Algebra Help Online


Algebra can be tough subject to understand. Students struggling with Algebra in class usually turn to a variety of sources to get help. They might ask their parents or friends for help, opt for tutoring in learning centres, join a study group or struggle with learning the subject themselves. Online Algebra tutoring has grown popular over the past few years as more and more students are using the internet for help and information. Students are very comfortable learning, playing and interacting in an online environment so it makes sense to move the tutoring also online. Algebra Help online is available from the comfort of home. A student can connect with a highly trained, and well-qualified tutor and get to learn the subject online. The tutor will take the student through the basics and then work with him/her on solving more complex problems in a step-by-step manner.


Help with Algebra Word Problems

Algebra help available online makes the learning process easy and hassle-free and it is the best resource to learn the subject. Students who have tried it have seen a remarkable improvement in their grades and are additionally also able to get their homework and assignments done accurately and on time. The Internet now offers a variety of options for getting Algebra tutoring online. At its most basic, there are sites which have a free online Algebra calculator that you can input your problem into and it gives you an answer instantly. The more detailed sites also show you the steps while comprehensive sites assign each student to an algebra problem solver who will help with any problems that they may need help with.

Solved Examples

Question 1: The length of a rectangle exceed its width by 4 feet. If its area is 45 sq. feet. Find its dimensions.

Solution:
Step 1:
Let the width of the rectangle = x

Therefore length of the rectangle = x + 4
(length exceed its width by 4 feet)

Area of the rectangle = 45

Step 2:


Area of the rectangle = Length * Width

=> 45 = x(x + 4)

=> 45 = $x^2$ + 4x

=> $x^2$ + 4x - 45 = 0

Step 3:


Solve for x,

=> $x^2$ + 4x - 45 = 0

=> $x^2$ + 9x - 5x - 45 = 0

=> x(x + 9) - 5(x + 9) = 0

=> (x - 5)(x + 9) = 0

Step 4:

either x - 5 = 0    or    x + 9 = 0

=> x = 5    or    x = - 9

The solution x = - 9 must be discarded because a rectangle can't have a negative width.

=> x = 5, is the width of the rectangle.

Step 5:


Length of the rectangle = x + 4 = 5 + 4 = 9

Hence the dimensions of the rectangle are: Length = 9 feet and Width = 5 feet. answer
 

Question 2: Find the largest of three consecutive odd integers if the sum of the second and the third integers is 9 less than the five times the first integer.
Solution:
Let the three consecutive odd integer are, x, x + 2 and x + 4.
That is,
First integer = x
Second integer = x + 2
Third integer = x + 4

Step 1:

The problem states that:

Sum of the second and third integers = 5(first integer) - 9.

=> (x + 2) + (x + 4) = 5x - 9

=> x + 2 + x + 4 = 5x - 9

Step 2:

Combine the like terms:

=> 2x + 6 = 5x - 9

Subtract 2x from both side

=> 2x - 2x + 6 = 5x - 9 - 2x

=> 6 = 3x - 9

Subtract 6 from both sides

=> 6 - 6 = 3x - 9 - 6

=> 0 = 3x - 15

=> 3x = 15

Divide each side by 3

=> $\frac{3x}{3} = \frac{15}{3}$

=> x = 5, first integer.

Step 3:

Largest or third integer = x + 4 = 5 + 4 = 9

Hence the largest integer is 9. answer