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It would be nice if you could clarify exactly what you're trying to find, but I'll give it my educated guess.

Isolate 'y' from everything else first to put it into slope-intercept form.

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{y}} = {\mathtt{1}}$$

$${\mathtt{\,-\,}}\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{y}}\right) = {\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}$$

$${\mathtt{y}} = {\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$

The slope is 3/4 and the y-intercept is -1/4

Now, just plug it in to check!

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}} = {\frac{{\mathtt{7}}}{{\mathtt{2}}}} = {\mathtt{3.5}}$$

Since the point (5,8) does not work with the equatino, (5,8) is not a point on the line.

Good answer Anon, only thing awry was a lack of formatting. Remember, we're taking a percent of money. Therefore, the formatted answer would be $2114.50

$${\mathtt{\,-\,}}{\left|-{\mathtt{23}}\right|}{\mathtt{\,-\,}}{\left|{\mathtt{\,-\,}}{\mathtt{21}}{\mathtt{\,-\,}}{\mathtt{17}}\right|}$$

Remember to simplify the numbers first.

$${\mathtt{\,-\,}}{\left|-{\mathtt{23}}\right|}{\mathtt{\,-\,}}{\left|-{\mathtt{38}}\right|}$$

Once all terms are simplified into a single number, take the absolute value of the numbers.

$${\mathtt{\,-\,}}{\mathtt{23}}{\mathtt{\,-\,}}{\mathtt{38}}$$

Subtract.

$${\mathtt{\,-\,}}{\mathtt{23}}{\mathtt{\,-\,}}{\mathtt{38}} = -{\mathtt{61}}$$

Are you just looking to simplify the expression? If so:

8 - 6(7 - 4t) + 4t = 8 - 6*7 + 6*4t + 4t = 8 - 42 +24t +4t = -34 + 28t

$${\mathtt{10\,572.51}}{\mathtt{\,\times\,}}{\mathtt{0.2}} = {\mathtt{2\,114.502}}$$

$$$$(-1+i)^7=(-1+i)^2(-1+i)^2(-1+i)^2(-1+i)$$\\
$$(-1+i)^2 = 1-2i+i^2=1-2i-1=-2i$$\\
Therefore $$(-1+i)^7=(-2i)^3(-1+i)=(-2)^3i^3(-1+i)=8i(-1+i)=-8-8i$$

What do you mean by solve here?  You can derive a power series for the expression (the first 7 terms of which are shown below - there are an infinite number of terms), but it doesn't look very interesting!

A = 20,000

1 scruple = 1.2959782 grams

4 because 40 / 10 = 4 if thats what you are looking for

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x, y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}={\mathtt{25}}\\
{\mathtt{x}}=\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right){\mathtt{\,\times\,}}{\mathtt{y}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{\mathtt{25}}}{{\mathtt{7}}}}\\
{\mathtt{y}} = {\frac{{\mathtt{150}}}{{\mathtt{7}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{3.571\: \!428\: \!571\: \!428\: \!571\: \!4}}\\
{\mathtt{y}} = {\mathtt{21.428\: \!571\: \!428\: \!571\: \!428\: \!6}}\\
\end{array} \right\}$$

well usually mass is translated to kilagrams so 1 kilagram would equal about 771.617917647
scruples, or 1000 grams......

1*584/25=23.36 because 1*584=584

and 584/25=23.36 but i dont know how you could put that as a fraction though....

Hey, it's not my fault you and your family is gonna be eaten by the lizard illuminati.

maths sucks no matter what you say. it doesnt change the facts. s***w maths

eh thats nothing new....

You missed the part about being sandwiched in between 2 pieces of bread and being served to school children.

oh no im gonna get my head........photochopped!!!!!!

its infinite