# Thread: coin-word problems in algebra

1. ## coin-word problems in algebra

Hi everyone,

I am doing lots of word problems and so far I was doing pretty good. For some reason I can't figure this one out. I am hoping someone is able to help me.

In a two-digit number , the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the number.

ten's digit= x+3 one's digit = x
10(x+3) + x
reversed would be 10x + x+3 -27 = 10(x+3)+x ??? I don't get it.

2. Originally Posted by Christine Magruder
Hi everyone,

I am doing lots of word problems and so far I was doing pretty good. For some reason I can't figure this one out. I am hoping someone is able to help me.

In a two-digit number , the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the number.

ten's digit= x+3 one's digit = x
10(x+3) + x
reversed would be 10x + x+3 -27 = 10(x+3)+x ??? I don't get it.
number: 10x + 30 + x
reversed : 10x + x + 3

10x+ 30 + x - 10x - x -3 = 27
0x = 0

in this case, x would be all the positive real values but check the equation by substituting values.
x = 1 (possible)
x = 2 (possible)
x = 3 (possible)
x = 4 (possible)
x = 5 (possible)
x = 6 (possible)
x = 7 (rejected since it gives a three digit number)

so x = + {1,2,3,4,5,6}