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Sub 505 Level|   Algebra|                     
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carcass
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S
(x-2)(y+2) = xy +2x - 2y -4
=xy +2(x-y) -4
=S + 2R - 4

Answer D
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(x-2) (y+2) = xy - 2y + 2x - 4

x-y = R, xy = S

xy - 2y + 2x - 4 = S + 2(x-y) - 4 = S + 2R - 4. Ans - D.
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If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S

x-y = R
xy = S
(x-2)(y+2) = xy - 2y + 2x -4 = xy + 2(x-y) - 4
= S + 2R - 4

Answer D
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carcass
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S

(x-2)(y+2)

= xy + 2x - 2y - 4

= xy + 2(x - y) - 4

= S + 2R - 4

Answer: D
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I made a frivulous mistake in hurry and picked B.

I solved xy+2(x-y)-4
But in hurry picked R+2S-4 Instead of S+2R-4.
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1. (x-2)(y+2)
2. xy + 2x -2y - 4 (order of operations/multiplication/parenthesis)

3. xy = S; 2x - 2y = 2R, for R = x - y; and - 4 stays the same
4. xy + 2x - 2y - 4 ----> S + 2R - 4

Ans = (D) 2R + S - 4
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\((x-2)(y+2)=xy+2x-2y-4\)
\(xy+2(x-y)-4\)
\(S+2R-4\)
\(2R+S-4\)

Answer D
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Let x=2 and y=1. Then: 2-1=1 (R) and 2*1=2 (S).

(x−2)(y+2) = (2−2)(1+2) = 0.

So we know that when 1 = (R) and 2 = (S), the equation = 0. Option D: 2R+S−4 = 2*1+2-4 = 0.
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Why are (y=2 and x=1) not working?
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carcass
If x-y= R and xy=S, then (x-2)(y+2)=

A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S

Plug in and solve

Let x = 3 & y = 1 , So R = 2 & S = 3

Thus, (x-2)(y+2) = 1*3 => 3

Now, check the options -

(A) R+S-4 = 1
(B) R+2S -4 = 4
(C) 2R - S -4 = -3
(D) 2R + S - 4 = 3
(E) R + S = 5

Thus, Answer must be (D) 3
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carcass



A. R+S-4

B. R+2S -4

C. 2R - S -4

D. 2R + S - 4

E. R + S

Asked: If x-y= R and xy=S, then (x-2)(y+2)=

(x-2)(y+2)= xy + 2x - 2y - 4 = S + 2R - 4

IMO D
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