Consecutive Integers

[lmward78727]

Need help on how to write out a word problem regarding consecutive integers. Have solved other problems before but this one has me confused.

Three consecutive integers are such that he first plus one-half the second plus seven less than twice the third is 2101. What are the integers? Priciples understood x= first integer, (x + 1) is second integer and and (x + 2) is third integer.

Another tutor said the problem shoudl be written as x + 1/2 + (x +1) + 2(x +2) - 7. Is this correct?

Questions:
What do you do to get rid of the fraction so collecting like terms is easier?
Please show the steps needed to simplify and solve for x so I can find the consecutive integers.

Thanks!
transplantedtexan

 

[soroban]

Hello, lmward78727!

Three consecutive integers are such that the first plus one-half the second plus seven less than twice the third is 2101.
What are the integers?

Principles understood: $\displaystyle x$ = first integer, $\displaystyle (x + 1)$ = second integer. and $\displaystyle (x + 2)$ = third integer. . Good!

Another tutor said the problem shoudl be written as: .$\displaystyle x + 1/2 + (x +1) + 2(x +2) - 7$
Is this correct? . no

Consider each phrase . . .

. . $\displaystyle \text{The first: }x$

. . $\displaystyle \text{plus one-half the second: }+ \tfrac{1}{2}(x+1)$

. . $\displaystyle \text{plus seven less than twice the third: }+ 2(x+2) - 7$

. . $\displaystyle \text{is 2101: }\;= \;2101$


The equation is: .$\displaystyle x + \tfrac{1}{2}(x+1) + 2(x+2) - 7 \;=\;2101$


Multiply through by 2:

. . $\displaystyle 2x + (x+1) + 4(x+2) - 14 \;=\;4202$

. . . . .$\displaystyle 2x + x + 1 + 4x + 8 - 14 \;=\;4202$

. . . . . . . . . . . . . . . . .$\displaystyle 7x - 5 \;=\;4202$

. . . . . . . . . . . . . . . . . . . $\displaystyle 7x \;=\;4207$

. . . . . . . . . . . . . . . . . . . . $\displaystyle x \;=\;601$


$\displaystyle \text{The three integers are: }\:601,\,602,\,603$