# Download Algebra II (Cliffs Quick Review) by Edward Kohn, David Alan Herzog PDF

By Edward Kohn, David Alan Herzog

In terms of pinpointing the belongings you actually need to understand, not anyone does it larger than CliffsNotes. This quickly, powerful instructional is helping you grasp middle algebraic recommendations -- from linear equations, family and features, and rational expressions to radicals, quadratic structures, and factoring polynomials -- and get the very best grade.

At CliffsNotes, we're devoted to assisting you do your most sensible, regardless of how tough the topic. Our authors are veteran lecturers and proficient writers who know the way to chop to the chase -- and nil in at the crucial details you must prevail.

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**Extra resources for Algebra II (Cliffs Quick Review)**

**Example text**

Midpoint Formula M = d x1 + x 2 , 2 y1 + y 2 2 n Example 2: In Figure 2-3, R is the midpoint between Q(–9,–1) and T(–3,7). Find its coordinates. F 26 4/19/01 8:50 AM Page 26 CliffsQuickReview Algebra II Figure 2-3 Finding the midpoint. y T (−3,7) 7 6 5 4 3 R 2 1 −9 −8 −7 −6 −5 −4 −3 −2 −1 Q (−9,−1) −1 1 2 3 x −2 −3 By the midpoint formula, M = e x 1 +2 x 2 , R=e y1 + y 2 2 o - 9 + ^ - 3h - 1 + 7 , 2 o 2 = ` -212 , 26 j = ^ - 6, 3h Slope of a Line The slope of a line is a measurement of the steepness and direction of a nonvertical line.

F 4/25/01 8:39 AM Page 43 Chapter 3: Linear Sentences in Two Variables 43 Linear Equations: Solutions Using Substitution To solve systems using substitution, follow this procedure. 1. Select one equation and solve for one of its variables. 2. In the other equation, substitute for the variable just solved. 3. Solve the new equation. 4. Substitute the value found into any equation involving both variables and solve for the other variable. 5. Check the solution in both original equations. Example 2: Solve this system of equations by using substitution.

Figure 2-11 The boundary is dashed. y 4 3 2 1 −3 −2 −1 1 2 3 −1 x 4 3x + 4y = 12 −2 −3 Now, select a point not on the boundary, say (0, 0). Substitute this into the original inequality: 3x + 4y < 12 ? 3(0) + 4 (0) < 12 ? 0 + 0 < 12 0 < 12 ✓ This is a true statement. This means that the “(0, 0) side” of the boundary line is the desired region to be shaded. Now, shade that region as shown in Figure 2-12. F 4/19/01 8:50 AM Page 37 Chapter 2: Segments, Lines, and Inequalities 37 Figure 2-12 The shading is below the line.