Question

One number is 5 less than twice another. If the sum of the two numbers is 49, find the two numbers?

Answer 1

`18, 31`

Explanation:

Given: one number is 5 less than twice another number. The sum of the two numbers = 49.

Define the variables: `n_1, n_2`

Create two equations based on the given info:

`n_2 = 2n_1 - 5; " "n_1 + n_2 = 49`

Use substitution to solve:

`n_1 + 2n_1 - 5 = 49`

`3n_1 - 5 = 49`

`3n_1 = 54`

`(3n_1)/3 = 54/3`

`n_1 = 18`

`n_2 = 49 - 18 = 31`

Answer 2

One number is `18`
The other number is `31`

Explanation:

Problems like this are confusing because it's hard to know how to write a math expression for statements like "One number is 5 less than twice another."

The trick is to go one step at a time.

Let `x` represent "one of the numbers"

A number . . . . . . . . . .. `x` `larr` one of the numbers
Twice as much . . . . . `2x`
5 less than that . . . . .`2x - 5` `larr` the other number

Together, these two amounts add up to `49`

[ one number] + [the other number] `= 49`
[ . . . . . `x`.. . . . .] + [ . . . . `2x - 5`. . . . .] `= 49`

Now solve for `x`

`(x )+ (2x - 5) = 49`

1) Combine like terms
`3x - 5 = 49`

2) Add `5` to both sides to isolate the `3x` term
`3x = 54`

3) Divide both sides by `3` to isolate `x`
`x = 18` `larr` answer for "one of the numbers"

One number = `18`

The other number has already been defined as
`2x - 5`

Sub in `18` for `x` and calculate "the other number"
`2(18) - 5`
`color(white)(m)` `36`   `- 5`
`color(white)(mnn)` `31` `larr` answer for "the other number"

Answer
One number is `18` and the other number is `31`

`color(white)(mmmmmmmm)`――――――――

Check

Using the original equation, sub in `18` in the place of `x`
to verify that the equation is still true.

`color(white)(.)` `x + 2  x   - 5 = 49`
`18 + 2(18) - 5` should still equal `49`

1) Clear the parentheses by distributing the `2`
`18 + 36 - 5` should equal `49`

2) Combine like terms
`18 + 31` should equal `49`
`color(white)(nn)` `49`       does equal   `49`

`Check`

Answer 3

The numbers are `18 and 31`

Explanation:

We need to define our numbers first.

Let one number be `x`

If the numbers add to `49` then the other number is `49-x`

One number `(49-x)` is `5` less than twice the other `(x)`.

`2x-5 = 49-x`

`2x+x = 49+5`

`3x = 54`

`x=18" "larr` this is one number

`49-18=31" "larr` this is the other number

Check:
`2(18) -5 = 31`
`31+18 = 49`