Answer:
[tex]\textsf{3.} \quad \dfrac{5}{6}x+ 10=y[/tex]
4. See below.
[tex]\textsf{5.} \quad \dfrac{10}{3}x-2=y[/tex]
Step-by-step explanation:
Question 3
Given equation:
[tex]\dfrac{1}{3}x-\dfrac{2}{5}y=-4[/tex]
[tex]\textsf{Add \;$\dfrac{2}{5}y$ \; to both sides}:[/tex]
[tex]\implies \dfrac{1}{3}x-\dfrac{2}{5}y+\dfrac{2}{5}y=-4+\dfrac{2}{5}y[/tex]
[tex]\implies \dfrac{1}{3}x=\dfrac{2}{5}y-4[/tex]
Add 4 to both sides:
[tex]\implies \dfrac{1}{3}x+4=\dfrac{2}{5}y-4+4[/tex]
[tex]\implies \dfrac{1}{3}x+4=\dfrac{2}{5}y[/tex]
[tex]\textsf{Multiply both sides by \; $\dfrac{5}{2}$}:[/tex]
[tex]\implies \dfrac{5}{2} \cdot \dfrac{1}{3}x+ \dfrac{5}{2} \cdot4=\dfrac{5}{2} \cdot \dfrac{2}{5}y[/tex]
[tex]\implies \dfrac{5}{6}x+ 10=y[/tex]
[tex]\boxed{\begin{minipage}{3cm}$\dfrac{1}{3}x-\dfrac{2}{5}y=-4\\\\\underline{\phantom{\dfrac{1}{3}x}+\dfrac{2}{5}y\phantom{=-4}\dfrac{2}{5}y}\\\\\dfrac{1}{3}x=\dfrac{2}{5}y-4\\\underline{+4 \phantom{)=3y}+4\phantom{)))))}}\\\\\dfrac{1}{3}x+4=\dfrac{2}{5}y\\\\\underline{\phantom{)}\times\dfrac{5}{2}\phantom{))))}\times \dfrac{5}{2}}\\\\\dfrac{5}{6}x+10=y$ \end{minipage}}[/tex]
Question 4
See the attachment for the highlighted mistake.
The mistake Mike made is that he did not divide 10x by 3.
Question 5
Given equation:
[tex]10x-3y=6[/tex]
Add 3y to both sides:
[tex]\implies 10x-3y+3y=6+3y[/tex]
[tex]\implies 10x=6+3y[/tex]
Subtract 6 from both sides:
[tex]\implies 10x-6=6+3y-6[/tex]
[tex]\implies 10x-6=3y[/tex]
Divide both sides by 3:
[tex]\implies \dfrac{10x}{3}-\dfrac{6}{3}=\dfrac{3y}{3}[/tex]
[tex]\implies \dfrac{10}{3}x-2=y[/tex]
[tex]\boxed{\begin{minipage}{3cm}$10x-3y=6\\\underline{\phantom{10x}+3y\phantom{=6}+3y}\\\\10x=3y+6\\\underline{-6 \phantom{)=3y}-6\phantom{)))))}}\\\\\dfrac{10x-6}{3}=\dfrac{3y}{3}\\\\\dfrac{10}{3}x-2=y$ \end{minipage}}[/tex]
