# How do you solve 2x+3>7 or 2x+9>11?

##### 1 Answer
Dec 31, 2016

The solution set is $\left\{x | x > 1\right\}$.

#### Explanation:

For each of these inequalities, there will be a set of $x$-values that make them true. For example, it's pretty clear that large values of $x$ (like 1,000) work for both, and negative values (like -1,000) will not work for either.

Since we're asked to solve a "this OR that" pair of inequalities, what we'd like to know are all the $x$-values that will work for at least one of them. To do this, we solve both inequalities for $x$, and then overlap the two solution sets.

Inequality 1:

$2 x + 3 > 7 \text{ "=>" "2x>4" }$(subtract 3 from both sides)
$\textcolor{w h i t e}{2 x + 3 > 7} \text{ "=>" "x>2" }$(divide both sides by 2)

Inequality 2:

$2 x + 9 > 11 \text{ "=>" "2x>2" }$(subtract 9 from both sides)
$\textcolor{w h i t e}{2 x + 9 > 11} \text{ "=>" "x>1" }$(divide both sides by 2)

So we need to list all the $x$-values that satisfy either $x > 2$ or $x > 1$.

In this case, if an $x$-value is greater than 2, it will automatically be greater than 1. Thus, the solution set for $2 x + 3 > 7$ is a subset of the one for $2 x + 9 > 11$. That means, all we need to do here is list the solution set for $2 x + 9 > 11$, and we're done.

The solution set we need is simply "all $x$ such that $x$ is greater than 1", or (in math terms):

$\left\{x | x > 1\right\}$
or
$x \in \left(1 , \infty\right)$