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For any two real numbers $x$ and $y,$ define \[x \star y = ax + by + cxy,\]where $a,$ $b,$ and $c$ are constants. It is known that $1 \star 2 = 3,$ $2 \star 3 = 4,$ and there is a non-zero real number $d$ such that $x \star d = x$ for any real number $x.$ What is the value of $d$?

 May 4, 2021
 #1
avatar+2209 
+1

Well, what do we know?

a + 2b + 2c = 3

2a + 3b + 6c = 4

I'm just going to assume a = 1.

1 + 2b + 2c = 3

2b + 2c = 2

2 + 3b + 6c = 4

3b + 2c = 2

b = 0

c = 1

 

We're looking for a d, where 1x + 0d + 1xd = x. 

x + xd = x

d = 0 hopefully. :))

 

=^._.^=

 May 4, 2021
 #2
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0

Sorry it says: "and there is a non-zero real number $d$"

 May 4, 2021
 #3
avatar+2209 
+1

Woah, I am really messing up lately. 

Any idea how to do this problem?

 

=^._.^=

catmg  May 4, 2021

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