Math Solver

Math online solver is one of the easiest ways for students to get help in math. Its growing popularity among students and parents is a testament to how effective it has been in helping students get math help in a fun and easy way. Math solver is very convenient as it give you a lot of flexibility in terms of time and place, which tutoring centers or after school programs cannot beat. The math solvers on the internet are experienced and qualified and being online, you get to select the best of them, wherever they may be. Math solver have solutions and lessons for students from K-12 to college.

Math Problem Solver

With math solver, the focus is on getting each student to understand the subject and what they need to do to solve math problems. So once you learn the lessons, you’ll be able to solve any problems with ease. Irrespective of what your curriculum is like or which state you are in, you will find lessons and math solvers who follow your curriculum. Get access to lots of solved practice material. Math question solver is the easiest ways for students to get help in solving math problems.

Solved Examples

Question 1: Three years ago, father's age was the square of his son's age. Twelve years hence, his age will be twice that of his son's age. Find the present age of the father.

Solution:
Let the age of the son 3 years ago = x years

Therefore age of his father = $x^2$

Step 1:

Twelve years hence: 

Age of Son = x + 15

Therefore age of his father  = $x^2$ + 15

Step 2:

The problem states :

Age of father = 2(age of his son)

=>  $x^2$ + 15 = 2(x + 15)

=> x2 + 15 = 2x + 30

=> x2 - 2x = 30 - 15

=> x2 - 2x = 15

=> x2 - 2x - 15 = 0

Step 3:
Solve for x,
x2 - 2x - 15 = 0

=> x2 - 5x + 3x - 15 = 0

=> x(x - 5) + 3(x - 5) = 0

=> (x + 3)(x - 5) = 0

Step 4:

either x + 3 = 0   or x - 5 = 0

=> x = -3    (neglecting negative value)

or   x = 5

=> x = 5, son's age 3 years ago

Therefore father's age 3 years ago = 52 = 25

So father's present age = 25 + 3 = 28

Thus, the present age of the father = 28 years.
 

Question 2: Solve the system of linear equations
6x + 15y = 50  and 3x + 10y = 50

Solution:
Step 1:
Given system of linear equations
6x + 15y = 50                                   ................................(1)

and

3x + 10y = 50                                  ..................................(2)

Step 2:

Multiply equation (2) by 2

=> 2(3x + 10y) = 2 * 50

=> 6x + 20y = 100                          ...................................(3)

Step 3:

Solve (1) and (3):
Subtract equation( 3) from equation ( 1)

=> 6x + 15y - 6x - 20y = 50 - 100

=> - 5y = - 50

Divide each side by -5

=> y = 10

Step 4:

Put y = 10 in (1)

=> 6x + 15 * 10 = 50

=> 6x + 150 = 50

=> 6x = 50 - 150

=> 6x = - 100

Divide each side by 6, to isolate x,

=> x = $\frac{-100}{6}$

=> x = $\frac{-50}{3}$

Hence, the solution to the system, (x, y) = ($\frac{-50}{3}$, 10).