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

(Solution 2)
(Correct labelling)
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== Problem ==
+
==Problem==
 
 
 
How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations?
 
How many ordered pairs of real numbers <math>(x,y)</math> satisfy the following system of equations?
 
<cmath>x+3y=3</cmath>  
 
<cmath>x+3y=3</cmath>  
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\textbf{(E) } 8 </math>
 
\textbf{(E) } 8 </math>
  
== Solution ==
+
==Solution 1==
 
 
 
 
 
The graph looks something like this:
 
The graph looks something like this:
 
<asy>
 
<asy>
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dot((0,1));
 
dot((0,1));
 
</asy>
 
</asy>
 +
Now, it becomes 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)
+
==Solution 2==
 
 
== Solution 2 ==
 
 
 
 
 
 
<math>x+3y=3</math> can be rewritten to <math>x=3-3y</math>. Substituting <math>3-3y</math> for <math>x</math> in the second equation will give <math>||3-3y|-y|=1</math>. Splitting this question into casework for the ranges of y will give us the total number of solutions.
 
<math>x+3y=3</math> can be rewritten to <math>x=3-3y</math>. Substituting <math>3-3y</math> for <math>x</math> in the second equation will give <math>||3-3y|-y|=1</math>. Splitting this question into casework for the ranges of y will give us the total number of solutions.
 
 
 
<math>\textbf{Case 1:}</math> <math>y>1</math>
 
<math>\textbf{Case 1:}</math> <math>y>1</math>
 
 
<math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math>
 
<math>3-3y</math> will be negative so <math>|3-3y| = 3y-3.</math>
 
 
<math>|3y-3-y| = |2y-3| = 1</math>
 
<math>|3y-3-y| = |2y-3| = 1</math>
 
     Subcase 1: <math>y>\frac{3}{2}</math>
 
     Subcase 1: <math>y>\frac{3}{2}</math>
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     Subcase 2: <math>1<y<\frac{3}{2}</math>
 
     Subcase 2: <math>1<y<\frac{3}{2}</math>
 
<math>2y-3</math> is negative so <math>|2y-3| = 3-2y = 1</math>. <math>2y = 2</math> and so there are no solutions (<math>y</math> can't equal to <math>1</math>)
 
<math>2y-3</math> is negative so <math>|2y-3| = 3-2y = 1</math>. <math>2y = 2</math> and so there are no solutions (<math>y</math> can't equal to <math>1</math>)
 
 
 
<math>\textbf{Case 2:}</math> <math>y = 1</math>
 
<math>\textbf{Case 2:}</math> <math>y = 1</math>
 
It is fairly clear that <math>x = 0.</math>
 
It is fairly clear that <math>x = 0.</math>
 
 
 
<math>\textbf{Case 3:}</math> <math>y<1</math>
 
<math>\textbf{Case 3:}</math> <math>y<1</math>
 
 
<math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math>
 
<math>3-3y</math> will be positive so <math>|3-3y-y| = |3-4y| = 1</math>
 
     Subcase 1: <math>y>\frac{4}{3}</math>
 
     Subcase 1: <math>y>\frac{4}{3}</math>
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     Subcase 2: <math>y<\frac{4}{3}</math>
 
     Subcase 2: <math>y<\frac{4}{3}</math>
 
<math>3-4y</math> will be positive so <math>3-4y = 1</math> \rightarrow <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>.
 
<math>3-4y</math> will be positive so <math>3-4y = 1</math> \rightarrow <math>4y = 2</math>. <math>y = \frac{1}{2}</math> and <math>x = \frac{3}{2}</math>.
 
 
Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>3,</math> or <math>\boxed{\textbf{(C)}}</math>
 
Thus, the solutions are: <math>(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)</math>, and the answer is <math>3,</math> or <math>\boxed{\textbf{(C)}}</math>
 
 
Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel.
 
Solution by Danny Li JHS, <math>\text{\LaTeX}</math> edit by pretzel.
  
== See Also ==
+
==See Also==
 
 
 
{{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:17, 9 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 1

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((3/2,1/2)); dot((0,1)); [/asy] Now, it becomes clear that there are $\boxed{3}$ intersection points. (pinetree1)

Solution 2

$x+3y=3$ 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. $\textbf{Case 1:}$ $y>1$ $3-3y$ 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 $1$) $\textbf{Case 2:}$ $y = 1$ It is fairly clear that $x = 0.$ $\textbf{Case 3:}$ $y<1$ $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$

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

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

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

$3-4y$ will be positive so $3-4y = 1$ \rightarrow $4y = 2$. $y = \frac{1}{2}$ 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 $3,$ or $\boxed{\textbf{(C)}}$ Solution by Danny Li JHS, $\text{\LaTeX}$ edit by pretzel.

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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