Need help pls! @ganeshie8 @Jhannybean @Directrix @mathmath333
i got n=3 and m=-2
How did you get those answers? Using x and y-intercepts maybe as points on the graph, maybe?
please, notethat, your equation can be rewritten as below: \[y=mx+n\] if mx+n is greater or equal to 0 and \[y=-mx-n\] if mx+n is less than zero. Now from first case we have to solve this inequality: mx+n>=0, from your graph I see that the solution has to be x>0 because x=3/2. So the above inequality has solution if and only if m>0 because after i dived by m, I get: \[x \ge \frac{ -n }{ m }\] SImilar discussion is going for the second case, namely mx+n<0, or mx<-n which is possible if and only if m is positive, so we can get \[x <\frac{ -n }{ m }\] in conclusion, all that we can say is: \[\frac{ -n }{ m }=\frac{ 3 }{ 2 }, m>0\]
@MARC_
even if m<0 can be a solution too!
Using the points, (1.5, 0) and (0,3) from the graph, I set equations 0 = abs(m* (1.5) + n) 3 = abs(0 + n) ---------------- to solve simultaneously.
Realizing, of course, that absolute value is tricky when variables are involved.
right! If I use the point (0,3), I can stes completely the solution, because, now I know that n=3 and m=-2
oops ...I can state...
I have two sets of answers for (m,n) but I have not yet attempted to interpret them in the context of this problem. (m,n) ==== (-2, 3) (2, -3) I am not saying that they are correct. @MARC_ should take a look at our work and ideas.
@MARC_ posted these: >>i got n=3 and m=-2
please note that n=3 is the only solution acceptable, even if there is the other solution, namely n=-3. Nevertheless in the graph there is no line with n=-3
Right. That is exactly what I want @MARC_ to realize: the what and the why.
if this was not 'abs' function then you can see that the intercept would be -3, and the slope 2 This function becomes negative when 2x=3, i.e. x=1.5 so when the 'abs' is taken into account it becomes the graph oyu see
so @Michele_Laino I do not agree that n=3 is the correct soluyion
Thanks! @Directrix
plot y= 2x-3 then plot y = abs(2x-3)
abs(2x-3) = abs(-2x+3)
@MARC_ I got the same
This was my working... \[y=mx+n\] \[y=m(0)+n\]
\[y=n\]
\[n=3\]
\[y=mx+n\]
\[0=mx+n\] \[-n=mx\]
\[-(3)=m(\frac{ 3 }{ 2 })\]
\[-6=3m\] \[m=-2\]
but the real answer is m=2 and n=-3 @Directrix @MrNood @mathmath333 @Michele_Laino @triciaal
Thnx @Directrix @Michele_Laino @MrNood @mathmath333 @triciaal
Thanks! @MARC_
@MARC_ m=-2 n=3 is an equally valid solution - as shown above they produce exactly the same result
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