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INTRODUCTION TO ALGEBRA • Algebra is the practice of using expressions with letters or variables that represent numbers. Words can be changed into a mathematical expression by using the words, plus, exceeds, diminished, less, times, the product, divided, the quotient and many more. • When given an algebraic expression, it can be solved by filling in a number for the variable. • Word problems can be turned into variable expressions by changing the words to mathematical terms. • If an expression has more than one variable expression, it can be combined as long as both have the same variable factor; this is called combining like terms. • With algebra, inverse operations can be used to solve equations. Inverse operations are used to isolate a variable. Inverse operations undo an operation; addition is the inverse operation of subtraction and vice versa as well as multiplication being the inverse operation of division and vice versa. How to use algebra: • An example of an algebraic expression is 3x + 5. Here the x represents a number that is to be multiplied by three. If x = 2, then the expression equals 3 · 2 + 5 = 11. By filling in 2 for the x, the expression can be solved. • Words can be changed into mathematical terms. Look at the following words and translate them into mathematical terms: Ex. Five times a number minus three → 5 · n - 3 = 5n - 3 • Each word represents a mathematical term. Once this is done, the expression can either be left this way or solved if given a value for n. © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.

• Word problems are also changed into variable expressions in the same way. Look at this word problem: Jack rented a movie. The store charged $1.99 for the first day and $.50 for each day after that. If Jack had the movie for d days, what expression could be used to represent the cost of renting a movie in terms of d? $1.99 for the first day and $.50 for each day after that (.50 · d ) → the expression is 1.99 + .50d This expression can be solved when 3 (or any other number) is substituted for d, the number of days Jack had the movie. So, 1.99 + .50 · 3 = 1.99 + 1.50 = 3.49 or $3.49. • If an expression or equation has more than one variable term, the terms may be combined if the terms have the same variable factor. Example: 5x + 4 - 2x → 3x + 4 8x - 6y + x - 2y → 9x - 8y • To solve an equation using inverse operations, the variable must be isolated first and then the variable can be solved. Example: Solve for x: x + 17 = 27 - 17 -17 x = 10 Seventeen is subtracted from both sides to solve for x. On the left side, the numbers cancel out and on the right side 27 - 17 = 10, the answer. © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.

Try This! 1. Solve if n = 3: 7 - n 2n + 8 4n ÷ 6 2. Translate into an algebraic expression: o Six times a number minus two o A number plus seven 3. Translate the word problem into a variable expression in terms of d, days: Sharon rents a car that costs $89 for the first day and $50 for every day after that. She rents the car for d, days. 4. Combine like terms: 7x - 4 x 11x + 14y - 5x + 2y 6x - 64 – 3x 5. Solve by using inverse operations: x + 14 = 67 5x = 45 x/2 = 42 © Copyright NewPath Learning. All Rights Reserved. Permission is granted for the purchaser to print copies for non-commercial educational purposes only. Visit us at www.NewPathLearning.com.