find the value of x (x-y)^2 when x=2 and y= 8
Answers
Answer:
x=89
y=94 . .. . . . .. . . .
...
Step-by-step explanation:
that x=y is not a solution since the second equation would imply x2=y2=8 which in turn implies x2+y2=16 contradicting the first equation’s assertion that this sum is 20 . So either x2<y2 or x2>y2 .
Next, notice that if we switch the roles of x and y in the pair of equations, they remain unchanged. This symmetry implies that for any solution (x=a,y=b) in which a2>b2 there is another solution (x=b,y=a) in which a2<b2 .
So I’ll begin by assuming x2>y2 . To the first equation, I’ll add 2xy to the left hand side and 16 to the right hand side (which has the same value by the second equation).
x2+2xy+y2=36 ⟹ (x+y)2=36 ⟹ x+y=±6
Similarly, I’ll subtract 2xy and 16 from the two sides of the first equation.
x2−2xy+y2=4 ⟹ (x−y)2=4 ⟹ x−y=±2
Multiplying these resulting equations, we see that either
(x+y)(x−y)=x2−y2=12 or (x+y)(x−y)=x2−y2=−12
Since we are in the case x2>y2 , we know that 12 is the correct product. But from our first analysis, we conclude that there must be corresponding values of x and y in which y2>x2 which give x2−y2=−12 .
We conclude that 12 and −12 are both possible values for x2−y2 and that they are the only two possible values