Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 8"

Revision as of 15:52, 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

Revision as of 15:06, 11 July 2021 (view source) Samrocksnature (talk \| contribs) (→‎Problem) ← Older edit		Revision as of 15:52, 11 July 2021 (view source) Samrocksnature (talk \| contribs) (→‎Solution) Newer edit →
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	==Solution==		==Solution==
−	~~asdf~~	+	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

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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 8"

Revision as of 15:52, 11 July 2021

Problem

Solution