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Coming up: Today: Section 2.4 Next class period: Review for Online Quiz 2 Class period after that: Review for Gateway Quiz 2(taken right after Spring Break) Online Quiz 2 on sections 2.1-2. 4 Take Practice Quiz 2 before then

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Any questions on the homework for sections 2.2/2.3?

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NOW CLOSE YOUR LAPTOPS

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Section 2.4 Recall: Steps for solving linear equations in one variable: 1) Multiply to clear fractions (if necessary). 2) Use the distributive property as needed. 3) Simplify each side of the equation by combining like terms. 4) Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5) Simplify each side again by combining like terms. 6) Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7) Simplify the answer if needed. 8) Check solution by substituting into original problem.

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4a + 1 + a – 11 = 0 Example: 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

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4a + 1 + a – 11 = 0 5a – 10 = 0 (simplify left side) 5a = 10 (simplify both sides) a = 2 (simplify both sides) 5a – 10 + 10 = 0 + 10 (add 10 to both sides) (divide both sides by 5) NOW CHECK! Example:

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12x + 30 + 8x – 6 = 10 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

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Example: 12x + 30 + 8x – 6 = 10 20x + 24 = 10 (simplify left side) 20x = -14 (simplify both sides) 20x + 24 + -24 = 10 + -24 (add –24 to both sides) (divide both sides by 20)(simplify both sides) NOW CHECK!

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Example with a fraction: 1.Multiply to clear fractions (if necessary). 2.Use the distributive property as needed. 3.Simplify each side of the equation by combining like terms. 4.Get all the variable (letter) terms on one side and the constants (pure number terms) on the other side of equation by using the addition property of equality. 5.Simplify each side again by combining like terms. 6.Get the variable completely by itself (get rid of the coefficient (the number in front of the letter) by using the multiplication property of equality. 7.Simplify the answer if needed. 8.Check solution by substituting into original problem.

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Example with a fraction: (multiply both sides by 5) (simplify, then use distributive property) (add –3y to both sides) (simplify, then add –30 to both sides) (simplify, then divide both sides by 7) (simplify both sides) NOW CHECK!

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Answer: x = 21Don’t forget to check it!

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New type of outcome when solving an equation: Solve 3x – 7 = 3(x + 1) 3x – 7 = 3x + 3 (use distributive property) 3x + (-3x) – 7 = 3x + (-3x) + 3 (add –3x to both sides) -7 = 3 (simplify both sides) Since no value for the variable x can be substituted into this equation that will make this a true statement, there is “no solution.”

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Solve 5x – 5 = 2(x + 1) + 3x – 7 5x – 5 = 2x + 2 + 3x – 7 (use distributive property) 5x – 5 = 5x – 5 (simplify the right side) All the variables and all of the numbers cancel out, leaving “0 = 0”. Both sides of the equation are identical. Since this equation will be true for every x that is substituted into the original equation, the solution is “all real numbers.” Another new type of outcome when solving an equation: -5x + 5 -5x + 5 (add 5 to both sides and subtract 5x from both sides) 0 = 0 (simplify both sides)

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Notice the difference between the previous problem and this one: 5(x + 2) = 2(x + 5) 5x + 10 = 2x + 10 -2x -10 3x = 0 Now what? Divide by 3: 3x = 0 3 3 x = 0 = 0 3 So the answer is just x = 0 (CHECK this in original equation)

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Summary: Types of possible outcomes when solving linear equations in one variable: 1. One solution (nonzero). Example: 2x + 4 = 4(x + 3) Solution: x = -4 2. One solution (zero). Example: 2x + 4 = 4(x + 1) Solution: x = 0 3. Solution = “All real numbers”. Example: 2x + 4 = 2(x + 2) Solution ends up with 0 = 0, so answer is “all real numbers”. (“R” on computer) 4. No solutions. Example: 2x + 4 = 2(x + 3) Solution ends up with -2 = 0, so answer is “no solution” (“N” on computer)

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Problem from today’s homework:

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Before next class period: Finish Homework 2.4 Reminders:

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Math TLC Open Lab Hours: Next door in room 203 Monday - Thursday 8:00 a.m. – 6:30 p.m. Teachers and tutors available for one-on-one help on homework and practice quiz/test problems. NO APPOINTMENTS NECESSARY – JUST DROP IN.

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