The sum of a two digit number and the number obtained by interchanging the digits is 132. If the two digits differ by 2, find the number​

A two-digit number can be written as:

a*10 + b*1

Where a and b are single-digit numbers, and a ≠ 0.

We know that:

"The sum of a two-digit number and the number obtained by interchanging the digits is 132."

then:

a*10 + b*1 + (b*10 + a*1) = 132

And we also know that the digits differ by 2.

then:

a = b + 2

or

a =  b - 2

So let's solve this:

We start with the equation:

a*10 + b*1 + (b*10 + a*1) = 132

(a*10 + a) + (b*10 + b) = 132

a*11 + b*11 = 132

(a + b)*11 = 132

(a + b) = 132/11 = 12

Then:

a + b = 12

And remember that:

a = b + 2

or

a = b - 2

Then if we select the first one, we get:

a + b = 12

(b + 2) + b = 12

2*b + 2 = 12

2*b = 12 -2 = 10

b = 10/2 = 5

b = 5

then a = b + 2= 5 + 2 = 7

The number is 75.

And if we selected:

a = b - 2, we would get the number 57.

Both are valid solutions because we are changing the order of the digits, so is the same:

75 + 57

than

57 + 75.

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

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