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Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Algebra 1 Review.

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Presentation on theme: "Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Algebra 1 Review."— Presentation transcript:

1 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Algebra 1 Review

2 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Quiz 1. Squares from 1 to 12? 2. What is i? 3. Define slope? 4. What is a coordinate?

3 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

4 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. Example 1: Solving Equations with Variables on Both Sides To collect the variable terms on one side, subtract 5n from both sides. 7n – 2 = 5n + 6 –5n 2n – 2 = 6 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. 2n = 8 + 2 n = 4

5 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 4b + 2 = 3b. Check It Out! Example 1a To collect the variable terms on one side, subtract 3b from both sides. 4b + 2 = 3b –3b b + 2 = 0 b = –2 – 2 – 2

6 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 0.5 + 0.3y = 0.7y – 0.3. Check It Out! Example 1b To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication.

7 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Distribution Property 5(x + 7) 5(x) + 5(7) 5x + 35

8 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2: Simplifying Each Side Before Solving Equations Combine like terms. Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +36 40 – 2a = 10a + 2a +2a 40 = 12a Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides.

9 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Example 2 Continued 40 = 12a Since a is multiplied by 12, divide both sides by 12.

10 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve. Check It Out! Example 2a Since 1 is subtracted from b, add 1 to both sides. Distribute to the expression in parentheses. 1 2 + 1 3 = b – 1 To collect the variable terms on one side, subtract b from both sides. 1 2 4 = b

11 Holt McDougal Algebra 1 1-5 Solving Equations with Variables on Both Sides Solve 3x + 15 – 9 = 2(x + 2). Check It Out! Example 2b Combine like terms. Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 –2x x + 6 = 4 – 6 – 6 x = –2 To collect the variable terms on one side, subtract 2x from both sides. Since 6 is added to x, subtract 6 from both sides to undo the addition.

12 Holt McDougal Algebra 1 4-4 The Slope Formula Example 1: Finding Slope by Using the Slope Formula Find the slope of the line that contains (2, 5) and (8, 1). Use the slope formula. Substitute (2, 5) for (x 1, y 1 ) and (8, 1) for (x 2, y 2 ). Simplify. The slope of the line that contains (2, 5) and (8, 1) is.

13 Holt McDougal Algebra 1 4-6 Slope-Intercept Form Additional Example 1: Graphing by Using Slope and y-intercept Graph the line given the slope and y-intercept. Step 1 The y-intercept is 4, so the line contains (0, 4). Plot (0, 4). Step 3 Draw the line through the two points. Run = 5 Rise = –2 Step 2 Slope = Count 2 units down and 5 units right from (0, 4) and plot another point. y ; y intercept = 4 Slope =-


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