Here's the question you clicked on:
hugsandkisses
Solve the equation for y. Then find the value of y for the given value of x. 3/5y-4x=3-2y; x=9
What is it that you don't understand?
So you plug in the value of X and then you solve it. \[3/5y-36=3-2y\] \[13/5y=39 \]\[Multiply 5 \to each side\] \[13y=195\] \[y=195/13\] \[y = 15\]
Yup :). But I don't see what you did on your 2nd line... I really don't understand how you got that answer am getting something completely different your question is... \[\frac{3}{5y}-4x=3-2y\] right?
It ends up being a quadratic that has to be solved using the quadratic formula.
\[\frac{3}{5y}-4(9)=3-2y\] \[\frac{3}{5y}-36=3-2y\] Add 36 to both sides \[\frac{3}{5y}=39-2y\] Multiply both sides by 5y \[3=(39-2y)5y\] \[3=195y-10y^2\] Put the equation in standard form \[0=-10y^2+195y-3\]
You do know how to use the quadratic formula right?
Okay let me know if you have any problems! and You're welcome