Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Pythagorean theorem wikipedia , lookup
BKL singularity wikipedia , lookup
Unification (computer science) wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Schwarzschild geodesics wikipedia , lookup
Maxwell's equations wikipedia , lookup
Navier–Stokes equations wikipedia , lookup
Euler equations (fluid dynamics) wikipedia , lookup
Differential equation wikipedia , lookup
Equations of motion wikipedia , lookup
Solving Equations with 11-3 Variables on Both Sides Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 x=1 2. –4 = 6x + 22 – 4x x = -13 3. 2 + x = 5 1 x = 34 7 7 7 4. 9x – 2x = 3 1 16 4 8 Course 3 x = 50 Solving Equations with 11-3 Variables on Both Sides 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. Course 3 Solving Equations with 11-3 Variables on Both Sides TB P. 593-597 Learn to solve equations with variables on both sides of the equal sign. Course 3 Solving Equations with 11-3 Variables on Both Sides 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. Course 3 Solving Equations with 11-3 Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x 6 = –3x 6 = –3x –3 –3 –2 = x Course 3 Subtract 4x from both sides. Divide both sides by –3. Solving Equations with 11-3 Variables on Both Sides Helpful Hint Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Course 3 Solving Equations with 11-3 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 4b = 24 4b = 24 4 4 b=6 Course 3 Add 6 to both sides. Divide both sides by 4. Solving Equations with 11-3 Variables on Both Sides Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w 3≠ Subtract 9w from both sides. 7 No solution. There is no number that can be substituted for the variable w to make the equation true. Course 3 Solving Equations with 11-3 Variables on Both Sides Helpful Hint if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Course 3 Solving Equations with 11-3 Variables on Both Sides 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. Course 3 Solving Equations with 11-3 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 + 15 8z 8z 8 z Course 3 = =8 = 8 8 =1 –7 +15 Add 15 to both sides. Divide both sides by 8. Solving Equations with 11-3 Variables on Both Sides y 3y 3 7 + – =y– 5 5 10 4 y 3y 7 + – 3 =y– 5 5 10 4 y 3y 7 3 + – = 20 y – 10 5 5 4 by the LCD, ( ) ( ) Multiply 20. 7 y 3 3y 20(5 ) + 20( 5 ) – 20(4 )= 20(y) – 20( 10) 20 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Course 3 Combine like terms. Solving Equations with 11-3 Variables on Both Sides 16y – 15 = 20y – 14 – 16y – 16y –15 = 4y – 14 + 14 –1 = 4y –1 = 4y 4 4 –1= y 4 Course 3 + 14 Subtract 16y from both sides. Add 14 to both sides. Divide both sides by 4. Solving Equations with 11-3 Variables on Both Sides 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. Course 3 Solving Equations with 11-3 Variables on Both Sides 39.95 + 2.95r = 26.00 + 4.50r – 2.95r 39.95 – 2.95r = – 26.00 13.95 Subtract 2.95r from both sides. 26.00 + 1.55r Subtract 26.00 from both sides. – 26.00 = Let r represent the price of one rose. 1.55r 13.95 1.55r Divide both sides by = 1.55 1.55 1.55. 9=r The two services would cost the same when purchasing 9 roses. Course 3 Solving Equations with 11-3 Variables on Both Sides 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? Course 3 Solving Equations with 11-3 Variables on Both Sides First solve for the price of one doughnut. Let d represent the price 1.25 + 2d = 0.50 + 5d of one doughnut. – 2d – 2d Subtract 2d from both sides. 1.25 = 0.50 + 3d Subtract 0.50 from both – 0.50 – 0.50 sides. 0.75 = 3d Course 3 0.75 = 3d 3 3 Divide both sides by 3. 0.25 = d The price of one doughnut is $0.25. Solving Equations with 11-3 Variables on Both Sides Now find the amount of money Jamie spends each morning. Choose one of the original 1.25 + 2d expressions. 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 0.25n = 1.75 number of doughnuts. 0.25n 1.75 Divide both sides by 0.25. 0.25 = 0.25 n = 7; Jamie bought 7 doughnuts on Tuesday. Course 3 Solving Equations with 11-3 Variables Insert Lesson Title Here on Both Sides Lesson Quiz Solve. 1. 4x + 16 = 2x x = –8 2. 8x – 3 = 15 + 5x x = 6 3. 2(3x + 11) = 6x + 4 no solution 1 1 x = 36 4. 4 x = 2 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? An orange has 45 calories. An apple has 75 calories. Course 3