SOLUTION: Find two consecutive odd integers such that 5 times the first integer is 12 more then 3 times the second

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Question 112004: Find two consecutive odd integers such that 5 times the first integer is 12 more then 3 times the second
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Let u=first #, w=second #

Remember, consecutive odd integers follow the form 2x%2B1, 2x%2B3, etc.

So the statement "5 times the first integer is 12 more then 3 times the second" translates to

5u=3w%2B12

5%282x%2B1%29=3%282x%2B3%29%2B12 Let u=2x%2B1 and w=2x%2B3


10x%2B5=6x%2B9%2B12 Distribute


10x%2B5=6x%2B21 Combine like terms on the right side


10x=6x%2B21-5Subtract 5 from both sides


10x-6x=21-5 Subtract 6x from both sides


4x=21-5 Combine like terms on the left side


4x=16 Combine like terms on the right side


x=%2816%29%2F%284%29 Divide both sides by 4 to isolate x



x=4 Divide

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Answer:
So our answer is x=4


Now plug in x=4 into 2x%2B1 to get

2%284%29%2B1=8%2B1=9

So our first number is 9


Now plug in x=4 into 2x%2B3 to get

2%284%29%2B3=8%2B3=11

So our second number is 11


Check:
5u=3w%2B12 Start with the given equation

5%2A9=3%2A11%2B12 Plug in the given numbers

45=33%2B12 Multiply

45=45 Add. So these numbers work.