Question 112004
Let u=first #, w=second #


Remember, consecutive odd integers follow the form {{{2x+1}}}, {{{2x+3}}}, etc. 


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


{{{5u=3w+12}}} 


{{{5(2x+1)=3(2x+3)+12}}} Let {{{u=2x+1}}} and {{{w=2x+3}}}



{{{10x+5=6x+9+12}}} Distribute



{{{10x+5=6x+21}}} Combine like terms on the right side



{{{10x=6x+21-5}}}Subtract 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=(16)/(4)}}} 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+1}}} to get 


{{{2(4)+1=8+1=9}}}


So our first number is 9



Now plug in {{{x=4}}} into {{{2x+3}}} to get 


{{{2(4)+3=8+3=11}}}


So our second number is 11



Check:

{{{5u=3w+12}}} Start with the given equation


{{{5*9=3*11+12}}} Plug in the given numbers


{{{45=33+12}}} Multiply


{{{45=45}}} Add. So these numbers work.