Solve the equation $|x-7|-3|2x+1|=0$ This equation is very unfamiliar with me, I never seen things like that because I always solved equations of the form $|\text{something}|=\text{things}$ but never seen equations that look like $|\text{something}|=|\text{things}|$. So if I learn how to solve it I will be able to solve questions that looks like it. Thank you;
 A: You can turn it into form
$$\left|\dfrac{x-7}{2x+1}\right | = 3$$
(assuming $2x + 1 \neq 0$).
A: An alternative method would be to let $f(x)=|x-7|-3|2x+1|$ and then writing $f(x)$ without the absolute value using $3$ different cases. Then solving the equation $f(x)=0$ using what we did earlier.
A: Consider the cases $x<-\frac12,-\frac12\le x<7,x\ge 7$.
Case I: $x<-\frac12$
$7-x-3(-2x-1)=0$
$7-x+6x+3=0$
$5x+10=0$
$x=-2$
Case II: $-\frac12\le x<7$
$7-x-3(2x+1)=0$
$7-x-6x-3=0$
$4-7x=0$
$x=\frac47$
Case III: $x\ge 7$
$x-7-3(2x+1)=0$
$x-7-6x-3=0$
$-5x-10=0$
$x=-2$
Hence there is no solution for $x$.
Therefore, $x=-2$ or $x=\frac47$.
A: $$
x-7 = \pm3(2x+1)
$$
So choose "$+$" and find the solutions, and then choose "$-$" and find the solutions.
A: $x-7=3(2x+1)$ or $x-7=-3(2x+1)$
if $x-7=3(2x+1)$, then $x=-2$.
if $x-7=-3(2x+1)$, then $x=-\frac{10}7$. drop it
A: |x-7|=|6x+3|
then either 
x-7=6x+3
or 
x-7=-6x-3
solve these two equations.
A: One way to handle this kind of equation is to notice that any solution to that equation is a solution to one of
$$
\frac{x-7}{2x+1}=\pm3
$$
then solve each equation and toss answers that don't fit.
$$
\frac{x-7}{2x+1}=3\implies x=-2
$$
and
$$
\frac{x-7}{2x+1}=-3\implies x=\frac47
$$
Both of these fit the original equation and are therefore all of the solutions.
Note that if $2x+1=0$ we have $x-7=-\frac{15}{2}$ so the equation is not satisfied if the denominator is $0$.
A: $|x-7|-3|2x+1|=0\qquad$ iff $\qquad|x-7|=3|2x+1|\qquad$ iff $\qquad|x-7|^2=3^2|2x+1|^2$
iff $\qquad(x-7)^2=3^2(2x+1)^2\qquad$ iff $\qquad\cdots$
A: $$ |x-7|-3|2x+1|=0 $$
Here's a trick that avoids cases.
\begin{align}
   |x-7|-3|2x+1| &= 0 \\
   |x-7| &= 3|2x+1| \\
   (|x-7|)^2 &= 9(|2x+1|)^2 \\
   x^2-14x + 49 &= 36x^2 +36x + 9 \\
   35x^2+50x-40 &= 0 \\
   7x^2+10x - 8 &= 0 \\
   (7x - 4)(x + 2) &= 0 \\
   x &\in \left\{ \dfrac 47, -2  \right\}
\end{align}
