Solution:

 \(\begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix}\)

= \(3\ \times\)   5   \(\:-\:\)  2  \(\ \times 4\ \)

=  15  \(\:-\: 8\)

=  7 

Solution:

 \(\begin{vmatrix} - 1 & 7 \\ 2 & 4 \end{vmatrix}\)

\(= - 1 \times {4} - 7 \times {2}\)

\(= - 4 - 14\)

\(= - 18\)

Solution:

 \(\begin{vmatrix} 5 & 3 \\ - 7 & 0 \end{vmatrix}\)

\(= 5 \times {0} - 3 \times {- 7}\)

\(= 0 + 21\)

\(= 21\)

Solution:

 \(\LARGE {\begin{vmatrix} \frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2} \end{vmatrix}}\)

\(\displaystyle = \frac{7}{3} \times {\frac{1}{2}} - \frac{5}{3} \times {\frac{3}{2}}\)

\(\displaystyle = \frac{7}{6} - \frac{15}{6}\)

\(\displaystyle = \frac{7 - 15}{6}\)

\(\displaystyle = - \frac{\cancelto{4}{8}}{\cancelto{3}{6}}\)

\(\displaystyle = - \frac{4}{3}\)

Solution:

 \(3x - 4y = 10\)
and \(4x + 3y = 5\)

Comparing with the standard form:
\(a_{1} = 3, b_{1} = -\:4, c_{1} = 10,\\a_{2} = 4, b_{2} = 3, c_{2} = 5\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 3 & -\:4 \\ 4 & 3 \end{vmatrix}\)

∴ \(D = 3 \times 3 - (- 4) \times 4\)
∴ \(D = 9 + 16\)
∴ \(D = 25\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{x} = \begin{vmatrix} 10 & - 4 \\ 5 & 3 \end{vmatrix}\)

∴ \(D_{x} = 10 \times 3 -\:(-\:4) \times 5\)
∴ \(D_{x} = 30 + 20\)
∴ \(D_{x} = 50\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{y} = \begin{vmatrix} 3 & 10 \\ 4 & 5 \end{vmatrix}\)

∴ \(D_{y} = 3 \times 5 -\:10 \times 4\)
∴ \(D_{y} = 15 -\:40\)
∴ \(D_{y} = -\:25\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{2}{50}}{\cancelto{1}{25}}\)

∴ \(\displaystyle x = 2\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{- 1}{-25}}{\cancelto{1}{25}}\)

∴ \(\displaystyle y = -\:1\) ... (v)

∴ The solution is \((x, y) = (2, -\:1)\)

First let’s express the given equations in the standard form,

 \(4x + 3y\:-\:4 = 0\)
∴ \(4x + 3y = 4\)

and \(6x = 8\:-\:5y\)
∴ \(6x + 5y = 8\)

Now, comparing with the standard form:
\(a_{1} = 4, b_{1} = 3, c_{1} = 4,\\a_{2} = 6, b_{2} = 5, c_{2} = 8\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 4 & 3 \\ 6 & 5 \end{vmatrix}\)

∴ \(D = 4 \times 5 - 3 \times 6\)
∴ \(D = 20 - 18\)
∴ \(D = 2\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{x} = \begin{vmatrix} 4 & 3 \\ 8 & 5 \end{vmatrix}\)

∴ \(D_{x} = 4 \times 5 - 3 \times 8\)
∴ \(D_{x} = 20 - 24\)
∴ \(D_{x} = - 4\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{y} = \begin{vmatrix} 4 & 4 \\ 6 & 8 \end{vmatrix}\)

∴ \(D_{y} = 4 \times 8 -\:4 \times 6\)
∴ \(D_{y} = 32 -\:24\)
∴ \(D_{y} = 8\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{- 2}{- 4}}{\cancelto{1}{2}}\)

∴ \(\displaystyle x = - 2\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{4}{8}}{\cancelto{1}{2}}\)

∴ \(\displaystyle y = 4\) ... (v)

∴ \((x, y) = (\:-\:2, 4)\)

Solution:

 \(x + 2y = -\:1\)
and \(2x\:-\:3y = 12\)

Comparing with the standard form:
\(a_{1} = 1, b_{1} = 2, c_{1} = \:-\:1,\\a_{2} = 2, b_{2} = \:-\:3, c_{2} = 12\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 1 & 2 \\ 2 & \:-\:3 \end{vmatrix}\)

∴ \(D = 1 \times (\:-\:3)\:-\:2 \times 2\)
∴ \(D = \:-\:3 \:-\:4\)
∴ \(D = \:-\:7\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{x} = \begin{vmatrix} \:-\:1 & 2 \\ 12 & \:-\:3 \end{vmatrix}\)

∴ \(D_{x} = (\:-\:1) \times( \:-\:3)\:-\:2 \times 12\)
∴ \(D_{x} = 3\:-\:24\)
∴ \(D_{x} = \:-\:21\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{y} = \begin{vmatrix} 1 & \:-\:1 \\ 2 & 12 \end{vmatrix}\)

∴ \(D_{y} = 1 \times 12\:-\:(\:-\:1) \times 2\)
∴ \(D_{y} = 12 + 2\)
∴ \(D_{y} = 14\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{3}{\:-\:21}}{\cancelto{1}{\:-\:7}}\)

∴ \(\displaystyle x = 3\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{- 2}{14}}{\cancelto{1}{\:-\:7}}\)

∴ \(\displaystyle y = -\:2\) ... (v)

∴ The solution is \((x, y) = (3, -\:2)\)

Solution:

 \(6x\:-\:4y =\:-\:12\)
and \(8x\:-\:3y =\:-\:2\)

Comparing with the standard form:
\(a_{1} = 6, b_{1} =\:-\:4, c_{1} =\:-\:12,\\a_{2} = 8, b_{2} =\:-\:3, c_{2} =\:-\:2\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 6 & \:-\:4 \\ 8 & \:-\:3 \end{vmatrix}\)

∴ \(D = 6 \times (\:-\:3)\:-\:(\:-\:4) \times 8\)
∴ \(D = \:-\:18 + 32\)
∴ \(D = 14\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{x} = \begin{vmatrix} \:-\:12 & \:-\:4 \\ \:-\:2 & \:-\:3 \end{vmatrix}\)

∴ \(D_{x} = (\:-\:12) \times(\:-\:3)\:-\:(\:-\:4) \times (\:-\:2)\)
∴ \(D_{x} = 36\:-\:8\)
∴ \(D_{x} = 28\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{y} = \begin{vmatrix} 6 & \:-\:12 \\ 8 & \:-\:2 \end{vmatrix}\)

∴ \(D_{y} = 6 \times (\:-\:2)\:-\:(\:-\:12) \times 8\)
∴ \(D_{y} = \:-\:12 + 96\)
∴ \(D_{y} = 84\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{2}{28}}{\cancelto{1}{14}}\)

∴ \(\displaystyle x = 2\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{6}{84}}{\cancelto{1}{14}}\)

∴ \(\displaystyle y = 6\) ... (v)

∴ The solution is \((x, y) = (2, 6)\)

Solution:

 \(4m + 6n = 54\)
and \(3m + 2n = 28\)

Comparing with the standard form:
\(a_{1} = 4, b_{1} = 6, c_{1} = 54,\\a_{2} = 3, b_{2} = 2, c_{2} = 28\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 4 & 6 \\ 3 & 2 \end{vmatrix}\)

∴ \(D = 4 \times 2\:-\:6 \times 3\)
∴ \(D = 8\:-\:18\)
∴ \(D =\:-\:10\) ... (i)

Also, \(D_{m} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{m} = \begin{vmatrix} 54 & 6 \\ 28 & 2 \end{vmatrix}\)

∴ \(D_{m} = 54 \times 2\:-\:6 \times 28\)
∴ \(D_{m} = 108\:-\:168\)
∴ \(D_{m} =\:-\:60\) ... (ii)

And, \(D_{n} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{n} = \begin{vmatrix} 6 & \:-\:12 \\ 8 & \:-\:2 \end{vmatrix}\)

∴ \(D_{n} = 4 \times 28\:-\:54 \times 3\)
∴ \(D_{n} = 112\:-\:162\)
∴ \(D_{n} =\:-\:50\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle m = \frac{D_{m}}{D}\)

∴ \(\displaystyle m = \frac{\cancelto{6}{\:-\:60}}{\cancelto{1}{\:-:10}}\)

∴ \(\displaystyle m = 6\) ... (iv)

Also, \(\displaystyle n = \frac{D_{n}}{D}\)

∴ \(\displaystyle n = \frac{\cancelto{5}{\:-\:50}}{\cancelto{1}{\:-\:10}}\)

∴ \(\displaystyle n = 5\) ... (v)

∴ The solution is \((m, n) = (6, 5)\)

Solution:

 \(\displaystyle 2x + 3y = 2\)

and \(\displaystyle x\:-\:\frac{y}{2} = \frac {1}{2}\)

Multiplying both sides by 2,

 \(\displaystyle x \times 2 \:-\:\frac{y}{\cancel{2}} \times \cancel{2} = \frac {1}{\bcancel{2}} \times \bcancel{2}\)

∴ \(2x - y = 1\)

Comparing with the standard form:
\(a_{1} = 2, b_{1} = 3, c_{1} = 2,\\a_{2} = 2, b_{2} =\:-\:1, c_{2} = 1\)

 \(D = \begin{vmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\)

∴ \(D = \begin{vmatrix} 2 & 3 \\ 2 & \:-\:1 \end{vmatrix}\)

∴ \(D = 2 \times (\:-\:1)\:-\:3 \times 2\)
∴ \(D = \:-\:2\:-\:6\)
∴ \(D =\:-\:8\) ... (i)

Also, \(D_{x} = \begin{vmatrix} c_{1} & b_{1} \\ c_{2} & b_{2} \end{vmatrix}\)

∴ \(D_{x} = \begin{vmatrix} 2 & 3 \\ \:-\:1 & \:-\:1 \end{vmatrix}\)

∴ \(D_{x} = 2 \times(\:-\:1)\:-\:3 \times 1\)
∴ \(D_{x} =\:-\:2\:-\:3\)
∴ \(D_{x} =\:-\:5\) ... (ii)

And, \(D_{y} = \begin{vmatrix} a_{1} & c_{1} \\ a_{2} & c_{2} \end{vmatrix}\)

∴ \(D_{y} = \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\)

∴ \(D_{y} = 2 \times 1\:-\:2 \times 2\)
∴ \(D_{y} = 2\:-\:4\)
∴ \(D_{y} =\:-\:2\) ... (iii)

Now, using Cramer’s Rule:

 \(\displaystyle x = \frac{D_{x}}{D}\)

∴ \(\displaystyle x = \frac{\cancelto{5}{\:-\:5}}{\cancelto{8}{\:-\:8}}\)

∴ \(\displaystyle x = \frac {5}{8}\) ... (iv)

Also, \(\displaystyle y = \frac{D_{y}}{D}\)

∴ \(\displaystyle y = \frac{\cancelto{1}{\:-\:2}}{\cancelto{4}{\:-\:8}}\)

∴ \(\displaystyle y = \frac {1}{4}\) ... (v)

∴ The solution is \(\displaystyle (x, y) = \left(\frac {5}{8},\frac {1}{4}\right)\)