topic graphing linear equations? in PTC Mathcad
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398605#M156505
<HTML><HEAD></HEAD><BODY>O.K.? well...we did something like...get rid of x and somehow find y? i don't know..but, what about a problem such as... 2x-5y=12?...i just made that up! so i'm not sure if it will work?</BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMTangels-disabled2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398603#M156503
??? how do you do that? here's an example:8x-y=16 the only thing i think i know is that...i think you only get one point and it wil just make one straight line? we're having a test on it i think next friday....but i have work on it due tomorrow!Fri, 04 May 2018 06:00:56 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398603#M156503angels-disabled2018-05-04T06:00:56Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398604#M156504
<HTML><HEAD></HEAD><BODY>In the case of 8x-y=16, the ordered pair (3,8) satisfies the equation. Can you find another pair that also satisfies it? You then plot the two points and draw the line.<BR />PKA<BR /></BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398604#M156504pkalford-disabl2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398605#M156505
<HTML><HEAD></HEAD><BODY>O.K.? well...we did something like...get rid of x and somehow find y? i don't know..but, what about a problem such as... 2x-5y=12?...i just made that up! so i'm not sure if it will work?</BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398605#M156505angels-disabled2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398606#M156506
<HTML><HEAD></HEAD><BODY><B><I>Look</I></B>, ordered pairs are points in the plane. Lines are collections of <B>points</B> in the plane. These points, (p,q), satisfy the equations of lines, ax+by=c, that is a*p+b*q=c. All I have to do is find two points (u,v) & (t,s) such that a*u+b*x=c & a*t+b*s=c. Two points determine a line! The way we find these points is <B>not fixed</B>. So you may solve for one of the variables, either x or y. But, this is not necessary! To <B>graph</B> a line, all we need is <B>two points</B>.<BR />PKA<BR /></BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398606#M156506pkalford-disabl2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398607#M156507
<HTML><HEAD></HEAD><BODY>??? how do you do that? here's an example:<BR />8x-y=16 the only thing i think i know is that...i think you only get one point and it wil just make one straight line? we're having a test on it i think next friday....but i have work on it due tomorrow! <BR /><BR />your question:<BR /> first you isolate the "y" to get it by itself ( almost )<BR /><BR />8x-y=16<BR /> +y +y<BR />8x=16+y<BR /><BR />subtract 16<BR /><BR />8x-16=y <BR />then you just pick points:<BR />lets say you want s to be 3<BR />substitute:<BR /><BR />8(3)-16=y<BR />24-16=y<BR />8=y<BR />so your point is (3,8)<BR /></BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398607#M156507jackal_1500-dis2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398608#M156508
<HTML><HEAD></HEAD><BODY>sorry, jackal_1500<BR />this is from Laura</BODY></HTML>Thu, 07 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398608#M156508jackal_1500-dis2002-11-07T08:00:00Zgraphing linear equations?
https://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398609#M156509
<HTML><HEAD></HEAD><BODY>Angel,<BR /><BR />When you mentioned you were confused about your teacher asking you to get rid of the x term, I think you were remembering about the y-intercept, which is where a line crosses the y-axis. At that point x = 0.<BR /><BR />So in your equation, 8x - y = 16<BR />Set x = 0.<BR /><BR />8*0 - y = 16<BR />0 - y = 16<BR /><BR />- y = 16<BR /><BR />y = -16<BR /><BR />So one point on your line is (0,-16).<BR />Now you have two points, with the one found earlier.<BR /><BR />For any line, you can try a value for x in the equation, then find a matching value for y to find a point on the line.<BR /><BR />Mona</BODY></HTML>Fri, 08 Nov 2002 08:00:00 GMThttps://community.ptc.com/t5/PTC-Mathcad/graphing-linear-equations/m-p/398609#M156509mzeftel2002-11-08T08:00:00Z