Problem 1:

a+b+c = 11

a = b+c+1

a×100+b×10+c = c×100+b×10+a+396

The third one can be reduced to: a×100+c = c×100+a+396

And then to: a×99 = c×99+396

And then to: a=c+4

(Yet to be solved ... anyone?)

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Problem 2:

A bit tricky deciding what length is which. Working on the clues backwards:

The perimeter tells us: a+b+c = 29

And if we call "c" the third (and longest) side we can say: c = a+b-1

Now we know one side is 3 longer than another side. But we have to decide is it:

* c = b+3, or

* a = b+3

(a or b can be interchanged, but we have already said c is the side that equals a+b-1)

Now a bit of logic about triangles could probably help us work out which of those two to use. Perhaps if we solved for both possibilities, one of them will be shown to be right and the other wrong.

(Yet to be solved ... anyone?)

]]>Write 3 equations in 3 variable for each problem

The sum of the digits of a three digit number is 11. The hundreds digt exceeds the sum of the tens digit and the units digit by 1. When the digits are reversed, the new number is 396 less than the original number. Find the number. Write the three equations

One side of a triangle is 3 inches longer than another side of the riangle. The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle. If the perimeter is 29 inches, find the length of each side. Writge the three equations

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