# steps to simplify a problem

• Dec 23rd 2009, 02:22 PM
integral
steps to simplify a problem
Hello, can someone show me the steps to simplifying:
\$\displaystyle \Delta y=(t+\Delta t)^2 - t^2 + 8(t+\Delta t)-8t\$
to
\$\displaystyle \Delta y=2t(\Delta t)\$\$\displaystyle +(\Delta t)^2\$\$\displaystyle +8\$\$\displaystyle \Delta\$\$\displaystyle t\$

• Dec 23rd 2009, 02:44 PM
pickslides
Hi there integral, all you need to do here is expand the expression and group the like terms.
• Dec 23rd 2009, 04:41 PM
integral
\$\displaystyle \Delta y=(t+\Delta t)^2 -t^2 +8(t+\Delta t)-8t\$
\$\displaystyle \Delta y=(t+\Delta t)(t+\Delta t) -t^2 +8(t+\Delta t)-8t\$
\$\displaystyle \Delta y=t^2+t\Delta t+ (\Delta t(t))+\Delta t^2 -t^2 +8(t+\Delta t)-8t\$
\$\displaystyle \Delta y=t^2+t\Delta t+ (\Delta t(t))+\Delta t^2 -t^2 +8\Delta t\$
\$\displaystyle \Delta y=t\Delta t+ t\Delta t+\Delta t^2 +8\Delta t\$
\$\displaystyle \Delta y=2(t\Delta t)+\Delta t^2 +8\Delta t\$

right? (Smirk) I feel so dumb now. lol
• Dec 23rd 2009, 04:49 PM
pickslides
Looks ok, I would do it like this.

\$\displaystyle \Delta y=(t+\Delta t)^2 - t^2 + 8(t+\Delta t)-8t\$

Grouping

\$\displaystyle \Delta y=[(t+\Delta t)^2 - t^2] + [8(t+\Delta t)-8t]\$

Using the difference of 2 squares in the first part and expanding the 2nd part we get

\$\displaystyle \Delta y=((t+\Delta t) - t)((t+\Delta t)+ t) + [8t+8\Delta t)-8t]\$

And simplifying

\$\displaystyle \Delta y=\Delta t (2t+\Delta t) + 8\Delta t\$

And expanding again

\$\displaystyle \Delta y=2t(\Delta t)+(\Delta t)^2 + 8\Delta t\$
• Dec 23rd 2009, 05:07 PM
integral
That works as well (Thanks for the idea).
Thank you :D