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# College Algebra Solver

College algebra covers a range of topics like linear equations, quadratic equations, matrices, logarithms and polynomials. It is preferred that college level students have a sound understanding of algebra 1 and algebra 2. But even the best of us stumble at times, so there is no reason to let your apprehensions about algebra close that path for you. Help is available to all who ask and its much closer than you think. Free online college algebra solvers are readily available and cover all the topics that are covered in college algebra courses. Starting with basic refresher lessons and moving on to more advanced topics, free college algebra solvers have proven to be a great help to college students everywhere.

## Free College Algebra Solver

So how exactly do college algebra problem solvers work and what can you expect from them? One of the best things about them is that you can start from whichever level of algebra you are comfortable with. Most online college algebra solvers cover high school algebra as well so that you can go back from time to time to anything that needs more clarity. Free online college algebra solvers have step by step solutions to your problems. College algebra equation solvers give step by step solutions to each problem. They also explain how each step leads to the next and what operations or rules were involved. Both college algebra equation solvers and college algebra word problem solvers provide practice questions and worksheets, mock test and exam prep help to make learning Algebra easy.

### Solved Examples

Question 1: The sum of the square of the two consecutive positive even integers is 164. Find the integers.
Solution:
Let two consecutive even integers, x and x + 2

Step 1:

sum of the square of the two consecutive positive even integers = 164

=> $x^2$ + $(x + 2)^2$ = 164

=> $x^2$ + $x^2$ + $2^2$ + 2 * $x$ * 2 = 164

=> 2$x^2$ + 4 + 4x = 164

=> 2$x^2$ + 4x = 164 - 4

=> 2$x^2$ + 4x = 160

=> 2$x^2$ + 4x - 160 = 0

Step 2:

Solve for x, 2$x^2$ + 4x - 160 = 0

=> 2$x^2$ + 20x - 16x - 160 = 0

=> 2x(x + 10) - 16(x + 10) = 0

=> (2x - 16)(x + 10) = 0

Step 3:

either 2x - 16 = 0   or    x + 10 = 0

=> 2x = 16     or    x = - 10

Since integers are consecutive positive, so neglect x = -10

=> 2x = 16

=> x = 8, first integer

and second integer = 8 + 2 = 10

Hence consecutive positive integer are, 8 and 10. answer

Question 2: Simplify the expression
$\frac{4i}{5 - i}$ * $\frac{3 - i}{5 + i}$
Solution:
Given $\frac{4i}{5 - i}$ * $\frac{3 - i}{5 + i}$

Step 1:
Multiply numerators and denominators of both the fractions

=> $\frac{4i}{5 - i}$ * $\frac{3 - i}{5 + i}$ = $\frac{(4i)(3 - i)}{(5 - i)(5 + i)}$

= $\frac{4i * 3 - 4i * i}{5^2 - i ^2}$

= $\frac{12 i - 4 i^2}{25 + 1}$

[(a - ib)(a + ib) = $a^2$ + $b^2$]

= $\frac{12 i + 4}{26}$

[$i^2$ = - 1]
Step 2:

Reduce the fraction

=> $\frac{12 i + 4}{26}$ = $\frac{2(6 i + 2)}{2 * 13}$

= $\frac{6 i + 2}{13}$

or $\frac{ 2 + 6i}{13}$

=> $\frac{4i}{5 - i}$ * $\frac{3 - i}{5 + i}$ = $\frac{2 + 6i}{13}$

Question 3: The ratio of land and water area in the earth is 1 : 2. In the northern hemisphere this ratio is 2 : 3. Find the land and water area in the southern hemisphere.

Solution:
Let the area of each hemisphere be x.

Therefore total area of the earth is 2x.

Area of land in the northern hemisphere = $\frac{2}{2 + 3}$ * x = $\frac{2}{5}$x

and area of water in northern hemisphere = $\frac{3}{2 + 3}$ * x = $\frac{3}{5}$x

Area of land in the whole earth = $\frac{1}{1+2}$ 2x = $\frac{2}{3}$x

and area of water in the whole earth = $\frac{2}{1+2}$ * 2x = $\frac{4}{3}$x

Therefore area of land in the southern hemisphere = $\frac{2x}{3} - \frac{2x}{5} = \frac{4x}{15}$

and area of water in the southern hemisphere = $\frac{4x}{3} - \frac{3x}{5} = \frac{11x}{15}$

Thus the ratio of land and area of water in the southern hemisphere = $\frac{4x}{15}$ : $\frac{11x}{15}$ = 4 : 11.