To isolate the x-variable, multiply both sides by the coefficient of the x-variable.

The answer is:  

Multiply both sides of the equation by  to eliminate the fraction:

Subtract  from both sides:

The only possible solution is , However, if this is substituted in the original equation, the expression at left is undefined, as seen here:

An expression with a denominator of 0 has an undefined value, so this statement is false. The equation has no solution.

 can be seen to fit the perfect square trinomial pattern:

The equation can therefore be rewritten as

Multiply both sides of the equation by the least common denominator of the expressions, which is :

This can be solved using the  method. We are looking for two integers whose sum is  and whose product is . Through some trial and error, the integers are found to be  and , so the above equation can be rewritten, and solved using grouping, as

By the Zero Product Principle, one of these factors is equal to zero:

Either: 

Or:

Both solutions can be confirmed by substitution; the solution set is .

To solve for x, multiply by negative one-third on both sides.

The answer is:  

Add nine on both sides.

Divide by negative six 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.

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 .

Subtract  on both sides.

Simplify both sides.

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

The answer is:  

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: