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Saturday 13th of July
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Slopes of Perpendicular Lines
Linear Equations
Roots - Radicals 1
Graph of a Line
Sum of the Roots of a Quadratic
Writing Linear Equations Using Slope and Point
Factoring Trinomials with Leading Coefficient 1
Writing Linear Equations Using Slope and Point
Simplifying Expressions with Negative Exponents
Solving Equations 3
Solving Quadratic Equations
Parent and Family Graphs
Collecting Like Terms
nth Roots
Power of a Quotient Property of Exponents
Adding and Subtracting Fractions
Solving Linear Systems of Equations by Elimination
The Quadratic Formula
Fractions and Mixed Numbers
Solving Rational Equations
Multiplying Special Binomials
Rounding Numbers
Factoring by Grouping
Polar Form of a Complex Number
Solving Quadratic Equations
Simplifying Complex Fractions
Common Logs
Operations on Signed Numbers
Multiplying Fractions in General
Dividing Polynomials
Higher Degrees and Variable Exponents
Solving Quadratic Inequalities with a Sign Graph
Writing a Rational Expression in Lowest Terms
Solving Quadratic Inequalities with a Sign Graph
Solving Linear Equations
The Square of a Binomial
Properties of Negative Exponents
Inverse Functions
Rotating an Ellipse
Multiplying Numbers
Linear Equations
Solving Equations with One Log Term
Combining Operations
The Ellipse
Straight Lines
Graphing Inequalities in Two Variables
Solving Trigonometric Equations
Adding and Subtracting Fractions
Simple Trinomials as Products of Binomials
Ratios and Proportions
Solving Equations
Multiplying and Dividing Fractions 2
Rational Numbers
Difference of Two Squares
Factoring Polynomials by Grouping
Solving Equations That Contain Rational Expressions
Solving Quadratic Equations
Dividing and Subtracting Rational Expressions
Square Roots and Real Numbers
Order of Operations
Solving Nonlinear Equations by Substitution
The Distance and Midpoint Formulas
Linear Equations
Graphing Using x- and y- Intercepts
Properties of Exponents
Solving Quadratic Equations
Solving One-Step Equations Using Algebra
Relatively Prime Numbers
Solving a Quadratic Inequality with Two Solutions
Operations on Radicals
Factoring a Difference of Two Squares
Straight Lines
Solving Quadratic Equations by Factoring
Graphing Logarithmic Functions
Simplifying Expressions Involving Variables
Adding Integers
Factoring Completely General Quadratic Trinomials
Using Patterns to Multiply Two Binomials
Adding and Subtracting Rational Expressions With Unlike Denominators
Rational Exponents
Horizontal and Vertical Lines
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Collecting Like Terms

In an algebraic expression, there are terms with identical symbolic or variable parts. We call them like terms. For example,

‘4x’ and ‘2x‘ are like terms because both of them contain the symbolic part ‘x’

‘3xyz 2’ and ‘11x yz 2’ are like terms because they both contain the symbolic part ‘xyz 2

‘2x 5 ’and ‘-2x’ are not like terms because even though the symbol present in both is ‘x’, the symbolic part ‘x 5 ’ is not identical to the symbolic part ‘x’.

We can simplify algebraic expressions containing like terms by combining each group of like terms into a single term.

In fact, it is quite easy to see why this is possible (and valid!). For instance, consider the expression

2x + 5x

which is the sum of two like terms, representing the accumulation of two x’s and another five x’s. It is no news that the result of this sum is 7x.

2x + 5x = (2 + 5)x = 7x

We can combine (or collect) like terms with each group of like terms that appear in an expression. The result will be an equivalente expression to the given one, which looks simpler (in the sense that it can be written in fewer terms).

This process of combining (or collecting ) like terms can be performed for each group of like terms that appear in an expression. The net effect will be that the original expression can now be written with fewer terms, yet which are entirely equivalent to the terms in the original expression.

Sometimes a bit more care must be taken to recognize like terms.


Simplify 5abc – 3bca + 9cba – 4cab


This expression has four terms altogether. At first glance, all four may seem to have different literal parts, since if we were reading the literal parts as English words, abc, bca, cba, and cab would certainly appear to be different.

However, keep in mind that these literal parts of the terms are not characters in a word, but each character or symbol represents a number. A literal part such as ‘abc’ then represents the result of multiplying together the three numbers represented by ‘a’, ‘b’, and ‘c’. When we multiply three numbers together, it doesn’t matter in what order we do the multiplication:

3 × 4 × 5 = 4 × 3 × 5 = 5 × 4 × 3, … etc.

Similarly, then

abc, bac, cba, and cab

really represent the same result or quantity because they are products of the same three numbers, but with the multiplications being done in different orders.

An easy way to be able to compare the literal parts of the terms in the example expression here is to write the symbols in each product in alphabetic order. Thus

5abc – 3bca + 9cba – 4cab = 5abc – 3abc + 9abc – 4abc

Now we see that all four terms here have exactly the same literal parts, and so all four are like terms. Thus,

5abc – 3bca + 9cba – 4cab = 5abc – 3abc + 9abc – 4abc

= (5 – 3 + 9 – 4)abc

= 7abc

So, the original four term expression can be simplified to just ‘7abc’. No matter what values are substituted for a, b, and c, the simplified expression will give the same result as the original, much lengthier expression (which is what we mean by saying that the two expressions are equivalent). Obviously then, it is preferable to retain the simplified expression, 7abc, in place of the original, much lengthier expression.

Some care must be taken to ensure that terms being combined in this was really are like terms.


Simplify: 5x 2 y – 3xy 2 + 7yx 2


All three terms in this expression involve x, and y, and a square. Following the strategy used in the previous example, we rewrite the expression so that the symbols in each term appear in alphabetic order:

5x 2 y – 3xy 2 + 7yx 2 = 5x 2 y – 3xy 2 + 7x 2 y

Now you can see that there are two like terms here – the first and the third. The second term, despite involving the same symbols, one squared and one not, is not actually “like” the other two because the symbol which is squared is not the same one:

x 2 y is not the same as xy 2

Thus, all we can do here is to combine the first and the third terms, which are like terms, to get

5x 2 y – 3xy 2 + 7yx 2

= 5x 2 y – 3xy 2 + 7x 2 y

= (5 + 7)x 2 y – 3xy 2

= 12x 2 y – 3xy 2

as a final answer (for now!).


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