Euc20205/Sub-Problem 2

Problem

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

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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Euc20205/Sub-Problem 2

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