First, we factor out $x$ in the expression on the left-hand side.

\begin{align*} 2kx - x &= 1 \\[5pt] x(2k-1) &=1 \end{align*}

Then, we divide both sides by $(2k-1)$ to isolate $x\mathbin{:}$

\begin{align*} \require{cancel} x(2k-1) &=1 \\[5pt] \dfrac{x(2k-1)}{2k-1} &=\dfrac{1}{2k-1} \\[5pt] \dfrac{x\cancel{(2k-1)}}{\cancel{2k-1}} &=\dfrac{1}{2k-1} \\[5pt] x &= \dfrac{1}{2k-1} \end{align*}

Note: This result assumes that $2k-1\neq 0.$

First, we factor out $p$ in the expression on the right-hand side. \begin{align*} 5 &= 3pq - p \\[5pt] 5 &=(3q-1)p \end{align*}

Then, we divide both sides by $(3q-1)$ to isolate $p\mathbin{:}$

\begin{align*} \require{cancel} 5 &= (3q-1)p \\[5pt] \dfrac{5}{3q-1} &= \dfrac{p(3q-1)}{3q-1}\\[5pt] \dfrac{5}{3q-1} &= \dfrac{p\cancel{(3q-1)}}{\cancel{3q-1}} \\[5pt] \dfrac{5}{3q-1} &= p\\[5pt] p &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{5}{3q-1}} \end{align*}

Note: This result assumes that $3q-1\neq 0.$

We need to isolate all terms with $m$ on one side of the equation. To do this, we subtract $4$ to both sides: \begin{align*} 2mn +m + 4&= 7 \\[5pt] 2mn +m + 4 - 4&= 7 - 4\\[5pt] 2mn + m &= 3 \end{align*}

Next, we factor out $m$ in the expression on the left-hand side. \begin{align*} 2mn + m&= 3 \\[5pt] m (2n+1)&= 3 \\[5pt] \end{align*}

Then, we divide both sides by $(2n+1)$ to isolate $m\mathbin{:}$ \begin{align*} \require{cancel} m (2n+1)&= 3 \\[5pt] \dfrac{m(2n+1)}{2n+1}&= \dfrac{3}{2n+1} \\[5pt] \dfrac{m\cancel{(2n+1)}}{\cancel{2n+1}} &=\dfrac{3}{2n+1} \\[5pt] m &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{3}{2n+1}} \end{align*}

Note: This result assumes that $2n+1\neq 0.$

We need to isolate all terms with $x$ on one side of the equation. To do this, we add $2x$ to both sides:

\begin{align*} 3xc &= 5-2x \\[5pt] 3xc+2x &= 5-2x+2x \\[5pt] 3xc+2x &= 5 \end{align*}

Notice that all terms containing $x$ are on the left-hand side, and all terms independent of $x$ are on the right-hand side.

Next, we factor out $x$ in the expression on the left-hand side.

\begin{align*} \require{cancel} 3xc+2x &= 5 \\[5pt] x(3c+2) &= 5 \\[5pt] \end{align*}

Finally, we divide both sides by $(3c+2)$ to isolate $x\mathbin{:}$

\begin{align*} x(3c+2) &= 5 \\[5pt] \dfrac{x(3c+2)}{3c+2} &= \dfrac{5}{3c+2} \\[5pt] \dfrac{x\cancel{(3c+2)}}{\cancel{3c+2}} &= \dfrac{5}{3c+2} \\[5pt] x &= \dfrac{5}{3c+2} \end{align*}

Note: This result assumes that $3c+2\neq 0.$

We need to isolate all terms with $y$ on one side of the equation. To do this, we start by adding $4$ to both sides:

\begin{align*} 3xy - 4&= 7y+2 \\[5pt] 3xy-4+4 &= 7y+2+4\\[5pt] 3xy &= 7y+6 \end{align*}

Next, we subtract $7y$ from both sides of the equation:

\begin{align*} 3xy &= 7y + 6 \\[5pt] 3xy - 7y &= 7y + 6 - 7y \\[5pt] 3xy-7y&= 6 \end{align*}

Notice that all terms containing $y$ are on the left-hand side, and all terms independent of $y$ are on the right-hand side.

Next, we factor out $y$ in the expression on the left-hand side.

\begin{align*} \require{cancel} 3xy-7y &= 6 \\[5pt] (3x-7)y &= 6 \end{align*}

Finally, we divide both sides by $(3x-7)$ to isolate $y\mathbin{:}$

\begin{align*} \require{cancel} (3x-7)y &= 6 \\[5pt] \dfrac{(3x-7)y}{3x-7} &= \dfrac{6}{3x-7} \\[5pt] \dfrac{\cancel{(3x-7)}y}{\cancel{3x-7}} &= \dfrac{6}{(3x-7)} \\[5pt] y &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{6}{3x-7}} \end{align*}

Note: This result assumes that $3x-7\neq 0.$

We need to isolate all terms with $m$ on one side of the equation. To do this, we start by subtracting $3$ from both sides: \begin{align*} 2mn + 3 &= n - 3m\\[5pt] 2mn + 3 - 3 &= n - 3m - 3 \\[5pt] 2mn &= n - 3m - 3 \\[5pt] \end{align*}

Next, we add $3m$ to both sides of the equation: \begin{align*} 2mn &= n - 3m - 3 \\[5pt] 2mn+3m &= n - 3m-3+3m \\[5pt] 2mn + 3m&= n- 3 \end{align*}

Notice that all terms containing $m$ are on the left-hand side, and all terms independent of $m$ are on the right-hand side.

Then, we factor out $m$ in the expression on the left-hand side. \begin{align*} \require{cancel} 2mn + 3m &= n - 3 \\[5pt] m(2n+3) &= n - 3\\[5pt] \end{align*}

Finally, we divide both sides by $(2n +3)$ to isolate $m\mathbin{:}$ \begin{align*} m(2n+3) &= n - 3 \\[5pt] \dfrac{m(2n +3)}{2n +3} &= \dfrac{n -3}{2n +3} \\[5pt] \dfrac{m\cancel{(2n +3)}}{\cancel{2n +3}} &= \dfrac{n-3}{2n +3} \\[5pt] m &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{n-3}{2n +3}} \end{align*}

Note: This result assumes that $2n+3\neq 0.$

We need to isolate all terms with $p$ on one side of the equation. To do this, we first cross-multiply:

\begin{align*} \dfrac{5pq}{2}&=\dfrac{4p+q}{3}\\[5pt] 3\cdot 5pq &= 2 \cdot (4p+q) \\[5pt] 15pq &= 8p+2q \end{align*}

Next, we subtract $8p$ from both sides of the equation: \begin{align*} 15pq &= 8p+2q\\[5pt] 15pq-8p &= 8p+2q-8p\\[5pt] 15pq -8p&= 2q \end{align*}

Notice that all terms containing $p$ are on the left-hand side, and all terms independent of $p$ are on the right-hand side.

Then, we factor out $p$ from the left-hand side: \begin{align*} \require{cancel} 15pq -8p&= 2q\\[5pt] (15q -8)p&= 2q \\[5pt] \end{align*}

Finally, we divide both sides by $(15q-8)$ to isolate $p\mathbin{:}$ \begin{align*} (15q-8)p &= 2q \\[5pt] \dfrac{\cancel{(15q-8)}p}{\cancel{15q-8}} &= \dfrac{2q}{15q-8} \\[5pt] p &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{2q}{15q-8}} \end{align*}

Note: This result assumes that $15q-8\neq 0.$

We need to isolate all terms with $m$ on one side of the equation. To do this, we first cross-multiply: \begin{align*} \dfrac{km-1}{2} &= \dfrac{m+4}{3} \\[5pt] 3\cdot (km - 1) &= 2 \cdot (m+ 4) \\[5pt] 3km - 3 &= 2m + 8 \end{align*}

Next, we subtract $2m$ from both sides of the equation: \begin{align*} 3km - 3 &= 2m + 8 \\[5pt] 3km - 3 - 2m &= 2m + 8 - 2m \\[5pt] 3km - 2m - 3 &= 8 \\[5pt] \end{align*}

Then, we add $3$ to both sides of the equation: \begin{align*} 3km - 2m - 3 &= 8 \\[5pt] 3km - 2m - 3 + 3 &= 8 + 3 \\[5pt] 3km - 2m &= 11 \\[5pt] \end{align*}

Notice that all terms containing $m$ are on the left-hand side, and all terms independent of $m$ are on the right-hand side.

Then, we factor out $m$ from the left-hand side: \begin{align*} \require{cancel} 3km - 2m &= 11\\[5pt] m(3k - 2) &= 11 \\[5pt] \end{align*}

Finally, we divide both sides by $(3k-2)$ to isolate $m\mathbin{:}$ \begin{align*} m(3k-2) &= 11 \\[5pt] \dfrac{m\cancel{(3k-2)}}{\cancel{3k-2}} &= \dfrac{11}{3k-2} \\[5pt] m &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{11}{3k-2}} \end{align*}

Note: This result assumes that $3k-2\neq 0.$