2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 12 Easy

How many ordered pairs of real numbers (x,y) satisfy the following system of equations?x+3y=3, \left|\left|x\right|-\left|y\right|\right|=1. (2018 AMC 10A Problem, Question#12)

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    8

Answer:C

We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.

The graph looks something like this:

Now, it becomes clear that there are \boxed{\rm (C)~3} intersection points.

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>\frac{3}{2}.

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

Subcase 2: 1<y<\frac{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: It is fairly clear that x=0.

Case 3: y<1: 3-3y will be positive so |3-3y-y|=|3-4y|=1.

Subcase 1: y>\frac{4}{3}.

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

Subcase 2: y<\frac{4}{3}.

3-4y will be positive so 3-4y=1\to 4y=2. y=\frac{1}{2} and x=\frac{3}{2}. Thus, the solutions are: (-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2}\right), and the answer is \boxed{\rm (C)~3}.

Note that ||x|-|y|| can take on either of four values: x+y, x-y, -x+y, -x-y. Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: (0,1), (-3,2), (1.5,0.5) so the answer is \boxed{\rm (C)~3}. One of those equations overlap into (0,1) so there's only 3 solutions.

Just as in solution 2, we derive the equation ||3-3y|-|y||=1. If we remove the absolute values, the equation collapses into four different possible values. 3-2y, 3-4y, 2y-3, and 4y-3, each equal to either 1 or -1. Remember that if P-Q=a, then Q-P=-a. Because we have already taken 1 and -1 into account, we can eliminate one of the conjugates of each pair, namely 3-2y and 2y-3, and 3-4y and 4y-3. Find the values of y when 3-2y=1, 3-2y=-1, 3-4y=1 and 3-4y=-1. We see that 3-2y=1 and 3-4y=-1 give us the same value for y, so the answer is \boxed{\rm (C)~3}.

Just as in solution 2, we derive the equation x=3-3y. Squaring both sides in the second equation gives x^2+y^2-2|xy|=1. Putting x=3-3y and doing a little calculation gives 10y^2-18y+9-2|3y-3y^2|=1. From here we know that 3y-3y^2 is either positive or negative.

When positive, we get 2y^2-3y+1=0 and then, y=\frac{1}{2} or y=1. When negative, we get y^2-3y+2=0 and then, y=2 or y=1. Clearly, there are 3 different pairs of values and that gives us \boxed{\rm (C)~3}.

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