how to solve x,y in R, |x-2|+|2x+y-4|=0
that is clearly, x=2,y=0.
I try to solve in maxima
eq1:x-2=0;
eq2:2*x+y-4=0;
solve(eq1);
(%o1) x-2=0
(%i2)
(%o2) y+2*x-4=0
(%o3) [x=2]
how to use the solve of x and input eq2 to solve y=0?
> From: maxima-request at math.utexas.edu
> Subject: Maxima Digest, Vol 39, Issue 30
> To: maxima at math.utexas.edu
> Date: Sat, 10 Oct 2009 12:00:05 -0500
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> Today's Topics:
>
> 1. Further code to improve the assume database (Dieter Kaiser)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Sat, 10 Oct 2009 17:43:53 +0200
> From: Dieter Kaiser <drdieterkaiser at web.de>
> Subject: Further code to improve the assume database
> To: "maxima at math.utexas.edu" <maxima at math.utexas.edu>
> Message-ID: <1255189433.3499.2.camel at dieter-laptop>
> Content-Type: text/plain
>
> We have the known bug ID: 1045920 "a>1 and b>1, is a+b>2?"
>
> (%i2) assume(a>1,b>1);
> (%o2) [a > 1,b > 1]
>
> (%i3) sign(a+b-2);
> (%o3) pnz
>
> I have a routine sign-shift which will correct this result:
>
> (%i4) load("sign-shift.lisp");
> (%o4) "sign-shift.lisp"
>
> (%i5) sign(a+b-2);
> (%o5) pos
>
> It will work for a lot of more cases too:
>
> (%i8) assume(a > 1,b > 1,c > 2)$
>
> (%i2) is(b+a > 2)
> (%o2) true
>
> (%i4) is(b+2*a > 3)
> (%o4) true
>
> (%i6) is(c^2+b+2*a > 7)
> (%o6) true
>
> It works for negative bounds too:
>
> (%i8) assume(x < -1,y < -2,z < -2)$
>
> (%i10) is(y+x < -3)
> (%o10) true
>
> (%i12) is(y+2*x < -4)
> (%o12) true
>
> z^2 is positive and the following sign is unknown:
>
> (%i14) is(z^2+y+2*x < -7)
> (%o14) unknown
>
> z^3 is negative and the test will give true again:
>
> (%i16) is(z^3+y+2*x < -12)
> (%o16) true
>
> I have no problems with the testsuite and the share_testsuite with the
> exception of one example:
>
> ********************** Problem 40 ***************
> Input:
> fourier_elim(fourier_elim(eqs, [y, x, z]), [z, y, x])
>
>
> Result:
> [0 < z, z < min(4, - y - x + 4), 0 < y, y < 1, 0 < x, x < 1]
>
> This differed from the expected result:
> [0 < z, z < 4 - x - y, 0 < y, y < 1, 0 < x, x < 1]
>
> 124/125 tests passed (not counting 5 expected errors).
>
>
> I know nothing about the code of fourier_elim. My question is, if the
> new result is really an error or if we can accept it. Perhaps
> fourier_elim can be improved to take advantage of the extended
> functionality of $sign?
>
> This is the code of sign-shift:
>
> (defun sign-shift (expr)
> (do ((l (append (cdr (facts1 '$learndata)) (cdr (facts1 $context)))
> (cdr l))
> (e expr)
> (flag) (fact) (num))
> ((null l) (if flag ($expand e) expr))
> (setq fact (car l))
> (when (eq (caar fact) 'mgreaterp)
> ;; Do we have something like a>2 or a<-1 on the list of facts?
> (cond ((and (symbolp (cadr fact))
> (not ($freeof (cadr fact) e))
> (not (member (cadr fact) '($%pi $%e $%gamma $%phi)))
> (mnump (setq num (caddr fact)))
> (not (zerop1 num)))
> (setq flag t)
> (if (mminusp num) (setq num (mul -1 num)))
> (setq e (maxima-substitute (add (cadr fact) num) (cadr
> fact) e)))
> ((and (symbolp (caddr fact))
> (not ($freeof (caddr fact) e))
> (not (member (caddr fact) '($%pi $%e $%gamma $%phi)))
> (mnump (setq num (cadr fact)))
> (not (zerop1 num)))
> (setq flag t)
> (if (mminusp num) (setq num (mul -1 num)))
> (setq e (maxima-substitute
> (sub (caddr fact) num) (caddr fact) e)))))))
>
>
> The routine sign-shift has to be called in the routine signsum in
> compar.lisp:
>
> (defun signsum (x)
> (setq x (sign-shift x))
> ;; x might be simplified to an atom in sign-shift
> (when (atom x) (setq x (cons '(mplus) (list x))))
>
>
> Dieter Kaiser
>
>
>
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> End of Maxima Digest, Vol 39, Issue 30
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