High School Exam Question: Straight Line Intersecting with Absolute Value Line(s), and Discriminants Here's a question I've encountered in a recent high school examination.  
Find the range of values of m such that the line $y=mx-3$ intersects with the graph of  $ y=2-|3x - 5|$ at exactly two points.
The answer is $-3 < m < 3$.  
Picture
The suggested method to solve this by the exam setter is:
1. Sketch the absolute value line(s).
2. Calculate the two gradients of the absolute value line(s).
3. Use logic to determine the range of values in which the straight line intersects with both absolute value line(s).  
Picture of another method
However, I tried using another method.
1. Simultaneously solving
2. Turning them into a single quadratic equation
3. Using the discriminant to find 2 intersections  
But I'm not getting the correct answer. What went wrong?  
I also tried splitting $y = 2-|3x - 5|$ into 2 different equations, $y=2 - 3x + 5$ and $y=2 + 3x - 5$, but that wasn't too successful either.Failed attempt
I have no idea why both methods are wrong and am confused.
 A: You want to find values of $m$ such that the line $y=mx-3$ intersects 
with the graph of  $ y=2-|3x - 5|$ at exactly two points.  
Note that for $x\le\dfrac53, y=2+3x-5=3x-3,$ so $y=mx-3$ intersects $y=3x-3$ only when $x=0\le\dfrac53$ (unless $m=3$, in which case there are infinitely many intersection points). 
Thus (when $m\ne3)$ we have one intersection point when $x\le\dfrac53,$ so we want exactly one intersection point when $x>\dfrac53$.  When $x>\dfrac53$, $y=2-3x+5=7-3x,$ and this intersects $y=mx-3$ when $x=\dfrac{10}{m+3}$ (unless $m+3=0$, in which case there is no intersection), as you correctly calculated.    Now we want $\dfrac{10}{m+3}\gt\dfrac53;$ i.e., $\dfrac6{m+3}>1.$
This happens when $m+3>0$ and $m+3<6$; i.e., $-3<m<3$.
A: Hint: You have to distinguish the cases:
$x\geq \frac{5}{3}$ and you get the function $$y=2-3x+5=-3x+7$$
And in the other case: $x<\frac{5}{3}$ then we have $$y=2+3x-5=3x-3$$
A: I'd first change the coordinate system doing $y\to y+2$, so the problem is to determine $m$ in such a way that the line $y=mx-5$ intersects with the graph of $y=-|3x-5|$.
This becomes finding the intersections
\begin{cases}
y=mx-5 \\
y^2=(3x-5)^2 \\
y\le0
\end{cases}
The resolvent becomes $m^2x^2-10mx+25=9x^2-30x+25$, so
$$
(m^2-9)x^2+(30-10m)x=0
$$
For $m=3$, the equation becomes $0=0$: the line $y=3x-5$ indeed includes infinitely many points of the given graph.
For $m\ne3$, there is no need to find the discriminant, because the equation becomes $(m-3)(m+3)x^2-10(m-3)x=0$; in the case $m=-3$ the only solution is $x=0$, otherwise the solutions are $x=0$ and $x=10/(m+3)$.
Note that we can so eliminate the cases $m=3$ and $m=-3$ from consideration.
For $x=0$ you find $y=-5$, which satisfies the requirement. For $m\ne\pm3$, the second intersection is at
$$
x=\frac{10}{m+3}
$$
and we get
$$
y=\frac{10m}{m+3}-5=\frac{10m-5m-15}{m+3}=5\frac{m-3}{m+3}
$$
which is $\le0$ if and only if $-3<m<3$ (remember we removed the cases $m=\pm3$ beforehand).
A: $$mx-3 = 2-|3x-5|\implies |3x-5|^2 = (5-mx)^2$$
So $$x^2(9-m^2)-10x(3-m)=0$$ So $x=0$ is always a soluton. Say $x\ne 0$ then we get
$$(3+m)\Big(x(3-m)-10x\Big)=0$$ 
If $m=-3$ we have no solution if $m=3$ every $x$ is a solution. 
So if $m\neq \pm 3$ $$x ={10\over 3+m}\neq 0$$ and thus $2$ solution. 
Now you have check when this value fits starting equation:
$$\Big|{3-m\over 3+m}\Big| ={3-m\over 3+m}$$
So $${3-m\over 3+m}\geq 0\implies m\in (-3,3)$$
