You have to make this more clear. Are we solving x in terms of y or vice versa? I am going to assume that it is in terms of y.
What is \(\frac{y-5}{2}<-\frac{x}{3}\) ?
Multiply 6 on both sides to get \(3y-15<-2x\)
Divide -2 on both sides, but we have to change the sign \(-\frac{3}{2}y+\frac{15}{2}>x\)
Hi Jeff
You maths is fine but you have not made either x or y the subject.
Yours is fine, you only had to put x on the left hand side
what is (y-5)/2<-x/3
Make x the subject:
\((y-5)/2<-x/3\\ \frac{3(y-5)}{2}<-x\\ \frac{-3(y-5)}{2}>x\\ x<\frac{-3(y-5)}{2}\\ or\\ x<\frac{3(5-y)}{2}\\\)
Make y the subject
\(\frac{y-5}{2}<\frac{-x}{3}\\ y-5<\frac{-2x}{3}\\ y<-\frac{2x}{3}+5\\ \)