Re: Hints for #1

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: : : Can you explain and show me how to do these problems? : : : 1. Find three consecutive odd intergers such that 7 times the sum of the first and the third is 120 less than 10 times the opposite of the second? : : : 2. Find three consecutive even intergers such that 6 times the sum of the first and the third is 8 less than 14 times the second? : : : WE are desperately seeking help. We are homeschooler and I, the mom, have been out of school for a long time! We would appreciate any help you can give us. : : : Thanks : : ***************************************** : : odd numbers are expressed as (2*n+1) where n can be any integer. : : Let the first odd integer (in the series) be = 2n + 1 : : So the next one(odd) would be 2n + 3 and the next one would be 2n + 5 : : According to the problem" : : sum of first and third = (2n + 1) + (2n + 5) = 4n + 6 : : 10*(2n + 3) - (4n + 6) = 120 : : 20n + 30 - 4n - 6 = 120 : : 20n - 4n = 120 - 30 + 6 : : 16n = 96 : : n = 6 : : so the numbers are (2*6+1 = )13, 15, 17 : : check: : : 13 + 17 = 30 : : 10 * 15 = 150 and : : 150 - 30 = 120 : : similarly you can do the second problem. Even integers are expressed as (2n). So the consecutive even integers would be (2n), (2n + 2) & (2n + 4). Carry on from there... : MR. Khan, : i recognize that your are a brilliant Mathematician, but you misread her problem. : You didnt notice that the problem said... the opposite of 10 time the OPPOSITE of the second. : here is how i did it. : let x equal first odd integer. : X+2 second odd integer : x+4 third : 7(x+x+4) is seven time the sum of first and second whis is 120 less than... : 7(x+x=4) + 120= : ten times the Opposite of the second : (10)(-1)(x+2) : all together : 7(x+x+4)+120=(10)(-1)(x+2) : simplify : 7(2x+4)+120=-10(x+2) : 14x+28+120=-10x-20 : 14x+148=-10x-20 : 24x=-168 : x=-7 : therefore : -7 -5 -3 are your consecutive odd ints

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