Question

How do you find two numbers such that their sum is `9` and the difference of their squares is also `9`?

Answer 1

4 and 5 (By trial and error)

Explanation:

You should try numbers such as 1 and 2 (their sum is 3 not 9), or 3 and 4 (their sum is 7 not 9). The other must is the difference of their squares. Therefore, you can reach your conclusion.

`4+5=9`

`5^2-4^2=25-16=9`

Your numbers are 4 and 5.

Answer 2

`5` and `4`

Explanation:

Suppose that the two numbers are `a` and `b`. Then, from the question, it states that `{(a+b=9),(a^2-b^2=9):}`.

Divide the second equation by the first equation, or `(a^2-b^2)/(a+b)=9/9`, or `a-b=1`.

Thus, we have `{(a+b=9),(a-b=1):}`. Add these two equations to get `2a=10`, or `a=5`. Since `a+b=9`, `b=4`.

The answer is `5` and `4`.

Answer 3

The numbers are `5 and 4`

Explanation:

Let the numbers be `x and y`

A: `x+y =9" "` their sum is `9`

B: `x^2 - y^2 =9" "` the difference of their squares is `9`

Solve for `x` in equation A

`color(blue)(x = (9-y))`

Substitute for `x` in equation B.

`" "color(blue)(x^2) - y^2 =9`
`" "darr`
`color(blue)((9-y))^2 - y^2 =9`

`81-18y +y^2 -y^2 =9`

`81-9 = 18y`

`72 = 18y`

`y=4`

If `y=4" "rarr x = 5`

Check:

`5+4=9" and "25-16 = 9`