# How do you solve 10x-6<=4x+42 and graph the solution on a number line?

Oct 8, 2017

$x \le 8$

#### Explanation:

First, subtract $4 x$ from both sides.

$10 x - 6 - \textcolor{b l u e}{4 x} \le 4 x + 42 - \textcolor{b l u e}{4 x}$

6x - 6 le cancel(4x) + 42 - cancel(color(blue)(4x)

$6 x - 6 \le 42$

Next, add 6 to both sides.

$6 x - 6 + \textcolor{red}{6} \le 42 + \textcolor{red}{6}$

$6 x - \cancel{6} + \cancel{\textcolor{red}{6}} \le 48$

$6 x \le 48$

Finally, divide both sides by 6.

$\frac{6 x}{\textcolor{\lim e g r e e n}{6}} \le \frac{48}{\textcolor{\lim e g r e e n}{6}}$

$\frac{\cancel{6} x}{\cancel{\textcolor{\lim e g r e e n}{6}}} \le 8$

$x \le 8$

This is the solution! To graph it on a number line, first consider what the solution is telling you: "x is less than or equal to 8"

We need to include 8 in our solution, and everything less than 8. Therefore, we will draw a solid dot at $x = 8$ on our number line, and then shade in everything to the left of it. The number line will look like this:

Oct 8, 2017

See a solution process below:

#### Explanation:

First, add $\textcolor{red}{6}$ and subtract $\textcolor{b l u e}{4 x}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:

$- \textcolor{b l u e}{4 x} + 10 x - 6 + \textcolor{red}{6} \le - \textcolor{b l u e}{4 x} + 4 x + 42 + \textcolor{red}{6}$

$\left(- \textcolor{b l u e}{4} + 10\right) x - 0 \le 0 + 48$

$6 x \le 48$

Now, divide each side of the inequality by $\textcolor{red}{6}$ to solve for $x$ while keeping the inequality balanced:

$\frac{6 x}{\textcolor{red}{6}} \le \frac{48}{\textcolor{red}{6}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} x}{\cancel{\textcolor{red}{6}}} \le 8$

$x \le 8$

To graph this we draw a solid circle at $8$ on the number line. The circle is solid because the inequality operator contains an "or equal to" clause.

We will draw an arrow to the left of $8$ because the inequality operator also contains a "less than" clause: