# How do you solve 5x + 7 > 17?

May 1, 2015

You can perform all basic operations (addition, subtraction, plus multiplication and division by values greater than zero) on both sides of an inequality without invalidating the inequality. Multiplying or dividing by values less than zero reverses the inequality (not an issue for this example)

$5 x + 7 > 17$

$5 x > 10 \text{ subtracting 7 from both sides}$

$x > 2 \text{ dividing both sides by 5}$

May 1, 2015

Solve a linear inequality (like this one) just like you would an equality (an equation) except for an inequality if you multiply or divide by a negative you have to change the direction of the inequality.
($5 < 9$ but $- 10 > - 18$)

So, to solve:

$5 x + 7 > 17$, subtract $7$ from both sides (or move the $7$ if that's how you think about equations).

$5 x > 17 - 7$

So $5 x > 10$

Now we need to divide by $5$. Since $5$ is positive, there's nothing more to think about:

$\frac{5 x}{5} > \frac{10}{5}$ Now simplify:

$x > 2$
I admit, $x > 2$ says "x is greater than 2", but it doesn't look like the picture on the number line. Thinking on the number line, it says "x is to the right of 2".
That is more clear to me if I write

$2 < x$

Solution set:
Set Builder form: $\left\{x | 2 < x\right\}$

Interval Notation: $\left(2 , \infty\right)$