How do you solve #9-2x \le 3 or 3x+10 \le 6-x#?

Question

How do you solve #9-2x \le 3 or  3x+10 \le 6-x#?

Nathan Axel · Accepted Answer

See explanation

We have two conditions that are combined to define the limit of values that may be assigned to #x#

Condition 1: #9-2x<=3#
Condition 2: #2x+10<=6-x#

Consider condition 1

Add #2x# to both sides

#9color(white)("dd")ubrace(-2x+2x)<=3+2x#

#9 color(white)("ddd")+ 0color(white)("dddd")<=2x+3#

Subtract 3 from both sides

#9-3<=2x + 0#

Divide both sides by 2

#(9-3)/2<=2/2color(white)(.)x#

#3<=x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider condition 2

Add #x# to both sides

#3x+10<=6#

Subtract 10 from both sides

#3x<=-4#

Divide both sides by 3

#x<=-4/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Combining these we have:

#-4/3 >= x>=3#

In other words #x# does not take on any values between and excluding #-4/3 and 3#

Ethan Adkins · Answer

undefined To solve $9 - 2x \leq 3$ or $3x + 10 \leq 6 - x$, we solve each inequality separately and then find the union of their solutions.

For $9 - 2x \leq 3$:
$$
\begin{align*}
9 - 2x & \leq 3 \
-2x & \leq 3 - 9 \
-2x & \leq -6 \
x & \geq \frac{-6}{-2} \
x & \geq 3
\end{align*}
$$

For $3x + 10 \leq 6 - x$:
$$
\begin{align*}
3x + 10 & \leq 6 - x \
3x + x & \leq 6 - 10 \
4x & \leq -4 \
x & \leq \frac{-4}{4} \
x & \leq -1
\end{align*}
$$

So, the solution to the compound inequality is $x \geq 3$ or $x \leq -1$.