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

\(\displaystyle \frac{2}{3}x \cdot \frac{3}{2} = \frac{5}{2}\cdot\frac{3}{2}\)

The answer is:  \(\displaystyle \frac{15}{4}\)

Multiply both sides of the equation by \(\displaystyle x^{2}-2x\) to eliminate the fraction:

\(\displaystyle \frac{ x^{2}-4x}{x^{2}-2x} = 2\)

\(\displaystyle \frac{ x^{2}-4x}{x^{2}-2x}\cdot (x^{2}-2x) = 2 \cdot (x^{2}-2x)\)

\(\displaystyle x^{2}-4x = 2 x^{2}-4x\)

Subtract \(\displaystyle x^{2}-4x\) from both sides:

\(\displaystyle x^{2}-4x- (x^{2}-4x) = 2 x^{2}-4x - (x^{2}-4x)\)

\(\displaystyle 2 x^{2} - x^{2} -4x + 4x = 0\)

\(\displaystyle x^{2} = 0\)

The only possible solution is \(\displaystyle x = 0\), However, if this is substituted in the original equation, the expression at left is undefined, as seen here:

\(\displaystyle \frac{ x^{2}-4x}{x^{2}-2x} = 2\)

\(\displaystyle \frac{ 0^{2}-4 \cdot 0}{0^{2}-2 \cdot 0} = 2\)

\(\displaystyle \frac{ 0}{ 0} = 2\)

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

\(\displaystyle x^{2}-6x+9\) can be seen to fit the perfect square trinomial pattern:

\(\displaystyle x^{2}-6x+9 = x^{2} - 2 \cdot x \cdot 3 + 3 ^{2} = (x-3) ^{2}\)

The equation can therefore be rewritten as

\(\displaystyle \frac{15 }{ (x-3) ^{2}} - \frac{11}{x-3}+ 2= 0\)

Multiply both sides of the equation by the least common denominator of the expressions, which is \(\displaystyle LCM (x-3 , (x-3) ^{2}) = (x-3)^{2}\):

\(\displaystyle \left [\frac{15 }{ (x-3) ^{2}} - \frac{11}{x-3}+ 2\right ](x-3)^{2}= 0(x-3) ^{2}\)

\(\displaystyle \frac{15 }{ (x-3) ^{2}} \cdot (x-3)^{2} - \frac{11}{x-3} \cdot (x-3)^{2} + 2\cdot (x-3)^{2} = 0\)

\(\displaystyle 15 -11(x-3 )+ 2\cdot (x^{2} -6x+9 ) = 0\)

\(\displaystyle 15 -11x+33 + 2 x^{2} -12x+18 = 0\)

\(\displaystyle 2 x^{2} -23x +66= 0\)

This can be solved using the \(\displaystyle ac\) method. We are looking for two integers whose sum is \(\displaystyle -23\) and whose product is \(\displaystyle 2 \cdot 66 = 132\). Through some trial and error, the integers are found to be \(\displaystyle -12\) and \(\displaystyle -11\), so the above equation can be rewritten, and solved using grouping, as

\(\displaystyle 2 x^{2} -12x - 11x +66= 0\)

\(\displaystyle (2 x^{2} -12x) - (11x -66)= 0\)

\(\displaystyle 2x (x-6 ) - 11 (x-6 )= 0\)

\(\displaystyle (2x - 11 )(x-6 )= 0\)

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

Either: 

\(\displaystyle 2x-11 = 0\)

\(\displaystyle 2x-11+ 11 = 0 + 11\)

\(\displaystyle 2x = 11\)

\(\displaystyle \frac{2x }{2}= \frac{11}{2}\)

\(\displaystyle x = 5 \frac{1}{2}\)

Or:

\(\displaystyle x - 6 = 0\)

\(\displaystyle x - 6 + 6 = 0 + 6\)

\(\displaystyle x = 6\)

Both solutions can be confirmed by substitution; the solution set is \(\displaystyle \left \{ 5 \frac{1}{2} , 6 \right \}\).

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

\(\displaystyle -3x \times -\frac{1}{3} = \frac{1}{5} \times -\frac{1}{3}\)

The answer is:  \(\displaystyle -\frac{1}{15}\)

Add nine on both sides.

\(\displaystyle -6x-9 +9= 11+9\)

\(\displaystyle -6x = 20\)

Divide by negative six on both sides.

\(\displaystyle \frac{-6x }{-6}=\frac{ 20}{-6}\)

The answer is:  \(\displaystyle -\frac{10}{3}\)

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

\(\displaystyle 4x-7 = 0\)

\(\displaystyle 4x= 7\)

\(\displaystyle x =1 \frac{3}{4}\)

\(\displaystyle 3x^{2}-13x + 12 = 0\)

\(\displaystyle (3x-4)(x-3)= 0\)

Either 

\(\displaystyle x- 3= 0\)

\(\displaystyle x= 3\)

or 

\(\displaystyle 3x-4 = 0\)

\(\displaystyle 3x= 4\)

\(\displaystyle x= 1\frac{1}{3}\)

Since the numerator may be equal to 0, \(\displaystyle x =1 \frac{3}{4}\) is included as a solution; , since the denominator may not be equal to 0, \(\displaystyle x= 1\frac{1}{3}\) and \(\displaystyle x= 3\) are excluded as solutions.

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

\(\displaystyle \left ( -\infty, 1 \frac{1}{3} \right )\)

Let's test \(\displaystyle x= 0\):

\(\displaystyle \frac{4x-7}{3x^{2}-13x + 12}\le 0\)

\(\displaystyle \frac{4(0)-7}{3(0)^{2}-13 (0) + 12}\le 0\)

\(\displaystyle \frac{ -7}{ 12}\le 0\)

\(\displaystyle -\frac{ 7}{ 12}\le 0\)

This is true, so \(\displaystyle \left ( -\infty, 1 \frac{1}{3} \right )\) is included in the solution set.

\(\displaystyle \left ( 1 \frac{1}{3}, 1 \frac{3}{4}\right )\)

Let's test \(\displaystyle x = 1\frac{1}{2} = 1.5\):

\(\displaystyle \frac{4x-7}{3x^{2}-13x + 12}\le 0\)

\(\displaystyle \frac{4(1.5)-7}{3(1.5)^{2}-13(1.5) + 12}\le 0\)

\(\displaystyle \frac{6-7}{3(2.25)-13(1.5) + 12}\le 0\)

\(\displaystyle \frac{6-7}{6.75-19.5 + 12}\le 0\)

\(\displaystyle \frac{-1}{-0.75}\le 0\)

\(\displaystyle \frac{4}{3} \le 0\)

This is false, so \(\displaystyle \left ( 1 \frac{1}{3}, 1 \frac{3}{4}\right )\) is excluded from the solution set.

\(\displaystyle \left ( 1 \frac{3}{4}, 3 \right )\)

Let's test \(\displaystyle x = 2\):

\(\displaystyle \frac{4x-7}{3x^{2}-13x + 12}\le 0\)

\(\displaystyle \frac{4 (2)-7}{3(2)^{2}-13(2) + 12}\le 0\)

\(\displaystyle \frac{8-7}{3(4) -13(2) + 12}\le 0\)

\(\displaystyle \frac{1}{12 -26 + 12}\le 0\)

\(\displaystyle \frac{1}{ -2}\le 0\)

\(\displaystyle -\frac{1}{ 2}\le 0\)

This is true, so \(\displaystyle \left ( 1 \frac{3}{4}, 3 \right )\) is included in the solution set.

\(\displaystyle \left ( 3, \infty \right )\)

Let's test \(\displaystyle x = 4\):

\(\displaystyle \frac{4x-7}{3x^{2}-13x + 12}\le 0\)

\(\displaystyle \frac{4 (4)-7}{3 (4)^{2}-13 (4) + 12}\le 0\)

\(\displaystyle \frac{16-7}{3 (16) -13 (4) + 12}\le 0\)

\(\displaystyle \frac{16-7}{48 -52+ 12}\le 0\)

\(\displaystyle \frac{9}{8}\le 0\)

This is false, so \(\displaystyle \left ( 3, \infty \right )\) is excluded from the solution set.

The solution set is therefore \(\displaystyle \left ( -\infty, 1 \frac{1}{3} \right ) \cup \left [1 \frac{3}{4}, 3 \right )\).

Put the inequality in standard form, then 

\(\displaystyle x^{3}+ 3x^{2} > 9x + 27\)

\(\displaystyle x^{3}+ 3x^{2} - 9x -27 > 0\)

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

\(\displaystyle x^{3}+ 3x^{2} - 9x -27 = 0\)

\(\displaystyle x^{2} (x+3) - 9 (x+3) = 0\)

\(\displaystyle \left (x^{2}-9 \right )(x+3) = 0\)

\(\displaystyle (x-3) (x+3 )(x+3) = 0\)

\(\displaystyle x= 3\) or \(\displaystyle x = -3\).

Since the inequality is exclusive (\(\displaystyle >\)), 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:

\(\displaystyle (-\infty, -3)\)

Let's test \(\displaystyle x = -4\):

\(\displaystyle x^{3}+ 3x^{2} > 9x + 27\)

\(\displaystyle (-4)^{3}+ 3(-4)^{2} > 9(-4) + 27\)

\(\displaystyle -64+ 3(16) > -36+ 27\)

\(\displaystyle -64+ 48> -36+ 27\)

\(\displaystyle -16> -9\)

This is false, so \(\displaystyle (-\infty, -3)\) is excluded from the solution set.

\(\displaystyle (-3, 3)\)

Let's test \(\displaystyle x = 0\):

\(\displaystyle x^{3}+ 3x^{2} > 9x + 27\)

\(\displaystyle 0^{3}+ 3(0)^{2} > 9 (0) + 27\)

\(\displaystyle 0 > 27\)

This is false, so \(\displaystyle (-3, 3)\) is excluded from the solution set.

\(\displaystyle (3, \infty)\)

Let's test \(\displaystyle x = 4\):

\(\displaystyle x^{3}+ 3x^{2} > 9x + 27\)

\(\displaystyle (4)^{3}+ 3(4)^{2} > 9(4) + 27\)

\(\displaystyle 64+ 3(16) > 36+ 27\)

\(\displaystyle 64+ 48> 36+ 27\)

\(\displaystyle 112> 63\)

This is true, so \(\displaystyle (3, \infty)\) is included in the solution set.

The solution set is the interval \(\displaystyle (3, \infty)\).

Subtract \(\displaystyle 7x\) on both sides.

\(\displaystyle 2x-7x< 7x-6-7x\)

Simplify both sides.

\(\displaystyle -5x< -6\)

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

\(\displaystyle \frac{-5x}{-5}< \frac{-6}{-5}\)

The answer is:  \(\displaystyle x>\frac{6}{5}\)

The first thing we can do is clean up the right side of the equation by distributing the \(\displaystyle 5\), and combining terms:

\(\displaystyle -2x - 40 < 5(6 + x) + 7x\)

\(\displaystyle -2x - 40 < 30 + 5x + 7x\)

\(\displaystyle -2x - 40 < 12x+30\)

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.

\(\displaystyle -14x< 70\)

\(\displaystyle x>-5\)

First, we distribute the \(\displaystyle -4\) and then collect terms:

\(\displaystyle -4(4 + 7x) + x > -6x + 5\)

\(\displaystyle -16 - 28x + x > -6x + 5\)

\(\displaystyle -16 - 27x > -6x + 5\)

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

\(\displaystyle -21 > 21x\)

\(\displaystyle -1>x\)