Question 342693
"The larger of 2 consecutive integers is eleven more than one-half of the smaller.
Find all of the integers." means that {{{x+1=(1/2)x+11}}} where 'x' is the first integer and 'x+1' is the second integer (ie 'x+1' is larger than 'x')



{{{x+1=(1/2)x+11}}} Start with the given equation.



{{{2(x+1)=2((1/cross(2))x+11)}}} Multiply both sides by the LCD {{{2}}} to clear any fractions.



{{{2x+2=x+22}}} Distribute and multiply.



{{{2x=x+22-2}}} Subtract {{{2}}} from both sides.



{{{2x-x=22-2}}} Subtract {{{x}}} from both sides.



{{{x=22-2}}} Combine like terms on the left side.



{{{x=20}}} Combine like terms on the right side.



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Answer:


So the solution is {{{x=20}}}  



This means that the first integer is 20 and the second integer is 21.



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