solve the system y=ax+b, y=absolute value cx+d

Question

anonymous · Answer

You should identify the fact that $$y=|cx+d|=\sqrt{(cx+d)^2}$$ by definition.

anonymous · Answer

Just wait - I want to see if there's an easier way...but this way will work.

anonymous · Answer

I'll just keep going.  From the definition above, expand the radicand and square both sides to get rid of the square root.  Then$$y^2=c^2x^2+2cdx+d^2$$

anonymous · Answer

Now, square both sides of the other equation,$$y^2=a^2x^2+2abx+b^2$$

anonymous · Answer

scrap that - I always default to that method...easier to do the following...

anonymous · Answer

Case-by-case method:

$$y=|cx+d|=cx+d _.or_.-(cx+d)$$

anonymous · Answer

Then, you go about it like you would with other linear equations.  You want to find all x such that the same y is yielded.  This occurs when,$$ax+b=cx+d_.or_.ax+b=-cx-d$$

anonymous · Answer

Solving the first:$$x(a-c)=d-b \rightarrow x=\frac{d-b}{a-c}$$

anonymous · Answer

and solving the second,$$x(a+c)=-d-b \rightarrow x= -\frac{b+d}{a+c}$$

snickers · Answer

cool thank you.

anonymous · Answer

You're welcome...feel free to 'fan' me ;)