## Snickers 5 years ago solve the system y=ax+b, y=absolute value cx+d

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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,$ax+b=cx+d_.or_.ax+b=-cx-d$

8. anonymous

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

9. anonymous

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

10. Snickers

cool thank you.

11. anonymous

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