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Difference between revisions of "2007 iTest Problems/Problem 22"

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<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \text{(A)}}</math>.
 
<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \text{(A)}}</math>.
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==See Also==
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{{iTest box|year=2007|num-b=21|num-a=23}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 21:05, 10 January 2019

Problem

Find the value of $c$ such that the system of equations \[|x+y|=2007 \\ |x-y|=c\] has exactly two solutions $(x,y)$ in real numbers.


$\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4 \quad \text{(F) } 5 \quad \text{(G) } 6 \quad \text{(H) } 7 \quad \text{(I) } 8 \quad \text{(J) } 9 \quad \text{(K) } 10 \quad \text{(L) } 11 \quad \text{(M) } 12 \quad$

$\text{(N) } 13 \quad \text{(O) } 14 \quad \text{(P) } 15 \quad \text{(Q) } 16 \quad \text{(R) } 17 \quad \text{(S) } 18 \quad \text{(T) } 223 \quad \text{(U) } 678 \quad \text{(V) } 2007 \quad$

Solution

$2007|x-y|=c$ must be one line; otherwise, the system would have four solutions. This occurs at $c=\boxed{0\ \text{(A)}}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 21 		Followed by:
Problem 23
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