Alan

The maximum value that the sine of an angle can have is 1.  Therefore it isn't possible for sinA to equal 8/6.

$$\left({{\mathtt{p}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\frac{{{\mathtt{4}}}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{p}}\right)}}\right){\mathtt{\,\times\,}}{\mathtt{1\,293}} = {\frac{{\mathtt{0.018\: \!5}}{\mathtt{\,\times\,}}{\mathtt{360}}}{{\mathtt{0.2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{27.44}}{\mathtt{\,\times\,}}{\mathtt{1\,293}}}{{\mathtt{4\,758}}}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{p}} = {\frac{{\mathtt{1\,616\,011}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{153\,802\,350}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)\\
{\mathtt{p}} = {\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1\,616\,011}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{153\,802\,350}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{p}} = {\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1\,616\,011}}}{\left({\mathtt{153\,802\,350}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{58\,211\,509\,993\,402\,437\,011\,390\,669}}}}}{\left({\mathtt{3\,375}}{\mathtt{\,\times\,}}{{\mathtt{683\,566}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{p}} = {\mathtt{\,-\,}}{\mathtt{1.002\: \!626\: \!759\: \!755\: \!284}}{\mathtt{\,-\,}}{\mathtt{1.727\: \!501\: \!105\: \!285\: \!170\: \!6}}{i}\\
{\mathtt{p}} = {\mathtt{\,-\,}}{\mathtt{1.002\: \!626\: \!759\: \!755\: \!284}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.727\: \!501\: \!105\: \!285\: \!170\: \!6}}{i}\\
{\mathtt{p}} = {\mathtt{2.005\: \!253\: \!519\: \!510\: \!568}}\\
\end{array} \right\}$$

Multiply through by y:

y² - 9 = 10y

Subtract 10y from both sides:

y² - 10y -9 = 0

This is now in standard quadratic equation form.

$${{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{5}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{34}}}}\\
{\mathtt{y}} = {\sqrt{{\mathtt{34}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = -{\mathtt{0.830\: \!951\: \!894\: \!845\: \!300\: \!5}}\\
{\mathtt{y}} = {\mathtt{10.830\: \!951\: \!894\: \!845\: \!300\: \!5}}\\
\end{array} \right\}$$

Ok, well, 25 is as simple as it gets (assuming you mean the dot to represent multiplication)!

Do you mean [23 - (4*2 + 5*2)]*[5*6 - 4 - 21] = (23 - 18)*(30 - 25) = 5*5 = 25 ? 

Curly brackets usually indicates a set or list.  Is this what you intended?  Or did you mean something like |-2|, which would be the absolute value of -2 (which is just 2)?

4 + 5t < 19

Subtract 4 from both sides

5t < 19 - 4 or 5t < 15

Divide both sides by 5

t < 3

$$$$\frac{1}{4}(2x-3)=8$$\\
Multiply both sides by 4 $$2x - 3=32$$\\
Add 3 to both sides $$2x = 35$$
Divide both sides by 2 $$x=17\frac{1}{2}$$

I've edited my previous post to replace the matrix method with a simpler one (and hopefully, used the correct data at the same time!).

Write this as 

m = 7 - 3*n

n = -2:   m = 7 - 3*(-2) = 13

n = 0:    m = 7 - 3*0  = 7

n = 1:    m = 7 - 3*1  = 4