The interval notation for this solution is 1 4 return

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Unformatted text preview: te values and so the only way the two sides are equal will be if the two quantities inside the absolute value bars are equal or equal but with opposite signs. Or in other words, we must have, 2 x - 1 = - ( 4 x + 9 ) = -4 x - 9 6 x = -8 8 4 x=- =6 3 2x -1 = 4x + 9 or or - 2 x = 10 or x = -5 Now, we won’t need to verify our solutions here as we did in the previous two parts of this problem. Both with be solutions provided we solved the two equations correctly. However, it will probably be a good idea to verify them anyway just to show that the solution technique we used here really did work properly. Let’s first check x = - 4 . 3 ? æ 4ö æ 4ö 2 ç - ÷ -1 = 4 ç - ÷ + 9 è 3ø è 3ø 11 ? 11 -= 3 3 11 11 = 33 OK In the case the quantities inside the absolute value were the same number but opposite signs. However, upon taking the absolute value we got the same number and so x = Now, let’s check x = -5 . 4 is a solution. 3 ? 2 ( -5 ) - 1 = 4 ( -5 ) + 9 ? -11 = -11 11 = 11 OK In the case we got the same...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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