Solving Equations with Variables on Both Sides 7-3 Warm Up Warm Up Lesson Presentation Lesson Presentation Problem of the Day Problem of the Day Lesson Quizzes Lesson Quizzes

Solving Equations with Variables on Both Sides 7-3 Warm Up Solve. 1. 2x + 9x – 3x + 8 = – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 x = 1 x = –13 x = x x9x 16 2x2x x = 50

Solving Equations with Variables on Both Sides 7-3 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.

Solving Equations with Variables on Both Sides 7-3 Learn to solve equations with variables on both sides of the equal sign.

Solving Equations with Variables on Both Sides 7-3 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.

Solving Equations with Variables on Both Sides 7-3 Solve. 4x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4x + 6 = x – 4x 6 = –3x Subtract 4x from both sides. Divide both sides by –3. –2 = x 6 –3 –3x –3 =

Solving Equations with Variables on Both Sides 7-3 Check your solution by substituting the value back into the original equation. For example, 4(2) + 6 = 2 or 2 = 2. Helpful Hint

Solving Equations with Variables on Both Sides 7-3 Solve. 9b – 6 = 5b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9b – 6 = 5b + 18 – 5b 4b – 6 = 18 4b4b = Subtract 5b from both sides. Divide both sides by 4. b = b = 24 Add 6 to both sides.

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

Solving Equations with Variables on Both Sides 7-3 If the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

Solving Equations with Variables on Both Sides 7-3 Solve. 5x + 8 = x Check It Out: Example 1A 5x + 8 = x – 5x 8 = –4x Subtract 5x from both sides. Divide both sides by –4. –2 = x 8 –4 –4x –4 =

Solving Equations with Variables on Both Sides 7-3 Solve. 3b – 2 = 2b + 12 – 2b b – 2 = 12 Subtract 2b from both sides. + 2 b = 14 Add 2 to both sides. Check It Out: Example 1B

Solving Equations with Variables on Both Sides 7-3 Solve. 3w + 1 = 3w ≠ 8 3w + 1 = 3w + 8 – 3w Subtract 3w from both sides. No solution. There is no number that can be substituted for the variable w to make the equation true. Check It Out: Example 1C

Solving Equations with Variables on Both Sides 7-3 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.

Solving Equations with Variables on Both Sides 7-3 Solve. 10z – 15 – 4z = 8 – 2z - 15 Additional Example 2: Solving Multi-Step Equations with Variables on Both Sides 10z – 15 – 4z = 8 – 2z – z – 15 = –2z – 7Combine like terms. + 2z Add 2z to both sides. 8z – 15 = – 7 8z = 8 z = 1 Add 15 to both sides. Divide both sides by 8. 8z =

Solving Equations with Variables on Both Sides 7-3 Solve. 12z – 12 – 4z = 6 – 2z + 32 Check It Out: Example 2 12z – 12 – 4z = 6 – 2z z – 12 = –2z + 38Combine like terms. + 2z Add 2z to both sides. 10z – 12 = 38 10z = 50 z = 5 Add 12 to both sides. Divide both sides by z =

Solving Equations with Variables on Both Sides 7-3 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.

Solving Equations with Variables on Both Sides 7-3 Additional Example 3 Continued r = r Let r represent the price of one rose. – 2.95r = r Subtract 2.95r from both sides. – Subtract from both sides = 1.55r r 1.55 = Divide both sides by = r The two services would cost the same when purchasing 9 roses.

Solving Equations with Variables on Both Sides 7-3 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.

Solving Equations with Variables on Both Sides 7-3 Check It Out: Example 3 Continued b = b Let b represent the price of one balloon. – 2.25b = b Subtract 2.25b from both sides. – Subtract from both sides = 0.75b b 0.75 = Divide both sides by = b The two services would cost the same when purchasing 8 balloons.

Solving Equations with Variables on Both Sides 7-3 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?

Solving Equations with Variables on Both Sides 7-3 Additional Example 4 Continued First solve for the price of one doughnut d = d Let d represent the price of one doughnut. – 2d 1.25 = d Subtract 2d from both sides. – 0.50 Subtract 0.50 from both sides = 3d d3d 3 = Divide both sides by = d The price of one doughnut is $0.25.

Solving Equations with Variables on Both Sides 7-3 Additional Example 4 Continued Now find the amount of money Jamie spends each morning d Choose one of the original expressions. Jamie spends $1.75 each morning (0.25) = n = Let n represent the number of doughnuts. Find the number of doughnuts Jamie buys on Tuesday. 0.25n = 1.75 n = 7; Jamie bought 7 doughnuts on Tuesday. Divide both sides by 0.25.

Solving Equations with Variables on Both Sides 7-3 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?

Solving Equations with Variables on Both Sides 7-3 Check It Out: Example 4 Continued First solve for distance around the track. 2x + 4 = 4x + 2 Let x represent the distance around the track. – 2x 4 = 2x + 2 Subtract 2x from both sides. – 2 Subtract 2 from both sides. 2 = 2x 2 2 2x2x 2 = Divide both sides by 2. 1 = x The track is 1 mile around.

Solving Equations with Variables on Both Sides 7-3 Check It Out: Example 4 Continued Now find the total distance Helene walks each day. 2x + 4 Choose one of the original expressions. Helene walks 6 miles each day.2(1) + 4 = 6 Let n represent the number of 1-mile laps. Find the number of laps Helene walks on Saturdays. 1n = 6 Helene walks 6 laps on Saturdays. n = 6

Solving Equations with Variables on Both Sides 7-3 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. Since m is added to 10e, subtract m from both sides. t = m + 10e –m –m t – m = 10e 10 t – m = e 10 Since e is multiplied 10, divide both sides by 10.

Solving Equations with Variables on Both Sides 7-3 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. Since gt is subtracted from i, add gt to both sides to undo the subtraction. f = i – gt + gt +gt f + gt = i

Solving Equations with Variables on Both Sides 7-3 Standard Lesson Quiz Lesson Quizzes Lesson Quiz for Student Response Systems

Solving Equations with Variables on Both Sides 7-3 Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = x 3. 2(3x + 11) = 6x 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 = 6 x = –8 no solution x = An orange has 45 calories. An apple has 75 calories.

Solving Equations with Variables on Both Sides Combine like terms. 4p + 14 = 11p A. p = 2 B. p = 7 C. p = 14 D. p = 15 Lesson Quiz for Student Response Systems

Solving Equations with Variables on Both Sides 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

Solving Equations with Variables on Both Sides 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

Solving Equations with Variables on Both Sides Solve for D. A. B. D = V C. D = mV D. Lesson Quiz for Student Response Systems