First, we multiply both sides of the equation by $10$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.2y &= 1 \\[3pt] 10 \cdot (0.2y ) &= 10 \cdot (1) \\[3pt] 2y &= 10 \end{align*}

Then, we solve the equation using the multiplication principle, as usual:

\begin{align*} \require{cancel} 2y &= 10 \\[5pt] \dfrac{2y}{2} &= \dfrac{10}{2} \\[5pt] \dfrac{\cancel{2}y}{\cancel{2}} &= \dfrac{10}{2} \\[5pt] y &= 5 \end{align*}

First, we multiply both sides of the equation by $10$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.3x &=-3.3\\[3pt] 10 \cdot (0.3x) & = 10 \cdot (-3.3) \\[3pt] 3x & = -33 \end{align*}

Then, we solve the equation using the multiplication principle, as usual:

\begin{align*} \require{cancel} 3x & = -33\\[5pt] \dfrac{3x}{3} &= \dfrac{-33}{3} \\[5pt] \dfrac{\cancel{3}x}{\cancel{3}} &= \dfrac{-33}{3} \\[5pt] x &= -11 \end{align*}

First, we multiply both sides of the equation by $10$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.4w&=-6\\[3pt] 10(0.4w) &=10(-6)\\[3pt] 4w & =-60 \end{align*}

Now, we solve the above equation using the multiplication principle, as usual:

\begin{align*} \require{cancel} 4w & =-60\\[5pt] \dfrac{4w}{4} &= \dfrac{-60}{4} \\[5pt] \dfrac{\cancel{4}w}{\cancel{4}} &= -\dfrac{60}{4}\\[5pt] w& =-15 \end{align*}

First, we multiply both sides of the equation by $100$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.15x&=4.5 \\[3pt] 100 \cdot (0.15x) & = 100 \cdot (4.5) \\[3pt] 15x & = 450 \end{align*}

Then, we divide by $15$ to isolate $x \mathbin{:}$

\begin{align*} \require{cancel} 15x & = 450 \\[5pt] \dfrac{15x}{15} &= \dfrac{450}{15} \\[5pt] \dfrac{\cancel{15}x}{\cancel{15}} &= 30 \\[5pt] x &= 30 \end{align*}

First, we multiply both sides of the equation by $100$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.16w & = -1\\[3pt] 100(0.16w) &=100(-1)\\[3pt] 16w & =-100 \end{align*}

Now, we solve the above equation using the multiplication principle, as usual:

\begin{align*} \require{cancel} 16w & =-100\\[5pt] \dfrac{16w}{16} &= \dfrac{-100}{16} \\[5pt] \dfrac{\cancel{16}w}{\cancel{16}} &= -\dfrac{25}{4}\\[5pt] w& =-6.25 \end{align*}

First, we multiply both sides of the equation by $100$ to get rid of the decimals:

\begin{align*} \require{cancel} -0.25t&=1\\[3pt] 100 \cdot (-0.25t) & = 100 \cdot (1) \\[3pt] -25t & = 100 \end{align*}

Then, we divide by $-25$ to isolate $t\mathbin{:}$

\begin{align*} \require{cancel} -25t &= 100 \\[5pt] \dfrac{-25t}{-25} &= \dfrac{100}{-25} \\[5pt] \dfrac{\cancel{-25}t}{\cancel{-25}} &= -4 \\[5pt] t &= -4 \end{align*}

First, we multiply both sides of the equation by $10$ to get rid of the decimals:

\begin{align*} \require{cancel} 1+1.5p&=4 \\[3pt] 10\cdot (1+1.5p) &=10\cdot (4) \\[3pt] 10 \cdot (1) + 10 \cdot (1.5) &= 40 \\[3pt] 10+15p & =40 \end{align*}

Then, we solve the equation using the addition and multiplication principles, as usual:

\begin{align*} \require{cancel} 10+15p & =40 \\[5pt] 15p+10 & =40 \\[5pt] 15p+10-10 & =40-10 \\[5pt] 15p +0& =30 \\[5pt] 15p & =30 \\[5pt] \dfrac{\cancel{15}p}{\cancel{15}} &= \dfrac{30}{15} \\[5pt] p& =2 \end{align*}

First, we multiply both sides of the equation by $100$ to get rid of the decimals:

\begin{align*} \require{cancel} 0.3+0.35x&=1 \\[3pt] 100 \cdot (0.3+0.35x) &= 100 \cdot (1) \\[3pt] 30+35x &= 100 \end{align*}

Then, we solve the equation using the addition and multiplication principles, as usual:

\begin{align*} \require{cancel} 30+35x &= 100 \\[5pt] 35x+30 &= 100 \\[5pt] 35x+30-30 &= 100-30 \\[5pt] 35x+0 &= 70 \\[5pt] 35x &= 70 \\[5pt] \dfrac{\cancel{35}x}{\cancel{35}} &= \dfrac{70}{35}\\[5pt] x&=2 \end{align*}