How do we solve questions with two modulus/absolute functions: $| 2x + 3 | - | x - 1 | = 6$

Question

How would we do this. it would have been easier if it was | | + | | but here its "-"

Answer 1

The LHS is a piecewise linear, continuous function. If you evaluate it at the "corner points", you will know in which interval(s) there are solutions.

$$\begin{array}{}x&-\infty&-\frac32&1&\infty\\\hline f(x)&\infty&-\frac52&5&\infty\end{array}.$$

Hence, with $x>1$,

$$2x+3-(x-1)=6$$

and with $x<-\frac32$,

$$-(2x+3)-(1-x)=6.$$

Answer 2

Hints:

$|2x+3|=\begin{cases}2x+3, & 2x+3 \ge 0 \iff x \ge -\frac{3}{2} \\-2x-3, & 2x+3 < 0 \iff x < -\frac{3}{2}\end{cases}$

$|x-1|=\begin{cases}x-1, & x-1 \ge 0 \iff x \ge 1 \\-x+1, & x-1 < 0 \iff x < 1\end{cases} \\$

Now consider each of cases (total number of cases is 3) and solve the equation for every case...

Answer 3

Rewrite the equation as $|2x+3|=|x-1|+6$ and sketch the graphs of $y=|2x+3|$ and $y=|x-1|+6$. You will see that the two branches of the steeper V shape of the first graph each intersect once with the left (downward sloping) branch of the less steep second graph. This tells us that the solutions satisfy the simpler equation

$$|2x+3|=7-x$$

at which point squaring both sides to $4x^2+12+9=49-14x+x^2$ leads to

$$3x^2+26x-40=(3x-4)(x+10)=0$$

so the two solutions are $x=4/3$ and $x=-10$. (Alternatively, and possibly more simply, $2x+3=7-x$ gives $x=4/3$ while $-2x-3=7-x$ gives $x=-10$.)

Taking a graphical approach does not help with all problems, but it's nonetheless worth keeping in mind that some equations can be solved by interpreting their solutions as the points where two curves intersect, especially when the curves are relatively easy to draw.