Download CliffsNotes Algebra I: Quick Review (2nd Edition) by Jerry Bobrow, Ed Kohn PDF

By Jerry Bobrow, Ed Kohn

Overview

Inside the Book:

Preliminaries and easy Operations
Signed Numbers, Frac-tions, and Percents
Terminology, units, and Expressions
Equations, Ratios, and Proportions
Equations with Vari-ables
Monomials, Polynomials, and Factoring
Algebraic Fractions
Inequalities, Graphing, and Absolute Value
Coordinate Geometry
capabilities and Variations
Roots and Radicals
Quadratic Equations
note Problems
evaluate Questions
source Center
Glossary

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Additional resources for CliffsNotes Algebra I: Quick Review (2nd Edition)

Example text

G. 5 = 2. 5 to both sides. g. Solve 4x = 14. x = 3 12 Dividing both sides by 4. e. equations) except that when multiplying or dividing by a negative number you must reverse the inequality sign. g. 4y + 6 < 26 4y < 20 Taking 6 from both sides. y<5 Dividing both sides by 4. g. Solve x = 7. g. Solve 6(3x – 5) = 42. 6 × 3x + 6 × –5 = 42 18x – 30 = 42 18x = 72 x=4 Alternatively: 6(3x – 5) = 42 3x – 5 = 7 3x = 12 x=4 32 Multiplying both sides by 3. y<5 0 1 2 3 4 5 6 The open circle means the value 5 is not included.

Expanding and factorising Brackets are used to group algebraic terms and the process of removing the brackets is called expanding. The process of rewriting an expression including the brackets is called factorising. g. Expand and simplify the following. 2a – 3(4b – 3a) = 2a – 3 × 4b – 3 × –3a = 2a – 12b + 9a = 11a – 12b Factorising into brackets To factorise an expression, look for terms that have common factors. g. Factorise the following. 10a – 15 = 5(2a – 3) 9rs + 12st = 3s(3r + 4t) 8ab2 – 16a2b = 8ab(b – 2a) NB You should check your answers by expanding the brackets.

E. either (x – 5) = 0 which implies that x = 5 or (x + 3) = 0 which implies that x = –3. So the solutions of the equation (x – 5)(x + 3) = 0 are x = 5 and x= –3. g. Solve the quadratic equation x2 – 6x – 27 = 0. To solve the equation, factorise the left-hand side of the equation (see Algebra card 3) and then solve as before. Factorising the left-hand side of the equation: x2 – 6x – 27 = (x )(x ) Now search for two numbers which multiply together to give –27. Try substituting: 1 × –27 gives 1 × +27 gives + 9 × –3 gives – 9 × +3 gives + – – – 27: 27: – 27: – 27: (x + 1)(x – 27) = x2 – 26x – 27 (x – 1)(x + 27) = x2 + 26x – 27 (x + 9)(x – 3) = x2 + 6x – 27 (x – 9)(x + 3) = x2 – 6x – 27 ✓ The quadratic equation can be written (x – 9)(x + 3) = 0 and since the product of the two brackets is zero then the expression inside one or both of them must be zero.

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