in order to multiply together to make \(24\) and add together to make \(-10\). The correct pair is \(-6\) and \(-4\), since \(-6 - 4 = -10\), so the expression can be factorised as \((a - 6)(a - 4)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-20\) and add together to make \(-1\). The factors of \(20\) are \(1\) & \(20\), \(2\) & \(10\) and \(4\) & \(5\), and one of each pair must be negative and one positive in order to multiply together to make \(-20\). The correct pair is \(4\) and \(-5\), since \(4 - 5 = -1\), so the expression can be factorised as \((b + 4)(b - 5)\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(1\) and add together to make \(2\). The only factors of \(1\) are \(1\) & \(1\), and we need both of these to be positive in order to multiply together to make \(1\) and add together to make \(2\). Hence the expression can be factorised as \((c + 1)(c + 1)\), which we would usually write as \((c + 1)^2\)

This is a quadratic so we need two sets of brackets, and two numbers that multiply together to make \(-30\) and add together to make \(1\). The factors of \(30\) are \(1\) & \(30\), \(2\) & \(15\), \(3\) & \(10\) and \(5\) & \(6\), and one of each pair must be negative and one positive in order to multiply together to make \(-30\). The correct pair is \(6\) and \(-5\), since \(6 - 5 = 1\), so the expression can be factorised as \((d + 6)(d - 5)\)