Question

(x-1)²+(y-1)²=0 find value of x and y .


plz help I want the correct answer ​

Answer 1

Answer:

Step-by-step explanation:

This is a single equation with multiple variables, so there are infinitely many possible (x, y) pairs of solutions.

Since your question has 2 variables, namely x and y, you will need TWO of these equations to get a definite solution (or a finite amount of solutions).

=>(x-1)²+(y-1)²=0 ,

=>[(x- 1)+(y- 1)]²+2[(x-1)+(y-1)]=0 ,

=>(x -1)+(y -1)+2=0 ,

=>x+y - 2+2=0 ,

=>x+y=0..... (i) ,

Again ,

=>(x -1)²+(y -1)²=0 ,

=>x² - 2x + 1 + y² - 2y +1 = 0 ,

=>x²+y²- 2(x+y) +2 = 0 ,

=>x²+y²= -2..... (ii) ,

Now multiply equation (i ) with x and subtract from equation ... ( ii) ,

=>x²+ y²- x²-xy = -2-0 ,

=>x - y = 2y.....( iii ) ,

Adding equation ( i) and ( iii ) ,

we get x= y ,

therefore when the value of x will be the same as the value of y will be the same.

formula s used here :

(a+b)²= a²+2ab+b ²

=>a²+b²=(a+b)²–2ab.

Answer 2

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(x+1)2+(y−5)2=0

Adding −(y−5)2 on both sides:

(x+1)2+(y−5)2−(y−5)2=0−(y−5)2

(x+1)2+0=0−(y−5)2

(x+1)2=−(y−5)2⟶ Equation E

The square of any number is always non-negative. So, we have

(x+1)2≥0

and

(y−5)2≥0

So, the equation E is valid if and only if

(x+1)2=0

and

(y−5)2=0

Solving the equations, we have

(x+1)2=0⟹x+1=0

⟹x=−1

and

(y−5)2=0⟹y−5=0

⟹y=5

So, we have

x+y=y+x=5+(−1)=5−1=4

x+y=4

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