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The sum of three integers is 193. The sum of the first and second integers exceeds the third by 13. The third integer is 11 less than the first. Find the three integers.

Apr 13, 2020

3+0 Answers

#1
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Error. Answer below..

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Apr 13, 2020
edited by Guest  Apr 13, 2020
edited by Guest  Apr 13, 2020
edited by Guest  Apr 13, 2020
#2
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let the three integers be \(x,y,z\)

It follows:

(1) \(x+y+z=193\)  (The sum of the three integers is 193)

(2) \(x+y=z+13\) (The sum of first and second integers are more than the third by 13)

(3) \(z+11=x\) (The third integer is 11 less than the first)

Substitute (3) into (2)

But first, rearrange (3) such that, \(z=x-11\)

\(x+y=x-11+13\)

\(y=2\)

Now the first equation (1)

\(x+y+z=193\)

\(x+z=191\)

We know from (3) that \(z=x-11\)

\(x+x-11=191\)\(=2x-11=191\)

\(2x=202\)

\(x=101\)

Again, the first equation (1)

\(x+y+z=193\)

\(101+2+z=193\)

\(z=90\)

So the three integers are:

\(x=101\)

\(y=2\)

\(z=90\)

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Apr 13, 2020
#3
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The sum of three integers is 193. The sum of the first and second integers exceeds the third by 13. The third integer is 11 less than the first. Find the three integers.

Die Summe von drei ganzen Zahlen ist 193. Die Summe der ersten und zweiten ganzen Zahlen überschreitet die dritte um 13. Die dritte ganze Zahl ist 11 kleiner als die erste. Finden Sie die drei ganzen Zahlen.

Hello Guest!

\(x+y+z=193\\ x+y=z+13\\ \underline{z=x-11}\)

\(x+y+x-11=193\\ \underline{x+y=x-11+13}\)

\(2x+y=193+11\)

\(y=-11+13\)

\(y=2\)

\(2x+2=204\\ 2x=202\)

\(x=101\)

\(z=101-11\)

\(z=90\)

!

Apr 13, 2020