How many methods are there to solve modulus problems having a negative sign inside its equations?

Nousher

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May 11, 2021
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In a video, a person discussed how to solve modulus problems with a negative sign. This is the link of that video lecture.

He showed two methods to solve the problem. The first method is commonly used. Later he showed another method where he used a number line and a graph.

Unfortunately, I couldn't understand his explanation properly. What is the original name of the second method he discussed in that video? If I could learn the actual name of the second method, I could googled to learn more about that.
 
I have not heard a specific name for this method. It's a kind of graphical method that uses the knowledge that this equation is linear between (and outside) the critical points.

He graphs the left hand side of the equation. A graph is produced of "a" vs "y" where y=abs(a+4) - abs(a) - abs(a-3). The equation is LINEAR and so draws straight lines between the critical points. NOTE this method won't work for non-linear cases containing a^2 etc.

To the right and left he adds extra points to work out if the graph goes up or down in those regions.

When complete, it can be seen that the graph does not cross the line y=12 (corresponding to the right hand side of the original equation). Therefore there are no solutions.

IF the graph had crossed, or touched, the line y=12 then there would be a solution at that point.

If you want to use a graphing method with non-linear equations, then you'd have to find any max/min values between critical points and use knowledge of the "intermediate value theorem". But it would probably be easier to use the traditional method in this circumstance!
 
I have not heard a specific name for this method. It's a kind of graphical method that uses the knowledge that this equation is linear between (and outside) the critical points.

He graphs the left hand side of the equation. A graph is produced of "a" vs "y" where y=abs(a+4) - abs(a) - abs(a-3). The equation is LINEAR and so draws straight lines between the critical points. NOTE this method won't work for non-linear cases containing a^2 etc.

To the right and left he adds extra points to work out if the graph goes up or down in those regions.

When complete, it can be seen that the graph does not cross the line y=12 (corresponding to the right hand side of the original equation). Therefore there are no solutions.

IF the graph had crossed, or touched, the line y=12 then there would be a solution at that point.

If you want to use a graphing method with non-linear equations, then you'd have to find any max/min values between critical points and use knowledge of the "intermediate value theorem". But it would probably be easier to use the traditional method in this circumstance!
Thanks!
 
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