Question with triangles and squares in it

[Love21]

Hello!

I'll first show you the question, then how far i've gotten....i'm not sure if i'm doing it right though : (
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Let the operations (triangle) and (square) be defined for all real numbers a and b as follows.
a (triangle) b = a + 3b
a (square) b = a + 4b
If 4 (triangle) (5y) = (5y) (square) 4, what is the value of y?
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So i substituted the two triangle equations together and the square...
4 (triangle) (5)(3)
4(triangle)15 = (5y)(square) 16....(i got 16 by multiplying the 2 fours)

4x15=60 60=(5y)(square)16

This is where I get stuck...I think i need to isolate the 5y...but i'm not sure, can someone tell me if i'm
doing this right, and how to proceed?
Thanks!

 

[soroban]

Hello, Love21!

\(\displaystyle \text{Let the operations }\Delta\text{ and }\square\text{ be defined for all real numbers }a\text{ and }b\text{ as follows.}\)

. . . \(\displaystyle a \Delta b \;= \;a + 3b\)

. . . \(\displaystyle a \square b \;= \;a + 4b\)

\(\displaystyle \text{If }4 \Delta (5y) \;= \;(5y) \square 4\text{, what is the value of }y?\)

\(\displaystyle \text{We have: }\;4 \Delta (5y) \;=\;4 + 3(5y) \;=\;4 + 15y\)

. . \(\displaystyle \text{and: }\;(5y) \square 4 \;=\;5y + 4(4) \;=\;5y + 16\)

\(\displaystyle \text{Since they are equal: }\;4 + 15y \:=\:5y + 16\quad\Rightarrow\quad10y \:=\:12\quad\Rightarrow\quad \boxed{y \:=\:\frac{6}{5}}\)

 

[pka]

\(\displaystyle \begin{gathered} 4\Delta \left( {5y} \right) = \left( {5y} \right)\square 4 \hfill \\ 4 + 15y = 5y + 1 \hfill \\ \end{gathered}\)
Just solve for y.

 

[Love21]

Thank you so much!