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Exercise 12.3 · Q6

Q.In the set Q\mathbb{Q} define a⊙b=a+b+aba\odot b = a+b+ab. For what value of yy, 3⊙(y⊙5)=73\odot(y\odot 5)=7?

(1) y=23y=\dfrac23
(2) y=−23y=\dfrac{-2}3
(3) y=−32y=\dfrac{-3}2
(4) y=4y=4
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We work from the inside out: first evaluate y⊙5y\odot5 in terms of yy, substitute into the outer 3⊙( )3\odot(\ ), then solve the resulting linear equation for yy.

Step 1. Compute the inner expression y⊙5y\odot5. Using a⊙b=a+b+aba\odot b=a+b+ab: y⊙5=y+5+5y=6y+5y\odot5=y+5+5y=6y+5.

Step 2. Substitute into 3⊙(y⊙5)3\odot(y\odot5). 3⊙(6y+5)=3+(6y+5)+3(6y+5)3\odot(6y+5)=3+(6y+5)+3(6y+5). …

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