# How do you solve and write the following in interval notation: 8x -3x + 2< 2(x + 7)?

Jun 2, 2016

$\left(- \infty , 4\right)$

#### Explanation:

We have the inequality

$8 x - 3 x + 2 < 2 \left(x + 7\right)$

On the left hand side, the $8 x$ and $- 3 x$ can be combined:

${\overbrace{8 x - 3 x}}^{8 x - 3 x = 5 x} + 2 < 2 \left(x + 7\right)$

$5 x + 2 < 2 \left(x + 7\right)$

Next, on the right hand side, distribute the $2$:

$5 x + 2 < {\overbrace{2 \left(x + 7\right)}}^{2 \left(x\right) + 2 \left(7\right)}$

$5 x + 2 < 2 x + 14$

Subtract $2 x$ from both sides.

${\overbrace{5 x - 2 x}}^{3 x} + 2 < {\overbrace{2 x - 2 x}}^{0} + 14$

$3 x + 2 < 14$

Subtract $2$ from both sides.

$3 x + {\overbrace{2 - 2}}^{0} < {\overbrace{14 - 2}}^{12}$

$3 x < 12$

Divide both sides by $3$.

$\frac{3 x}{3} < \frac{12}{3}$

$x < 4$

Now, we need to express $x < 4$ in interval notation. Since $x$ has to be less than $4$, we know that $4$ is the upper bound of the interval.

Since there is no lower bound, $x$ can extend all the way down, from $4$, to $0$, into the negative numbers, working its way left on the number line all the way to negative infinity.

So, the interval is $\left(- \infty , 4\right)$. Note that ( and ) are used instead of [ and ] because $x < 4$, not $x \le 4$. Additionally, ( and ) are always used with $\infty$ because $x$ can never equal $\infty$.