Algebra, STD2 A2 2022 HSC 14 MC v1 Which of the following correctly expresses \(x\) as the subject of \(y=\dfrac{mx-c}{3}\) ? \(x=\dfrac{3y}{m}+c\) \(x=\dfrac{y}{3m}+c\) \(x=\dfrac{y+c}{3m}\) \(x=\dfrac{3y+c}{m}\) Show Answers Only \(D\) Show Worked Solution \(y\) \(=\dfrac{mx-c}{3}\) \(3y\) \(=mx-c\) \(mx\) \(=3y+c\) \(\therefore\ x\) \(=\dfrac{3y+c}{m}\) \(\Rightarrow D\)