Objective Solve equations in one variable that contain variable terms on both sides.

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Presentation transcript:

Objective Solve equations in one variable that contain variable terms on both sides.

Identities and Contradictions WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2 + 1 = 2 + 1 3 = 3  ALGEBRA 2 + x = 2 + x –x –x 2 = 2 

Identities and Contradictions When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. WORDS x = x + 3 –x –x 0 = 3  1 = 1 + 2 1 = 3  ALGEBRA NUMBERS Identities and Contradictions

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. + 5x + 5x Add 5x to both sides. 11 = 11  True statement. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

Check It Out! Example 3b Solve 2c + 7 + c = –14 + 3c + 21. Identify like terms. 3c + 7 = 3c + 7 Combine like terms on the left and the right. –3c –3c Subtract 3c both sides. 7 = 7  True statement. The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.

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. –13x –13x Subtract 13x from both sides.  –3 = –4 False statement. The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

 Check It Out! Example 3a Solve 4y + 7 – y = 10 + 3y. Identify like terms. 3y + 7 = 3y + 10 Combine like terms on the left and the right. –3y –3y Subtract 3y from both sides.  7 = 10 False statement. The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.

Lesson Quiz Solve each equation. 1. 7x + 2 = 5x + 8 2. 2 = 4 – x 3. 7 + 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. -2(3x + 5) = 4x 3 2 -5 all real numbers -1