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

Latest revision as of 09:07, 12 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

Solution 2

Each number shares are factor of $6$ , which means $(x+y)(20x+21y)=6$ , or $x+y=2$ and $20x+21y=3$ . We see $y=-37$ and $x=39$ , so $39(21)-20(37)=\boxed{79}$

~Geometry285

Solution 3

Multiplying the equations together, we get $(x+y)^3(20x+21y)^3=2^3 \cdot 3^3 \implies (x+y)(20x+21y)=6$ Therefore, $x+y=2 \implies 20x+20y=40$ $20x+21y=3$ Subtracting the equations, we get $y=-37$ and $x=39$ , therefore, $21 (39) - 20 (37) =\boxed{79}$

- kante314 -

@@ Line 7: / Line 7: @@
 ==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
+==Solution 2==
+Each number shares are factor of <math>6</math>, which means <math>(x+y)(20x+21y)=6</math>, or <math>x+y=2</math> and <math>20x+21y=3</math>. We see <math>y=-37</math> and <math>x=39</math>, so <math>39(21)-20(37)=\boxed{79}</math>
+~Geometry285
+== Solution 3 ==
+Multiplying the equations together, we get
+<cmath>(x+y)^3(20x+21y)^3=2^3 \cdot 3^3 \implies (x+y)(20x+21y)=6</cmath>Therefore,
+<cmath>x+y=2 \implies 20x+20y=40</cmath><cmath>20x+21y=3</cmath>Subtracting the equations, we get <math>y=-37</math> and <math>x=39</math>, therefore, <math>21 (39) - 20 (37) =\boxed{79}</math>
+- kante314 -
 ==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]]
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