Question

Could someone help me with this please?

Find the values ​​of x and y in the equation below.
(x²-2x+1)⁷⁷+(3x-2y+5)⁸⁸=0​

Answer 1

(x, y) = (1, 4)

Explanation:

You want the solution to the equation (x²-2x+1)⁷⁷ +(3x-2y+5)⁸⁸ = 0​.

Even powers

The first term can be rewritten as ...

`(x^2-2x+1)^(77)=((x-1)^2)^(77)=(x-1)^(154)`

This means both of the terms are being raised to even powers. Any non-zero value of the base expression will result in a positive value for the term. That is, each term is non-negative.

Zero sum

The sum of two non-negative terms will be zero if and only if each of the terms individually is zero. That means the equation resolves to two equations:

• (x -1) = 0
• (3x -2y +5) = 0

Solution

The solution to the first equation is x=1. Using this value in the second equation, it becomes ...

3 -2y +5 = 0

8 = 2y . . . . . . . . add 2y

4 = y . . . . . . . divide by 2

The values of x and y that will make the equation true are ...

(x, y) = (1, 4)