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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?  
−
<cmath>x^2+3y=9</cmath>
+
<cmath>\begin{align*}
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<cmath>(|x|+|y|-4)^2 = 1</cmath>
+
x^2+3y&=9 \\
 +
(|x|+|y|-4)^2 &= 1
 +
\end{align*}</cmath>
 +
<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math>
  
−
<math>\textbf{(A )} 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math>
+
==Solution 1 (Graphing)==
  
 +
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:
 +
<asy>
  
−
==Solution (Graphing)==
+
Label f;
 +
f.p=fontsize(6);
 +
xaxis(-8,8,Ticks(f, 1.0,0.5));
 +
yaxis(-8,8,Ticks(f, 1.0,0.5));
  
−
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.
+
real f(real x)
 +
{
 +
return 3-x;
 +
}
 +
draw(graph(f,0,3));
  
−
<math>(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5</math>}.
+
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));
  
−
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:
+
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 <math>5</math> intersections, so the answer is <math>\boxed{\textbf{(D) } 5}</math>.
 +
 
 +
~KingRavi
  
 +
==Video Solution ==
 +
https://youtu.be/yASY-XL9vtI
  
−
<asy>
+
~Education, the Study of Everything
−
import graph;
 
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Label f;
 
−
f.p=fontsize(6);
 
−
xaxis(-8,8,Ticks(f, 2.0,0.5));
 
−
yaxis(-8,8,Ticks(f, 2.0,0.5))
 
  
 +
==Video Solution==
 +
https://youtu.be/zq3UPu4nwsE?t=974
  
 +
==Video Solution by WhyMath==
 +
https://youtu.be/5SVmxNrZUbY
  
−
</asy>
+
~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
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

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