Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 8"
Line 7: Line 7:
 
==Solution==
 
==Solution==
 
We let <math>a=(x+y)</math> and <math>b=(20x+21y)</math> to get the new system of equations <cmath>a^2b=12 \qquad (1)</cmath> <cmath>ab^2=18 \qquad(2).</cmath> Multiplying these two, we have <math>(ab)^3=12 \cdot 18</math> or <cmath>ab=6 \qquad (3).</cmath> We divide <math>(3)</math> by <math>(1)</math> to get <math>a=2</math> and divide <math>(2)</math> by <math>(1)</math> to get <math>b=3</math>. Recall that <math>a=x+y=2</math> and <math>b=20x+21y=3</math>. Solving the system of equations <cmath>x+y=2</cmath> <cmath>20x+21y=3,</cmath> we get <math>y=-37</math> and <math>x=39</math>. This means that <cmath>21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.</cmath> ~samrocksnature
 
We let <math>a=(x+y)</math> and <math>b=(20x+21y)</math> to get the new system of equations <cmath>a^2b=12 \qquad (1)</cmath> <cmath>ab^2=18 \qquad(2).</cmath> Multiplying these two, we have <math>(ab)^3=12 \cdot 18</math> or <cmath>ab=6 \qquad (3).</cmath> We divide <math>(3)</math> by <math>(1)</math> to get <math>a=2</math> and divide <math>(2)</math> by <math>(1)</math> to get <math>b=3</math>. Recall that <math>a=x+y=2</math> and <math>b=20x+21y=3</math>. Solving the system of equations <cmath>x+y=2</cmath> <cmath>20x+21y=3,</cmath> we get <math>y=-37</math> and <math>x=39</math>. This means that <cmath>21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.</cmath> ~samrocksnature
 +
 +
  
 
==See also==
 
==See also==
−
#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
+
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
−
#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
+
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
{{JMPSC Notice}}
 
{{JMPSC Notice}}

Revision as of 17:31, 11 July 2021

Problem

Let $x$ and $y$ be real numbers that satisfy \[(x+y)^2(20x+21y) = 12\] \[(x+y)(20x+21y)^2 = 18.\] Find $21x+20y$.

Solution

We let $a=(x+y)$ and $b=(20x+21y)$ to get the new system of equations \[a^2b=12 \qquad (1)\] \[ab^2=18 \qquad(2).\] Multiplying these two, we have $(ab)^3=12 \cdot 18$ or \[ab=6 \qquad (3).\] We divide $(3)$ by $(1)$ to get $a=2$ and divide $(2)$ by $(1)$ to get $b=3$. Recall that $a=x+y=2$ and $b=20x+21y=3$. Solving the system of equations \[x+y=2\] \[20x+21y=3,\] we get $y=-37$ and $x=39$. This means that \[21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.\] ~samrocksnature


See also

  1. Other 2021 JMPSC Invitationals Problems
  2. 2021 JMPSC Invitationals Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_8&oldid=158105"