If x > y and (x -1)^2 < (y -1)^2, which of the following inequalities

take x = -1/2 and y = -1/3

x > y
\(x^2\) > \(y^2\)

The algebraic method is shown above. Knowing that X > Y ——- we know (X - Y) > 0 —— and so after multiplying out the quadratics, we can Factor and divide each side of the inequality by (X + Y) and get:

E. (X + Y) < 2

Method 2: using the “Distance Interpretation” of Absolute Value

We can first take the square root of both sides and using the rule that the square root of a variable squared is = the absolute value of that variable:

[x - 1] < [y - 1]

Where: y < x

This can be interpreted as:

“On the number line, the distance from X to +1 is less than the distance from Y to +1.”

Essentially, wherever Y is placed on the number line, X must be placed in a spot in which

—-X is greater than Y

AND

—— X is closer to +1

This leads us to two possible cases:


[………Y……..X….+1…………] — no. line #1

Both X and Y are less than <1
But X is closer to +1 on the number line

OR

[………….Y………+1….X……..]— no. line #2

Y is less than <1
And
X is greater than >1 but, compared to Y, it is closer to +1 on the number line

no. line #1:
Both quantities inside the Modulus are negative —- we can open up the Modulus under this assumption and we get:

-(x - 1) < -(y - 1) ……. Multiply both sides by *(-1)

(x - 1) > (y - 1)………. Add +1 to both sides

x > y

from the assumption we made and the given fact that X is greater than Y:

1 > x > y

We are already given that X is greater than > Y , so this is no new information

no. line #2
the quantity inside the [x - 1] is non-negative (x > 1)and the quantity inside [y - 1] is negative (y < 1)

(x - 1) < -(y - 1)

x - 1 < -y + 1

x + y < 2

Answer E

We can also prove this by showing some cases under either scenario.

Under number line #1, both x and y are less than 1, so when you combine the inequalities: (x + y) < 2

Under number line #2,

If we made x a number greater than >1, for example:

X = 5

X is 4 units to the right of +1 on the number line

Thus, for the distance interpretation of the absolute value inequality to hold true, Y will have to be greater than 4 units to the left of +1 towards the negative side of the number line

This means Y must be: Y < -(3)

Y < -3

X = +5
_________

The SUM of X and Y will have to be some value less than < (-3 + 5) = 2

(X + Y) < 2

Hence, the sum of X and Y is always less than 2

E

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From the inequality: (x-1)^2 < (y-1)^2
Therefore: x^2 - 2x + 1 < y^2 - 2y +1
Move to one side: x^2 - y^2 - 2x - 2y < 0
=> (x-y) (x+y) - 2 (x-y) < 0
=> (x-y) (x+y-2) < 0
This leads to 2 solutions:
No.1: x-y<0 and x+y-2>0
Because x-y<0 leads to x<y, it is wrong since x>y
No.2: x-y>0 and x+y-2<0
x-y>0 then x>y ==> ACCEPT
x+y-2<0 then x+y<2 ==> Ans E