Algebraic word problem

Question

The sum of the digits of a 2-digit number is 7. If the digits are reversed the new number is 32 more than 10 times the tens' digit of the original number. What is the original number?

Answer 1

Let x is tens. Then once is 7 - x. Because given number is 10x + (7 - x) = 9x + 7.

If we digits reverse then new number is 10(7- x) + x = 70 - 9x.

Now we have : 70 - 9x = 10x + 32, 19x = 38 and x = 2

So, original number is 9·21 + ·7 = 25.·

Answer 2

We are beginning with a 2-digit number. Let's call it "xy". If we reverse the digit we have "yx"

We know x+y=7

We also know that yx = 10x + 32

Using the first equation and solving for y, we have y=7-x

So in the 2nd equation the tens digit, y, is 7-x

Since the 7-x is the 10s digit, it's value is 10(7-x) and we can add this to the x in the yx

and we have

10(7-x)+x = 10x+32

If this is a bit unclear, think of it this way...

In the number 31, 3 times 10 + 1, 10 times the tens digit plus the one's digit.

Back to the problem

Distributing,

70-10x+x=10x+32

70-9x=10x+32

Rearrange terms to solve for x

Subtract 32 from both sides

38 - 9x = 10x

Add 9x to both sides

38=19x

divide both sides by 19

x = 2

Since x+y=7, then y = 5

The original number is 25

The new number is 52 and 52 = 10(2) + 32