Presentation on theme: "Some problems produce equations that have variables on both sides of the equal sign."— Presentation transcript:
1Learn to solve equations with variables on both sides of the equal sign.
2Some 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.
3Additional Example 1A: Solving Equations with Variables on Both Sides Solve.4x + 6 = x4x + 6 = x– 4x – 4xSubtract 4x from both sides.6 = –3x6–3–3x=Divide both sides by –3.–2 = x
4Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2.Helpful Hint
5Additional Example 1B: Solving Equations with Variables on Both Sides Solve.9b – 6 = 5b + 189b – 6 = 5b + 18– 5b – 5bSubtract 5b from both sides.4b – 6 = 18Add 6 to both sides.4b = 244b424=Divide both sides by 4.b = 6
6Additional Example 1C: Solving Equations with Variables on Both Sides Solve.9w + 3 = 9w + 79w + 3 = 9w + 7– 9w – 9wSubtract 9w from both sides.3 ≠No solution. There is no number that can be substituted for the variable w to make the equation true.
7If the variables in an equation are eliminated and the resulting statement is false, the equation has no solution.Helpful Hint
8Check It Out: Example 1ASolve.5x + 8 = x5x + 8 = x– 5x – 5xSubtract 5x from both sides.8 = –4x8–4–4x=Divide both sides by –4.–2 = x
9Check It Out: Example 1BSolve.3b – 2 = 2b + 123b – 2 = 2b + 12– 2b – 2bSubtract 2b from both sides.b – 2 =Add 2 to both sides.b =
10Check It Out: Example 1CSolve.3w + 1 = 3w + 83w + 1 = 3w + 8– 3w – 3wSubtract 3w from both sides.1 ≠No solution. There is no number that can be substituted for the variable w to make the equation true.
11To 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.
12Additional Example 2: Solving Multi-Step Equations with Variables on Both Sides Solve.10z – 15 – 4z = 8 – 2z - 1510z – 15 – 4z = 8 – 2z – 156z – 15 = –2z – 7Combine like terms.+ 2z zAdd 2z to both sides.8z – 15 = – 7Add 15 to both sides.8z = 88z 88=Divide both sides by 8.z = 1
13Check It Out: Example 2Solve.12z – 12 – 4z = 6 – 2z + 3212z – 12 – 4z = 6 – 2z + 328z – 12 = –2z + 38Combine like terms.+ 2z zAdd 2z to both sides.10z – 12 =Add 12 to both sides.10z = 5010z10=Divide both sides by 10.z = 5
14Additional 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.
15Additional Example 3 Continued Let r represent the price of one rose.r = rSubtract 2.95r from both sides.– 2.95r – 2.95r= rSubtract from both sides.– – 26.00= r13.951.551.55r 1.55=Divide both sides by 1.55.9 = rThe two services would cost the same when purchasing 9 roses.
16Additional 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 + 10eLocate e in the equation.t = m + 10eSince m is added to 10e, subtract m from both sides.–m –mt – m = eSince e is multiplied 10, divide both sides by 10.t – m = 10e10t – m = e10
17Lesson QuizSolve.1. 4x + 16 = 2x2. 8x – 3 = x3. 2(3x + 11) = 6x + 44. x = x – 95. 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 = –8x = 6no solutionx = 361412An orange has 45 calories. An apple has 75 calories.