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# Solve for x x2 9x 36 0 a b c d e 12 and 3

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Unformatted text preview: of –4 and –5 is divided by the product of 9 and – (A) (B) (C) (D) (E) 4. –3 +3 –27 +27 – 1 3 1 , the result is 27 6. Solve for x: (A) (B) (C) (D) (E) 3 x + 2y = 5a + b 4x – 3y = a + 7b a+b a–b 2a + b 17a + 17b 4a – 6b 1 4 1 4 2. 1 1 7. Solve for x: 8x2 + 7x = 6x + 4x2 (A) (B) (C) (D) (E) – 0 and 0 3. 0 and – 1 4 none of these 8. Solve for x: x2 + 9x – 36 = 0 (A) (B) (C) (D) (E) –12 and +3 +12 and –3 –12 and –3 12 and 3 none of these x2 + 3 = x + 1 Solve for x: 7b + 5d = 5x – 3b (A) 2bd (B) 2b + d (C) 5b + d (D) 3bd (E) 2b Solve for y: 2x + 3y = 7 3x – 2y = 4 (A) 6 (B) (C) (D) (E) 5 2 1 5 1 3 4 5 9. Solve for x: (A) (B) (C) (D) (E) 5. ±1 1 –1 2 no solution 10. Solve for x: 2 x = –10 (A) (B) (C) (D) (E) 25 –25 5 –5 no solution www.petersons.com 128 Chapter 8 SOLUTIONS TO PRACTICE EXERCISES Diagnostic Test 1. 2. (D) (+4) + (–6) = –2 6. (C) (B) An odd number of negative signs gives a negative product. 1 1 ( –3 )(+4 2 ) – – = –2 2 3 Add the two equations. x+y=a x–y=b 2x = a + b 1 x = ( a + b) 2 3. (D) The product of (–12) and + 4 is –3. 1 1 7. (D) The product of (–18) and – 3 is 6. – 4. 3 1 =– 6 2 2x(2x – 1) = 0 2x = 0 2x – 1 = 0 1 x = 0 or 2 8. (D) (x – 7)(x + 3) = 0 x–7=0 x = 7 or –3 x+3=0 (C) ax + b = cx + d ax – cx = d – b (a – c)x = d – b x= d –b a–c 9. (E) x + 1 – 3 = –7 x + 1 = –4 x + 1 = 16 x = 15 5. (B) Multiply the first equation by 3, the second by 7, and subtract. 21x – 6y = 6 21x + 28y = 210 –34y = –204 y=6 Checking, 16 – 3 = –7, which is not true. 10. (B) x2 + 7 – 1 = x x2 + 7 = x + 1 x2 + 7 = x2 + 2x + 1 7 = 2x + 1 6 = 2x x=3 Checking, 16 – 1 = 3, which is true. www.petersons.com Concepts of Algebra—Signed Numbers and Equations 129 Exercise 1 1. 2. 3. (D) (B) (–4) + (+7) = +3 (29,002) – (–1286) = 30,288 Exercise 2 1. 2. (B) (D) 3x – 2 = 3 + 2x x=5 8 – 4a + 4 = 2 + 12 – 3a 12 – 4a = 14 – 3a –2 = a Multiply by 8 to clear fractions. y + 48 = 2y 48 = y 4. (B) Mul...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.

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