2
$\begingroup$

I can't solve this equation:

$$(xy + x'yz)(xz + x'y') = xyz$$

After applying distribution I got this:

$$xyz + yz + x'z = xyz$$

I can't find the answer and have been thinking for hours now.

$\endgroup$
  • $\begingroup$ Oh my god, facepalm I just scrapped both letters out of the equasion but leaved the rest in there, pretty retarded. Sorry for bothering you guys. $\endgroup$ – user3152069 Jan 11 '14 at 21:58
2
$\begingroup$

You should get four terms by distributivity:

\begin{align} (xy \vee \bar xyz)(xz \vee \bar x \bar y) &= xyxz \vee xy\bar x\bar y \vee \bar xyz xz \vee \bar xyz\bar x\bar y. \end{align}

Except from $xyz$ every term contains a variable and its negation, which results in $0$.

$\endgroup$
1
$\begingroup$

$(xy+x'yz)(xz+x'y')=xxyz+xx'yy'+xx'yzz+x'x'yy'z=xyz+0 \times 0+0 \times yz + x' \times 0 z = xyz$

$\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.