Sasini

why??? do u think am i wrong?

what's ur problem. tell me

I think it isn't boring.beacuse it is a very simple subject. plz don't think about that

$${\frac{{\mathtt{4}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{8}}}}$$

$${\frac{\left({\mathtt{32}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{40}}}}$$

$${\frac{{\mathtt{37}}}{{\mathtt{40}}}}$$

$${\left({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{2}}{i}\right)}^{{\mathtt{2}}}$$

$${\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{12}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}\left[{{i}}^{{\mathtt{2}}}\right]$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{y}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}\\ {\mathtt{y}}={\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\ {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}}={\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\ {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}={\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\\ {\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}={\mathtt{5}}\\ {\mathtt{x}}={\mathtt{1}}\end{array}\right)}$$

(X+3)^2 $${{\mathtt{X}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{X}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}}$$

it is true. we can write it anyway.

$${\frac{{\mathtt{x}}}{{\mathtt{8}}}} = {\mathtt{2}}$$

$${\frac{{\mathtt{x}}}{{\mathtt{8}}}}{\mathtt{\,\times\,}}{\mathtt{8}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{8}}$$

Now we solve this formula easily

x =16

This is the answer to this "x" variable

$${\frac{{\mathtt{x}}}{{\mathtt{8}}}} = {\mathtt{2}}$$

$${\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{8}}$$

$${\mathtt{x}} = {\mathtt{16}}$$

first solve the left side hand . bt, it's similar to right side hand