# How do you graph and solve |2x+3| <= 7?

Dec 4, 2015

$- 5 \le x < 2$

$x \in \left[- 5 , 2\right]$

#### Explanation:

Graphing

You can graph $| 2 x + 3 |$ first. How do you do this?

The graph of $| x |$ looks like this:

graph{|x| [-15, 15, -3, 12]}

To graph $| 2 x + 3 |$, we need to adapt the slope of the two "arms" and to shift the "elbow" appropriately.

• the slope is being described by the factor in front of $x$. Here, the slope is $2$.
• to find the "elbow", set $2 x + 3 = 0$ and solve for $x$. Thus, the "elbow" is at $\left(- \frac{3}{2} , 0\right)$

Thus, the graph of $| 2 x + 3 |$ is:

graph{|2x+3| [-15, 15, -3, 12]}

Now, the solution of the equation are all $x$ values for which the graph is underneath the line $y = 7$:

graph{(y - |2x+3|)(y-7) = 0 [-15, 15, -3, 12]}

So, you already see on the graph that the solution is $- 5 \le x < 2$ or $x \in \left[- 5 , 2\right]$.

How to find the same solution without the graph?

Solving

To solve the inequality, you need to evaluate the absolute value $| 2 x + 3 |$ .

Generally, for an absolute value, you have

$| x | = \left\{\begin{matrix}x & x \ge 0 \\ - x & x < 0\end{matrix}\right.$

In this case, you need to know for which $x$ the equation $2 x + 3 = 0$ has a solution.
We already know that this is the case for $x = - \frac{3}{2}$.

So, we have

$| 2 x + 3 | = \left\{\begin{matrix}2 x + 3 & x \ge - \frac{3}{2} \\ - \left(2 x + 3\right) & x < - \frac{3}{2}\end{matrix}\right.$

Thus, we need to consider the two cases:

1) $x \ge - \frac{3}{2}$:

$2 x + 3 \le 7$

$\iff x \le 2$

Don't forget to take a look at the condition $x \ge - \frac{3}{2}$ which needs to hold at the same time.
Here, the solution is $- \frac{3}{2} \le x \le 2$ or x in [-3/2; 2]

2) $x < - \frac{3}{2}$:

$- \left(2 x + 3\right) \le 7$

$\iff - 2 x - 3 \le 7$

$\iff - 2 x \le 10$

... divide by $- 2$ and don't forget to flip the inequality sign (this needs to be done if multiplying with or dividing by a negative number)...

$\iff x \ge - 5$

Don't forget to take a look at the condition $x < - \frac{3}{2}$ which needs to hold at the same time.
Thus, the solution for this case is $- 5 \le x < - \frac{3}{2}$ or x in [-5; -3/2)

In total, we have x in [-3/2; 2] or x in [-5; -3/2) which can be combined to

x in [-5; 2]