Difference between revisions of "2021 Fall AMC 10A Problems/Problem 14"
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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> | <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7</math> | ||
− | ==Solution (Graphing)== | + | ==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> | 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> | ||
Line 84: | Line 84: | ||
draw(graph(f,-5,5)); | draw(graph(f,-5,5)); | ||
</asy> | </asy> | ||
− | We see from the graph that there are 5 intersections, so the answer is <math>\boxed{\textbf{(D) } 5}</math>. | + | We see from the graph that there are <math>5</math> intersections, so the answer is <math>\boxed{\textbf{(D) } 5}</math>. |
~KingRavi | ~KingRavi | ||
− | ==Solution | + | ==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== | ==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
Contents
Problem
How many ordered pairs of real numbers satisfy the following system of equations?
Solution 1 (Graphing)
The second equation is . We know that the graph of
is a very simple diamond shape, so let's see if we can reduce this equation to that form:
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 see from the graph that there are
intersections, so the answer is
.
~KingRavi
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/zq3UPu4nwsE?t=974
Video Solution by WhyMath
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.