How to solve 1-(x-5)/(9) < 9+x

Q: How to solve 1-(x-5)/(9) < 9+x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 1-(x-5)/(9) < 9+x | Equatix

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Question

$$1- \frac{ x-5 }{ 9 } < 9+x$$

Solve for x

$x>-\frac{67}{10}$

Solution Steps

Multiply both sides of the equation by $9$. Since $9$ is positive, the inequality direction remains the same.

$$9-\left(x-5\right)<81+9x$$

To find the opposite of $x-5$, find the opposite of each term.

$$9-x-\left(-5\right)<81+9x$$

The opposite of $-5$ is $5$.

$$9-x+5<81+9x$$

Add $9$ and $5$ to get $14$.

$$14-x<81+9x$$

Subtract $9x$ from both sides.

$$14-x-9x<81$$

Combine $-x$ and $-9x$ to get $-10x$.

$$14-10x<81$$

Subtract $14$ from both sides.

$$-10x<81-14$$

Subtract $14$ from $81$ to get $67$.

$$-10x<67$$

Divide both sides by $-10$. Since $-10$ is negative, the inequality direction is changed.

$$x>-\frac{67}{10}$$

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