First, find the LCD, which is 6(z-1). Then, multiply the entire equation by this: 

\(\displaystyle 1\cdot 6(z-1)+\frac{5}{z-1}6(z-1)< \frac{7}{6}\cdot 6(z-1)\)

This simplifies to:

\(\displaystyle 6z-6+30< 7z-7\)

\(\displaystyle 6z+24< 7z-7\)

\(\displaystyle 31< z\)

Now, make a number line with points at 1 and 31, and vertical lines in between zones: 

Let's test three points, one in each zone:

f(0): \(\displaystyle 1+\frac{5}{-1} = 1-5=-6 < \frac{7}{6}\)

f(10):\(\displaystyle 1+\frac{5}{10-1}=1+\frac{5}{9}=\frac{14}{9}=\frac{28}{18}\nless \frac{7}{6}\)

f(35): \(\displaystyle 1+\frac{5}{35-1}=1+\frac{5}{34}=\frac{39}{34}< \frac{7}{6}\)

Therefore, the inequality is true for values of z such that z < 1 or z > 31. 

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

\(\displaystyle \frac{3p-1}{p^2-1}\)\(\displaystyle =\frac{3p-1}{(p+1)(p-1)}\)

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y. 

\(\displaystyle \frac{3p-1}{(p+1)(p-1)}=\frac{X}{p+1}+\frac{Y}{p-1}\)

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD. 

\(\displaystyle 3p-1=X(p-1) + Y(p+1)\)

Looking back to our original fraction, note that the values p = 1 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let p=1.

\(\displaystyle 3(1)-1=X(1-1) + Y(1+1)\)

\(\displaystyle 2=2Y\)

\(\displaystyle 1=Y\)

Second, let p=-1

\(\displaystyle 3(-1)-1=X(-1-1) + Y(-1+1)\)

\(\displaystyle -4=-2X\)

\(\displaystyle 2=X\)

Now, substitute these values in for X and Y. 

\(\displaystyle \frac{X}{p+1}+\frac{Y}{p-1}=\frac{2}{p+1}+\frac{1}{p-1}\)

Therefore, \(\displaystyle \frac{3p-1}{p^2-1}=\frac{2}{p+1}+\frac{1}{p-1}\).

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

\(\displaystyle \frac{2b+1}{b^2+b}=\frac{2b+1}{b(b+1)}\)

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y. 

\(\displaystyle \frac{2b+1}{b(b+1)}=\frac{X}{b}+\frac{Y}{b+1}\)

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD. 

\(\displaystyle 2b+1=X(b+1)+Yb\)

Looking back to our original fraction, note that the values b = 0 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let b=0.

\(\displaystyle 0+1=X(0+1)+0\)

\(\displaystyle 1=X\)

Second, let b=-1

\(\displaystyle 2(-1)+1=X(-1+1)+Y(-1)\)

\(\displaystyle -2+1=-Y\)

\(\displaystyle -1=-Y\)

\(\displaystyle 1=Y\)

Now, substitute these values in for X and Y. 

\(\displaystyle \frac{X}{b}+\frac{Y}{b+1}=\frac{1}{b}+\frac{1}{b+1}\)

Therefore, \(\displaystyle \frac{2b+1}{b^2+b}=\frac{1}{b}+\frac{1}{b+1}\).

To solve, first square both sides of the equation to reverse the square-rooting of the binomials, then simplify:

\(\displaystyle (\sqrt{x + 1})^{2} = (\sqrt{2x + 2})^{2}\)

\(\displaystyle x + 1 = 2x + 2\)

Now solve for \(\displaystyle x\):

\(\displaystyle -2 + 1 = 2x - x\)

\(\displaystyle -1 = x\)

\(\displaystyle x = -1\)

The x-coordinate for the intersection point is \(\displaystyle x = -1\).  

Choose one of the two radical functions that compose the equation, and set the function equal to y.  The more simple a function is, the easier it is to use:

\(\displaystyle y = \sqrt{x + 1}\)

Now substitute \(\displaystyle x = -1\) into the function.  

\(\displaystyle y = \sqrt{-1 + 1}\)

 \(\displaystyle y = \sqrt{0}\)

\(\displaystyle y = 0\)

The y-coordinate of the intersection point is \(\displaystyle y = 0\).

The intersection point of the two radical functions is \(\displaystyle (-1, 0)\).

Now graph the two radical functions:

\(\displaystyle f(x) = \sqrt{x + 1}\),  \(\displaystyle g(x) = \sqrt{2x + 2}\)

In order to solve this equation, we need to isolate the radical. In order to do so, we subtract 3 from both sides which leaves us with: 

\(\displaystyle \sqrt{4x}=10\)

To get rid of the radical, we square both sides:

\(\displaystyle (\sqrt{4x})^{2}=10^{2}\)

the radical is then canceled out leaving us with \(\displaystyle 4x=100\)

We solve for \(\displaystyle x\) by dividing by 4:

\(\displaystyle x=25\)

In order to get rid of the radical, we square both sides: 

\(\displaystyle (\sqrt{6x+1})^{2}=7^{2}\)

Since the radical cancels out, we're left with \(\displaystyle 6x+1=49\)

Subtracting both sides by 1 gives us \(\displaystyle 6x=48\)

We then divide both sides by 6 to get \(\displaystyle x=8\). 

We begin by sqaring both sides of the equation.  On the left side, the square root simply disappears, while on the right side we square the term.

\(\displaystyle \small \small \small \small x+6=(x+4)^2\)

\(\displaystyle \small \small \small x+6=x^2+8x+16\)

We then set the left side equal to 0 by subtracting everything on that side.

\(\displaystyle \small \small \small 0=x^2+7x+10\)

We then factor

\(\displaystyle \small \small 0=(x+2)(x+5)\)

Therefore, \(\displaystyle \small x=-2\:or\:-5\)

With problems of this type, it is always wise to double check for any extraneous roots (answers that don't actually work for some reason).  However, in this case both answers work.  Since\(\displaystyle \small -2\) is the only option among our choices, we should go with it.

When dealing with a radical equation, do the inverse operation to isolate the variable. In this case, the inverse operation of a square root is to square the expression. Thus we square both sides to continue. This yields the following. 

\(\displaystyle x + 2 = 10^2\)

\(\displaystyle x + 2 = 100\)

\(\displaystyle x = 98\)

1) To remove the radicals, raise both sides of the equation to the second power:

     \(\displaystyle (\sqrt{3\cdot \sqrt{2x + 3}})^{2} = (\sqrt{5x - 6})^{2}\)

2) To remove the radical, raise both side of the equation to the second power:

     \(\displaystyle (3\cdot \sqrt{2x + 3})^{2} = (5x - 6)^{2}\)

     \(\displaystyle 9(2x + 3) = 25x^{2} - 30x - 30x + 36\)

3) Now simplify, write as a quadratic equation, and solve:

     \(\displaystyle 0 = 25x^{2} - 78x + 9\)

     \(\displaystyle 0 = (25x - 3)(x - 3)\)

     \(\displaystyle 25x - 3 = 0\)      \(\displaystyle x - 3 = 0\)

     \(\displaystyle 25x = 3\)              \(\displaystyle x = 3\)

     \(\displaystyle x = \frac{3}{25}, 3\)

4) Checking for extraneous solutions.

Plugging in \(\displaystyle x=3\)

\(\displaystyle \\ \sqrt{3\cdot \sqrt{2(3) + 3}} = \sqrt{3\cdot \sqrt{9}}\\ \\=\sqrt{3\cdot 3}\\=3\)

\(\displaystyle \sqrt{5(3)-6}=\sqrt{9}=3\)

Plugging in  \(\displaystyle x=\frac{3}{25}\)

\(\displaystyle \\ \sqrt{3\cdot \sqrt{2(3/25) + 3}} = \sqrt{3\cdot \sqrt{3.24}}\\ \\=\sqrt{3\cdot 1.8}\\=5.4\)

\(\displaystyle \sqrt{5(3/25)-6}=\sqrt{0.6-6}=\sqrt{-5.4}\)

Since the square root of negative 5.4 gives us an imaginary solution we conclude that the only real solution is x=3.

\(\displaystyle \frac{\sqrt{x}}{\sqrt{2}}=\sqrt{4x+13}\)

Square both sides to eliminate all radicals:

\(\displaystyle \frac{x}{2}=4x+13\)

Multiply both sides by 2:

\(\displaystyle x=2(4x+13)\)

\(\displaystyle x=8x+26\)

Combine and isolate x:

\(\displaystyle 7x=-26\)

\(\displaystyle x=\frac{-26}{7}\)