Algebra problems tend to be worded a little differently from the typical 'Ann had a dozen apples' kind of arithmetic questions. Algebra deals with variables and numbers which are arranged in equations to form generalizations. Hence the terms used are understandably a bit different in order to convey the relation between different elements. Algebra word problems require students to decipher the meaning and form the correct equations. Algebra word problem solver is an online tool which will help users understand how algebra word problems are stated. With the word problem solver for algebra, students learn what the different terms mean and what kind of relation they imply. 

Word Problem Solver Algebra

Online algebra word problem solvers are available for all grades and levels of algebra. Algebra 1 word problem solvers and algebra 2 word problem solvers cover the beginner and intermediate levels of algebra. Free algebra word problem solvers also give you tips and hints on which terms or phrases to look out for. Learning to identify these will enable you to understand what you are expected to find out much quicker. College algebra word problem solvers are designed to handle the advanced course load and are very popular among college students who turn to the internet for a substantial amount of course help and research.

Solved Examples

Question 1: The base of a triangle is 2 m more than twice its altitude. If the area of the triangle is 30 square m. Find its altitude.

Solution:
Let the altitude of the triangle = x

therefore, base of the triangle = 2 + 2x

Given, area of the triangle = 30

Step 1:

Area of the triangle = $\frac{1}{2}$ Base * Height

=> 30 = $\frac{1}{2}$ * (2 + 2x) * x

=> 30 * 2 = (2 + 2x) * x

=> 60 = 2x + 2x2

=> 2x2 + 2x - 60 = 0

Step 2:


Solve for x, 2x2 + 2x - 60 = 0

=> 2x2 + 12x - 10x - 60 = 0

=> 2x(x + 6) - 10(x + 6) = 0

=> (2x - 10)(x + 6) = 0

Step 3:
either 2x - 10 = 0    or    x + 6 = 0

=> 2x = 10     or     x = - 6

Length must be positive, so neglect x = - 6.

=> 2x = 10

=> x = 5

Hence altitude of the triangle is 5 m. answer

 

Question 2: The sum of the ages of a father and his son is 45. 5 years ago, father's age is three years more than thrice the sons age. Find the age of the son 4 years hence.

Solution:
Let the age of the son = x years

Therefore the age of the father = 45 - x
(Sum of their ages is 45)

Step 1:
5 years ago:

Son's age = x - 5

and father's age  = 45 - x - 5 = 40 - x

Step 2:
The problem states:
 
Father's age = 3 + 3(Son's age)

=> 40 - x = 3 + 3(x - 5)

=> 40 - x = 3 + 3x - 15

=> 40 - x = 3x - 12

=> 3x + x = 40 + 12

=> 4x = 52

Divide each side by 4

=> x = $\frac{52}{4}$

=> x = 13

Son's present age = 13 years

=> After 4 years son will be of = 13 + 4 = 17 years.