Algebra - Absolute Value $\to$ Standard form

Question

Problem. Given that $x < 5$, rewrite $5x - |x - 5|$ without using absolute value signs.

I get that for positive $x$, this is $4x-5$ (this is because $|x-5|=5-x$, thus simplifying to $5x-5-x=4x-5$), but how does one deal with the negative case?

Answer 1

Since the absolute value signs are around $x - 5$, it's this value, not $x$, which is important regarding it being positive, $0$ or negative. In particular, the definition of absolute value for real numbers is

$$|z| = \begin{cases} z & \text{ if } \; z \ge 0 \\ -z & \text{ if } \; z \lt 0 \end{cases} \tag{1}\label{eq1A}$$

This is true for $z$ being any real constant, or it being any real value expression, such as $x - 5$. Next, from what you're given, subtracting $5$ from both sides gives

$$x \lt 5 \implies x - 5 \lt 0 \tag{2}\label{eq2A}$$

Thus, using \eqref{eq1A} gives

$$|x - 5| = -(x - 5) = -x + 5 \tag{3}\label{eq3A}$$

The expression to rewrite then becomes

$$\begin{equation}\begin{aligned} 5x - |x - 5| & = 5x - (-x + 5) \\ & = 5x + x - 5 \\ & = 6x - 5 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Answer 2

Think of it bro. In any case for z<5, x-5 would turn into 5-x. Considering about the case, the brackets would be used because the negative sign is on the whole of x-5. Now the answer should look clear. By calculating, the answer would come 6x-5. Thanks for the cooperation. Sorry if I misunderstood where you are facing problem in the question.

Answer 3

If $t$ is a real number, $|t|$ is the distance of $t$ from zero on the real number line. Thus, $$|t| = \begin{cases} t & \text{if $t \geq 0$}\\ -t & \text{if $t < 0$} \end{cases} $$

Let $t = x - 5$. Observe that $x - 5 \geq 0 \iff x \geq 5$. Hence, \begin{align*} |x - 5| & = \begin{cases} x - 5 & \text{if $x \geq 5$}\\ -(x - 5) & \text{if $x < 5$}\\ \end{cases}\\ & = \begin{cases} x - 5 & \text{if $x \geq 5$}\\ -x + 5 & \text{if $x < 5$}\\ \end{cases} \end{align*}

Therefore, \begin{align*} 5x - |x - 5| & = \begin{cases} 5x - (x - 5) && \text{if $x \geq 5$}\\ 5x - (-x + 5) && \text{if $x < 5$}\\ \end{cases}\\ & = \begin{cases} 4x + 5 && \text{if $x \geq 5$}\\ 6x - 5 && \text{if $x < 5$} \end{cases} \end{align*}
Notice that you made an error in the case when $x - 5 > 0$.