Equations with Variables on Both Sides Lesson 3
Equations with Variables on Both Sides Lesson 3 -3 Algebra 1 Additional Examples The measure of an angle is (5 x – 3)°. Its vertical angle has a measure of (2 x + 12)°. Find the value of x. 5 x – 3 = 2 x + 12 5 x – 3 – 2 x = 2 x + 12 – 2 x Vertical angles are congruent. Subtract 2 x from each side. 3 x – 3 = 12 Combine like terms. 3 x – 3 + 3 = 12 + 3 Add 3 to each side. 3 x = 15 Simplify. 3 x = 15 3 3 Divide each side by 3. x = 5 Simplify.
Equations with Variables on Both Sides Lesson 3 -3 Algebra 1 Additional Examples You can buy a skateboard for $60 from a friend and rent the safety equipment for $1. 50 per hour. Or you can rent all items you need for $5. 50 per hour. How many hours must you use a skateboard to justify buying your friend’s skateboard? Relate: cost of plus equipment equals friend’s rental skateboard and equipment rental Define: let h = the number of hours you must skateboard. Write: 60 + 1. 5 h = 5. 5 h
Equations with Variables on Both Sides Lesson 3 -3 Algebra 1 Additional Examples (continued) 60 + 1. 5 h = 5. 5 h 60 + 1. 5 h – 1. 5 h = 5. 5 h – 1. 5 h Subtract 1. 5 h from each side. 60 = 4 h Combine like terms. 60 = 4 h 4 4 Divide each side by 4. 15 = h Simplify. You must use your skateboard for more than 15 hours to justify buying the skateboard.
Equations with Variables on Both Sides Lesson 3 -3 Additional Examples Algebra 1 Solve 2. 5 x ± 1 ≠ 4 x – 2. 6 using a graphing calculator. Step 1: For Y 1= enter 2. 5 x + 1. For Y 2= enter 4 x – 2. 6. Step 2: Use the GRAPH feature to display the graph. You can adjust the window by using the ZOOM or WINDOW features. Step 3: Use the CALC feature. Select intersect to find the point where the lines intersect. The lines intersect at (2. 4, 7). The x-value 2. 4 is the solution of the equation.
Equations with Variables on Both Sides Lesson 3 -3 Additional Examples Algebra 1 Solve each equation. a. 4 – 4 y = – 2(2 y – 2) 4 – 4 y = – 4 y + 4 Use the Distributive Property. 4 – 4 y + 4 y = – 4 y + 4 y Add 4 y to each side. 4 = 4 Always true! The equation is true for every value of y, so the equation is an identity. b. – 6 z + 8 = z + 10 – 7 z – 6 z + 8 = – 6 z + 10 Combine like terms. – 6 z + 8 + 6 z = – 6 z + 10 + 6 z Add 6 z to each side. 8 = 10 Not true for any value of z! This equation has no solution
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