Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

Math Journal (5 Min) “What Do I Know About” – Each student will be given the title of the lesson that will be taught that day. They must then, at the beginning of class, write about what they know about the lesson before they have been taught the lesson, and at the end of class, write about what they now know about the lesson after they have been taught the lesson. Then, each student will discuss his/her answers within their group. Finally, to leave class, each student will have to give/write 1 sentence that they now know that pertained to the lesson.

Homework Review (5 Min)

Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. –4 = 6x + 22 – 4x 3. + = 5 3. + = 5 4. – = 3 x = 1 x = –13 2 7 x 7 7 1 x = 34 9x 16 2x 4 1 8 x = 50

Problem of the Day An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle? 22.5 in.

Textbook Examples (I Do) (5 Min)

Learn to solve equations with variables on both sides of the equal sign.

Some problems produce equations that have variables on both sides of the equal sign. Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

Additional Example 1A: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x Subtract 4x from both sides. 6 = –3x 6 –3 –3x = Divide both sides by –3. –2 = x

Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Helpful Hint

Additional Example 1B: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b – 5b Subtract 5b from both sides. 4b – 6 = 18 + 6 + 6 Add 6 to both sides. 4b = 24 4b 4 24 = Divide both sides by 4. b = 6

Additional Example 1C: Solving Equations with Variables on Both Sides Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w Subtract 9w from both sides. 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true.

If the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

Check It Out: Example 1A Solve. 5x + 8 = x 5x + 8 = x – 5x – 5x Subtract 5x from both sides. 8 = –4x 8 –4 –4x = Divide both sides by –4. –2 = x

Check It Out: Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = 12 + 2 + 2 Add 2 to both sides. b = 14

Check It Out: Example 1C Solve. 3w + 1 = 3w + 8 3w + 1 = 3w + 8 – 3w – 3w Subtract 3w from both sides. 1 ≠ 8 No solution. There is no number that can be substituted for the variable w to make the equation true.

To solve multi-step equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

Additional Example 2: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z - 15 10z – 15 – 4z = 8 – 2z – 15 6z – 15 = –2z – 7 Combine like terms. + 2z + 2z Add 2z to both sides. 8z – 15 = – 7 + 15 +15 Add 15 to both sides. 8z = 8 8z 8 8 = Divide both sides by 8. z = 1

Check It Out: Example 2 Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z + 2z Add 2z to both sides. 10z – 12 = 38 + 12 +12 Add 12 to both sides. 10z = 50 10z 50 10 = Divide both sides by 10. z = 5

Additional Example 3: Business Application Daisy’s Flowers sell a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florists’ bouquets cost the same price.

Additional Example 3 Continued Let r represent the price of one rose. 39.95 + 2.95r = 26.00 + 4.50r Subtract 2.95r from both sides. – 2.95r – 2.95r 39.95 = 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00 – 26.00 13.95 = 1.55r 13.95 1.55 1.55r 1.55 = Divide both sides by 1.55. 9 = r The two services would cost the same when purchasing 9 roses.

Check It Out: Example 3 Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both gift basket companies muffin baskets cost the same price.

Check It Out: Example 3 Continued Let b represent the price of one balloon. 22.00 + 2.25b = 16.00 + 3.00b Subtract 2.25b from both sides. – 2.25b – 2.25b 22.00 = 16.00 + 0.75b Subtract 16.00 from both sides. – 16.00 – 16.00 6.00 = 0.75b 6.00 0.75 0.75b 0.75 = Divide both sides by 0.75. 8 = b The two services would cost the same when purchasing 8 balloons.

Additional Example 4: Multi-Step Application Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

Additional Example 4 Continued First solve for the price of one doughnut. Let d represent the price of one doughnut. 1.25 + 2d = 0.50 + 5d – 2d – 2d Subtract 2d from both sides. 1.25 = 0.50 + 3d Subtract 0.50 from both sides. – 0.50 – 0.50 0.75 = 3d 0.75 3 3d = Divide both sides by 3. The price of one doughnut is $0.25. 0.25 = d

Additional Example 4 Continued Now find the amount of money Jamie spends each morning. Choose one of the original expressions. 1.25 + 2d 1.25 + 2(0.25) = 1.75 Jamie spends $1.75 each morning. Find the number of doughnuts Jamie buys on Tuesday. Let n represent the number of doughnuts. 0.25n = 1.75 0.25n 0.25 1.75 = Divide both sides by 0.25. n = 7; Jamie bought 7 doughnuts on Tuesday.

Check It Out: Example 4 Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

Check It Out: Example 4 Continued First solve for distance around the track. Let x represent the distance around the track. 2x + 4 = 4x + 2 – 2x – 2x Subtract 2x from both sides. 4 = 2x + 2 – 2 – 2 Subtract 2 from both sides. 2 = 2x 2 2x = Divide both sides by 2. The track is 1 mile around. 1 = x

Check It Out: Example 4 Continued Now find the total distance Helene walks each day. Choose one of the original expressions. 2x + 4 2(1) + 4 = 6 Helene walks 6 miles each day. Find the number of laps Helene walks on Saturdays. Let n represent the number of 1-mile laps. 1n = 6 n = 6 Helene walks 6 laps on Saturdays.

Additional Example 5: Solving Literal Equations for a Variable The equation t = m + 10e gives the test score t for a student who answers m multiple-choice questions and e essay questions correctly. Solve this equation for e. t = m + 10e Locate e in the equation. t = m + 10e Since m is added to 10e, subtract m from both sides. –m –m t – m = 10e Since e is multiplied 10, divide both sides by 10. t – m = 10e 10 t – m = e 10

Check It Out! Example 5 The formula for an object’s final velocity is f = i – gt, where i is the object’s initial velocity, g is acceleration due to gravity, and t is time. Solve for i. f = i – gt Locate i in the equation. f = i – gt + gt +gt Since gt is subtracted from i, add gt to both sides to undo the subtraction. f + gt = i

Class work Problems (We Do) (10 Min) Pg. 312-313 (1-9)

Small Group CW(Yall Do) (10 Min) Pg. 312-313 (10-32 EOE)

Homework (You Do) (10 Min) Pg. 312-313 (11, 17, 19, 23, 25 odd)

Math Journal (5 Min) “What Do I Know About” – Each student will be given the title of the lesson that will be taught that day. They must then, at the beginning of class, write about what they know about the lesson before they have been taught the lesson, and at the end of class, write about what they now know about the lesson after they have been taught the lesson. Then, each student will discuss his/her answers within their group. Finally, to leave class, each student will have to give/write 1 sentence that they now know that pertained to the lesson.

Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = 15 + 5x 3. 2(3x + 11) = 6x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = –8 x = 6 no solution x = 36 1 4 1 2 An orange has 45 calories. An apple has 75 calories.

Lesson Quiz for Student Response Systems 1. Combine like terms. 4p + 14 = 11p A. p = 2 B. p = 7 C. p = 14 D. p = 15

Lesson Quiz for Student Response Systems 2. Combine like terms. 3g – 6 = 4g – 7 A. g = 1 B. g = 7 C. g = 13 D. g = –13

Lesson Quiz for Student Response Systems 3. Combine like terms. 3(g – 2) = 7g – 18 A. g = 2 B. g = 3 C. g = –3 D. g = –2

Lesson Quiz for Student Response Systems 4. Solve for D. A. B. D = V C. D = mV D.