+118608 Hi Zac,
Thanks for answering these questions :)
There is a couple of things here.
1) I think it should be $$10t^2+5t-2=0\qquad (NOT -5t)$$
2) I am not sure what the asker INTENDED the question to be BUT this was the question.
1/2+2/5t-1=1/5t+t
Technically this should be interpreted as
$$\\\frac{1}{2}+\frac{2}{5}t-1=\frac{1}{5}t+t\\\\
\frac{1}{2}+\frac{2t}{5}-1=\frac{1t}{5}+t\qquad $The t is on the top$\\\\
\frac{-1}{2}+\frac{2t}{5}=\frac{6t}{5}\\\\
\frac{-1}{2}=\frac{4t}{5}\\\\
\frac{-1*5}{2*4}=t\\\\
t=\frac{-5}{8}\\\\$$
Also, I would like you to consider putting in some blank lines.
Sometimes mine is spread out too much, I realize that, but I personally find it a bit hard to read when there are no spaces. This may just be a peronal preference. I am not sure :/
+980 1.
$${\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}}$$
$${\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{9}}$$
$${\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{15}}$$
+980 I'm not sure if it's right, but I ended up with a quadratic by the end. Here goes:
$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{1}} = {\frac{{\mathtt{1}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{t}}$$ Take the 1/5t off both sides.
$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{t}}$$ Multiply everything by 10t.
$${\frac{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}} = {\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$ Simplify.
$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}} = {\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$
$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}}$$ Now using the quadratic formula,
$$t=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ or $$t=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
a=10, b=-5, c=-2
$$t=\frac{5+\sqrt{5^2-4\times{10}\times{-2}}}{2\times{10}}$$ or $$t=\frac{5+\sqrt{5^2-4\times{10}\times{-2}}}{2\times{10}}$$
$$t=\frac{5+\sqrt{25+80}}{20}$$ or $$t=\frac{5-\sqrt{25+80}}{20}$$
$$t=\frac{5+\sqrt{105}}{20}$$ or $$t=\frac{5-\sqrt{105}}{20}$$
which works out to $$\approx$$ 0.76235 or -0.26235.
P.S. I don't know how to say "plus minus" something.
+118608 Hi Zac,
Thanks for answering these questions :)
There is a couple of things here.
1) I think it should be $$10t^2+5t-2=0\qquad (NOT -5t)$$
2) I am not sure what the asker INTENDED the question to be BUT this was the question.
1/2+2/5t-1=1/5t+t
Technically this should be interpreted as
$$\\\frac{1}{2}+\frac{2}{5}t-1=\frac{1}{5}t+t\\\\
\frac{1}{2}+\frac{2t}{5}-1=\frac{1t}{5}+t\qquad $The t is on the top$\\\\
\frac{-1}{2}+\frac{2t}{5}=\frac{6t}{5}\\\\
\frac{-1}{2}=\frac{4t}{5}\\\\
\frac{-1*5}{2*4}=t\\\\
t=\frac{-5}{8}\\\\$$
Also, I would like you to consider putting in some blank lines.
Sometimes mine is spread out too much, I realize that, but I personally find it a bit hard to read when there are no spaces. This may just be a peronal preference. I am not sure :/
