Inequalities - Math

Two numbers of like sign have a positive quotient.

Therefore, has as its solution set the set of points at which and are both positive or both negative.

To find this set of points, we identify the zeroes of both expressions.

Since is nonzero we have to exclude ; is excluded anyway since it would bring about a denominator of zero. We choose one test point on each of the three intervals and determine where the inequality is correct.

Choose :

- True.

Choose :

- False.

Choose :

- True.

The solution set is

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, add 7 to each side.

Now, divide both sides by 2.

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, subtract 2from each side.

Now, divide both sides by 2.

To solve for an inequality, you solve like you would for a single variable expression and get by itself.

First, subtract from both sides to get,

.

Then divide both sides by and your final answer will be,

.

Simplify the left side.

The inequality becomes:

Divide by two on both sides.

The answer is:

Subtract on both sides.

Add 3 on both sides.

Divide by 7 on both sides.

The answer is:

First, find the zeroes of the numerator and the denominator. This will give the boundary points of the intervals to be tested.

;

Since the numerator may be equal to 0, and are included as solutions. However, since the denominator may not be equal to 0, is excluded as a solution.

Now, test each of four intervals for inclusion in the solution set by substituting one test value from each:

Let's test :

This is false, so is excluded from the solution set.

Let's test :

This is true, so is included in the solution set.

Let's test :

This is false, so is excluded from the solution set.

Let's test :

This is true, so is included in the solution set.

The solution set is therefore .

First, find the zeroes of the numerator and the denominator. This will give the boundary points of the intervals to be tested.

Either

or

Since the numerator may be equal to 0, is included as a solution; , since the denominator may not be equal to 0, and are excluded as solutions.

Now, test each of four intervals for inclusion in the solution set by substituting one test value from each:

Let's test :

This is true, so is included in the solution set.

Let's test :

This is false, so is excluded from the solution set.

Let's test :

This is true, so is included in the solution set.

Let's test :

This is false, so is excluded from the solution set.

The solution set is therefore .

Put the inequality in standard form, then

Find the zeroes of the polynomial. This will give the boundary points of the intervals to be tested.

or .

Since the inequality is exclusive (), these boundary points are not included.

Now, test each of three intervals for inclusion in the solution set by substituting one test value from each:

Let's test :

This is false, so is excluded from the solution set.

Let's test :

This is false, so is excluded from the solution set.

Let's test :

This is true, so is included in the solution set.

The solution set is the interval .

Start by finding the roots of the equation by changing the inequality to an equal sign.

Now, make a number line with the two roots:

Pick a number less than and plug it into the inequality to see if it holds.

For ,

is clearly not true. The solution set cannot be .

Next, pick a number between .

For ,

is true so the solution set must include .

Finally, pick a number greater than .

For ,

is clearly not the so the solution set cannot be .

Thus, the solution set for this inequality is .

Subtract on both sides.

Simplify both sides.

Divide by negative five on both sides. This requires switching the sign.

The answer is:

First, we distribute the through the equation:

Now, we collect and combine terms:

The first thing we can do is clean up the right side of the equation by distributing the , and combining terms:

Now we can combine further. At some point, we'll have to divide by a negative number, which will change the direction of the inequality.

First, we distribute the and then collect terms:

Now we solve for x, taking care to change the direction of the inequality if we divide by a negative number:

In order to solve this inequality, we need to apply each mathematical operation to all three sides of the equation. Let's start by subtracting from all the sides:

Now we divide each side by . Remember, because the isn't negative, we don't have to flip the sign:

Add 26 on both sides.

Divide by two on both sides.

The answer is:

Distribute the negative through the terms of the binomial.

Subtract on both sides.

Add 18 on both sides.

Divide by 13 on both sides.

The answer is:

The first thing we can do is distribute the and the into their respective terms:

Now we can start to simplify by gathering like terms. Remember, if we multiply or divide by a negative number, we change the direction of the inequality:

It can be tricky because there's a negative sign in the equation, but we never end up multiplying or dividing by a negative, so there's no need to change the direction of the inequality. We simply divide by and multiply by :

Remember, anything we do to one side of the inequality, we must also do to the other two sides. We can start by adding one to all three sides:

And now we divide each side by two: