# Step-by-step Solution

## Solve the inequality $\frac{\left(x-2\right)^2}{2}+\frac{5x+6}{6}<\frac{\left(x+3\right)\left(x-3\right)}{3}+6$

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### Videos

$x>-5.1429$

## Step-by-step explanation

Problem to solve:

$\frac{\left(x-2\right)^2}{2}+\frac{5x+6}{6}<\frac{\left(x+3\right)\left(x-3\right)}{3}+6$
1

Grouping terms

$\frac{\left(x-2\right)^2}{2}+\frac{5x+6}{6}-\left(\frac{\left(x+3\right)\left(x-3\right)}{3}\right)<6$
2

The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$, where:

• The first term ($a$) is $x$.
• The second term ($b$) is $3$.
Then:

$\frac{\left(x-2\right)^2}{2}+\frac{5x+6}{6}-\left(\frac{x^2-1\left(3^2\right)}{3}\right)<6$

$x>-5.1429$
$\frac{\left(x-2\right)^2}{2}+\frac{5x+6}{6}<\frac{\left(x+3\right)\left(x-3\right)}{3}+6$