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Algebra Word Problem Solver

Students looking for help with algebra can now get it online with the algebra word problem solver. The internet and online learning has grown very popular in the last few years. Algebra is one topic that students always need some extra help with. An algebra word problem solver makes solving problems easy with a step-by-step explanations. Easily accessible from anywhere at any time, algebra word problem solvers are quick and effective, especially before important tests and exams. There are also a number of online sites which offer free algebra word problem solvers where you can get the answers instantly. Besides, these sites also offer online algebra tutoring with expert online tutors. Many of these sites assign each student to an algebra problem solver tutor who will help students solve all the word problems they may have.

Solving Algebra Word Problems

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Help with algebra word problem solvers is highly effective as they go into the details of each problem. With concept explanations and examples students can very comfortable and familiar with solving algebra world problems. For additional help in algebra, students can opt for an online tutoring service and get help with homework and assignments and also get ready for tests and quizzes. Online algebra word problem solvers also have worksheets with plenty of problems that students can use to practice  hone their algebra skills. The best part is they can find out instantly whether a problem has been solved correctly. Use algebra word problem solvers to master algebra at your own pace and time.

Solved Examples

Question 1: Five times Madona's age plus four years is Stefen's age. Stefen is 24 years old. Find Madona's age.
Solution:

Step 1:
Let Madona's age = x

Stefen's age = 24

The statement states:

5 times Madona's age + 4 = Stefen's age

=> 5x + 4 = 24

Solve for x,

Subtract 4 from both sides of the equation

=> 5x + 4 - 4 = 24 - 4

=> 5x = 20

Divide each side by 5

=> $\frac{5x}{5} = \frac{20}{5}$

=> x = 4

Hence, Madona is 4 years old.
 

Question 2:

Find the numbers such that twice of the first number added to four times of the second number gives 800 and five times of first number added to the two times the second number gives 720.


Solution:
Let two numbers be x and y

Step 1:

Twice of the first number + four times of the second number = 800

=> 2x + 4y = 800                          ..........................(1)

Five times of first number + two times the second number = 720

=> 5x + 2y = 720                           .............................(2)

Step 2:

Multiply equation (1) by 5 and equation (2) by 2

=> 5(2x + 4y) = 800 * 5

=> 10x + 20y = 4,000                     ..............................(3)

and
2(5x + 2y) = 720 * 2

=> 10x + 4y = 1440                          .............................(4)

Step 3:
Subtract equation (4) from equation (3)

=> 10x + 20y - (10x + 4y) = 4000 - 1440

=> 10x + 20y - 10x - 4y = 2560

=> 16y = 2560

Divide both sides by 16

=> $\frac{16y}{16} = \frac{2560}{16}$

=> y = 160

Step 4:

Put y = 160 in equation (1)

=> 2x + 4 * 160 = 800


=> 2x + 640 = 800

=> 2x = 800 - 640 = 160

=> x = $\frac{160}{2}$ = 80

=> x = 80

Hence the numbers are 80 and 160.