Difference between revisions of "2018 AMC 10A Problems/Problem 12"

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Now it's clear that there are <math>\boxed{3}</math> intersection points. (pinetree1)
 
Now it's clear that there are <math>\boxed{3}</math> intersection points. (pinetree1)
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x+3y=3 can be rewritten to x=3-3y. Substituting 3-3y for x in the second equation will give ||3-3y|-y|=1. Splitting this question into casework for the ranges of y will give us the total number of solutions.
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Case 1: y>1
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3-3y will be negative so |3-3y| = 3y-3.
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|3y-3-y| = |2y-3| = 1
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    Subcase 1: y>3/2
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2y-3 is positive so 2y-3 = 1 and y = 2 and x = 3-3(2) = -3
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    Subcase 2: 1<y<3/2
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2y-3 is negative so |2y-3| = 3-2y = 1. 2y = 2 and so there are no solutions (y can't equal to 1)
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Case 2: y = 1
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x = 0
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Case 3: y<1
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3-3y will be positive so |3-3y-y| = |3-4y| = 1
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    Subcase 1: y>4/3
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3-4y will be negative so 4y-3 = 1 --> 4y = 4. There are no solutions (again, y can't equal to 1)
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    Subcase 2: y<4/3
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3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/2 and x = 3/2
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NOTE: Please fix this up using latex I have no idea how
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Solution by Danny Li JHS
  
 
== See Also ==
 
== See Also ==

Revision as of 17:57, 8 February 2018

Problem

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[x+3y=3\] \[\big||x|-|y|\big|=1\] $\textbf{(A) } 1 \qquad  \textbf{(B) } 2 \qquad  \textbf{(C) } 3 \qquad  \textbf{(D) } 4 \qquad  \textbf{(E) } 8$

Solution

The graph looks something like this: [asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3,0)); dot((0,1)); [/asy]

Now it's clear that there are $\boxed{3}$ intersection points. (pinetree1)

x+3y=3 can be rewritten to x=3-3y. Substituting 3-3y for x in the second equation will give ||3-3y|-y|=1. Splitting this question into casework for the ranges of y will give us the total number of solutions.


Case 1: y>1

3-3y will be negative so |3-3y| = 3y-3. |3y-3-y| = |2y-3| = 1

   Subcase 1: y>3/2

2y-3 is positive so 2y-3 = 1 and y = 2 and x = 3-3(2) = -3

   Subcase 2: 1<y<3/2

2y-3 is negative so |2y-3| = 3-2y = 1. 2y = 2 and so there are no solutions (y can't equal to 1)


Case 2: y = 1 x = 0


Case 3: y<1

3-3y will be positive so |3-3y-y| = |3-4y| = 1

   Subcase 1: y>4/3

3-4y will be negative so 4y-3 = 1 --> 4y = 4. There are no solutions (again, y can't equal to 1)

   Subcase 2: y<4/3

3-4y will be positive so 3-4y = 1 --> 4y = 2. y = 1/2 and x = 3/2

NOTE: Please fix this up using latex I have no idea how

Solution by Danny Li JHS

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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