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Solving Equations with Variables on Both Sides. 2-4. Contradictions & Identities. Holt Algebra 1. Vocabulary. identity contradiction. An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions.
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Solving Equations with Variables on Both Sides 2-4 Contradictions & Identities Holt Algebra 1
Vocabulary identity contradiction
An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. A contradiction is an equation that is not true for any value of the variable. It has no solutions.
Identities and Contradictions WORDS Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. NUMBERS 1 = 1 + 2 1 = 3 ALGEBRA x = x + 3 –x–x 0 = 3
+ 5x + 5x Example 3A: Infinitely Many Solutions or No Solutions Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 10– 5x+ 1 = 7x+ 11– 12x Identify like terms. 11 – 5x = 11 – 5x Combine like terms on the left and the right. Add 5x to both sides. 11 = 11 True statement. The equation 10 – 5x + 1 = 7x + 11 – 12xis an identity. All values of x will make the equation true. All real numbers are solutions.
–13x–13x Example 3B: Infinitely Many Solutions or No Solutions Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 12x– 3+ x = 5x– 4+ 8x Identify like terms. 13x – 3 = 13x – 4 Combine like terms on the left and the right. Subtract 13x from both sides. –3 = –4 False statement. The equation 12x – 3 + x = 5x – 4 + 8xis a contradiction. There is no value of x that will make the equation true. There are no solutions.
Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 82. 4(2x – 5) = 5x + 4 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. 6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same? 3 8 all real numbers 1 20 hours