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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 14"

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How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations?  
 
How many ordered pairs <math>(x,y)</math> of real numbers satisfy the following system of equations?  
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<cmath>x^2+3y=9</cmath>
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<cmath>\begin{align*}
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<cmath>(|x|+|y|-4)^2 = 1</cmath>
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x^2+3y&=9 \\
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(|x|+|y|-4)^2 &= 1
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\end{align*}</cmath>
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<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math>
  
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<math>\textbf{(A )} 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math>
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==Solution 1 (Graphing)==
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==Solution (Graphing)==
 
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The second equation is <math>(|x|+|y| - 4)^2 = 1</math>. We know that the graph of <math>|x| + |y|</math> is a very simple diamond shape, so let's see if we can reduce this equation to that form.
 
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<math>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5</math>}.
 
  
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The second equation is <math>(|x|+|y| - 4)^2 = 1</math>. We know that the graph of <math>|x| + |y|</math> is a very simple diamond shape, so let's see if we can reduce this equation to that form: <cmath>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.</cmath>
 
We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane:
 
We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane:
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<asy>
 
<asy>
  
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draw(graph(f,0,-5));
 
draw(graph(f,0,-5));
  
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real f(real x)
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{
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return 3-x;
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}
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draw(graph(f,0,3));
  
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real f(real x)
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{
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return 3+x;
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}
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draw(graph(f,0,-3));
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real f(real x)
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{
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return x-3;
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}
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draw(graph(f,0,3));
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real f(real x)
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{
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return (-x^2)/3+3;
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}
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draw(graph(f,-5,5));
 
</asy>
 
</asy>
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We see from the graph that there are <math>5</math> intersections, so the answer is <math>\boxed{\textbf{(D) } 5}</math>.
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~KingRavi
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==Video Solution ==
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https://youtu.be/yASY-XL9vtI
  
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==Solution 2 (Unrigorous but pretty standard)==
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~Education, the Study of Everything
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We can manipulate the first equation to get <math>y = -\frac{x^{2}}{3} + 3</math>. From the second equation, we have that <math>|x|+|y|-4 = 1</math> or <math>|x|+|y|-4 = -1</math>. We will consider each case separately.
 
  
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If <math>|x|+|y|-4 = 1</math>, then <math>|x|+|y| = 5</math>. The graph of this is a square with vertices <math>(5,0)</math>, <math>(-5,0)</math>, <math>(0,5)</math> and <math>(0,-5)</math>. The parabola from the first equation is downwards facing, and its vertex is inside this square; the parabola will clearly intersect the square twice. Therefore, this case gives us <math>\underline{2}</math> solutions.
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==Video Solution==
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https://youtu.be/zq3UPu4nwsE?t=974
  
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If <math>|x|+|y|-4 = -1</math>, then <math>|x|+|y| = 3</math>. The graph of this is a square with vertices <math>(3,0)</math>, <math>(-3,0)</math>, <math>(0,3)</math> and <math>(0,-3)</math>. The vertex of the parabola from the first equation is on one of the corners of this square (in particular, <math>(0,3)</math>). Also, at <math>y = 0</math>, the parabola has <math>x</math> intercepts of <math>\pm 3</math>; the square passes through both of those points. If we continue to move down, the square narrows in, while the parabola continues to expand. Therefore, these are our only <math>3</math> intersection points in this case: <math>(0,3)</math>, <math>(3,0)</math> and <math>(-3,0)</math>. This case gives us <math>\underline{3}</math> solutions.
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==Video Solution by WhyMath==
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https://youtu.be/5SVmxNrZUbY
  
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Adding these two cases together, we get our final answer of <math>\boxed{\textbf{(D) } 5}</math>.
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:20, 26 September 2023

Problem

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

Solution 1 (Graphing)

The second equation is $(|x|+|y| - 4)^2 = 1$. We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: \[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\] We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane: [asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy] We see from the graph that there are $5$ intersections, so the answer is $\boxed{\textbf{(D) } 5}$.

~KingRavi

Video Solution

https://youtu.be/yASY-XL9vtI

~Education, the Study of Everything

Video Solution

https://youtu.be/zq3UPu4nwsE?t=974

Video Solution by WhyMath

https://youtu.be/5SVmxNrZUbY

~savannahsolver

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AMC 10 Problems and Solutions

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