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Difference between revisions of "Euc20205/Sub-Problem 2"

(Created page with "== Problem == Find all pairs of (x,y,z) given: <math>(x-1)(y-2) = 0</math> <math>(x-3)(z+2) = 0</math> <math>x + yz = 9</math> == Video Solution == https://www.youtube.co...")
 
 
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  <math>(x-3)(z+2) = 0</math>
 
  <math>(x-3)(z+2) = 0</math>
 
  <math>x + yz = 9</math>
 
  <math>x + yz = 9</math>
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 +
== Solution ==
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From the first equation, we can get that <math>x=1</math> or <math>y=2</math>. Suppose that <math>x=1</math>. This means that the remaining equations become <math>-2(z+2)=0</math> and <math>yz=8</math>. From the first of these equations, <math>z=-2</math>, and from the second of these equations, because <math>z=-2</math> we get <math>y=-4</math>. So one triplet is <math>(1,-4,-2)</math>. Now suppose that <math>y=2</math>. The remaining equations become <math>(x-3)(z+2)=0</math> and <math>x+2z=9</math>. From the first equation of these equations, <math>x=3</math> or <math>x=-2</math>. If <math>x=3</math>, then <math>z=3</math> by the second of these equations, and similarly, if <math>z=-2</math>, then <math>x=13</math> by the second equation, so two more triplets are <math>(3,2,3)</math> and <math>(13,2,-2)</math>. After double checking to see if they work, our triplets are <math>\boxed{(1,-4,-2),(3,2,3)}</math> and <math>\boxed{(13,2,-2)}</math>
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 +
~Baihly2024
  
 
== Video Solution ==
 
== Video Solution ==

Latest revision as of 17:16, 12 October 2025

Problem

Find all pairs of (x,y,z) given: 

$(x-1)(y-2) = 0$
$(x-3)(z+2) = 0$
$x + yz = 9$

Solution

From the first equation, we can get that $x=1$ or $y=2$. Suppose that $x=1$. This means that the remaining equations become $-2(z+2)=0$ and $yz=8$. From the first of these equations, $z=-2$, and from the second of these equations, because $z=-2$ we get $y=-4$. So one triplet is $(1,-4,-2)$. Now suppose that $y=2$. The remaining equations become $(x-3)(z+2)=0$ and $x+2z=9$. From the first equation of these equations, $x=3$ or $x=-2$. If $x=3$, then $z=3$ by the second of these equations, and similarly, if $z=-2$, then $x=13$ by the second equation, so two more triplets are $(3,2,3)$ and $(13,2,-2)$. After double checking to see if they work, our triplets are $\boxed{(1,-4,-2),(3,2,3)}$ and $\boxed{(13,2,-2)}$

~Baihly2024

Video Solution

https://www.youtube.com/watch?v=Hx4-y_SipFc

~ NAMCG

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