Math problems are the bane of most students' existence but they can be solved quickly once students are familiar with the steps, rules, and formulas that need to be used. Students begin by solving arithmetic problems, before moving on to algebra, geometry, trigonometry, calculus and other topics in math. In order to become a good math problem solver, students need to be consistent and patient.

Practice is the key to learning and remembering math solutions and equations. One of the most effective ways to learn math is by practicing everyday. Class notes and examples act as a guide and are a useful reference anytime you get stuck. Once you have learned the basics well, work on more advanced problems and go through different sources like textbooks, class handouts, and the internet, so that you cover every type.

## Solved Examples

Question 1: Solve the system: 2x - y = 10 and 3x + 2y = 1
Solution:
Given system: 2x - y = 10 .............(i)

3x + 2y = 1 ...........(ii)

Step 1: Choose equation (i) and solve for y

2x - y = 10

y = 2x - 10

Step 2: Substitute the value of y in equation (ii)

3x + 2(2x - 10) = 1

3x + 4x - 20 = 1

7x - 20 = 1

7x = 21

x = 3  (Divide each side by 7)

Step 3: Again put the value of x in y = 2x - 10

y = 2 * 3 - 10 = 6 - 10 = -4

The solution of the system is (3, -4).

Question 2: Prove that $\frac{Cos\ x}{1 - Sin\ x}$ = $Sec\ x + Tan\ x$
Solution:

LHS =  $\frac{Cos\ x}{1 - Sin\ x}$

$\frac{Cos\ x}{1 - Sin\ x}$ *  $\frac{1 + Sin\ x}{1 + Sin\ x}$ (Rationalizing the denominator)

= $\frac{Cos\ x(1 + Sin\ x)}{(1 - Sin\ x)(1 + Sin\ x)}$

= $\frac{Cos\ x(1 + Sin\ x)}{1^2 - Sin^2 x}$

= $\frac{Cos\ x(1 + Sin\ x)}{Cos^2x}$

= $\frac{1 + Sin\ x}{Cos\ x}$  (Cancel common terms)

$\frac{1}{Cos\ x}$ $(1 + Sin\ x)$

= $\frac{1}{Cos\ x}$ + $\frac{1}{Cos\ x}$ * $Sin\ x$

= $(Sec\ x + Tan\ x)$  [$Sec\ x$ = $\frac{1}{Cos\ x}$ and $Tan\ x$ = $\frac{Sin\ x}{Cos\ x}$]

= RHS

Math word problems in particular, are challenging for most students. Math word problems use particular terms and phrases to explain the problem, and getting to know these is the key to solving them quickly. There are many useful resources and math sites on the internet which provide math calculators that can solve problems in minutes.
Solved Example
Question: John can do a piece of work in x days and Jacob can do it in (x + 16) days. If both working together can do it in 15 days. Find the value of x.

Solution:
John's one day work = $\frac{1}{x}$

Jacob's one day work = $\frac{1}{x + 16}$

Both can finished work in one day = $\frac{1}{x}$ + $\frac{1}{x + 16}$

Both can do the work in 15 days (Given)

So their one day work = $\frac{1}{15}$

$\therefore$ $\frac{1}{x}$ $\frac{1}{x + 16}$ =  $\frac{1}{15}$

=> $\frac{x + x + 16}{x(x + 16)}$ = $\frac{1}{15}$

=> $\frac{2x + 16}{x^2 + 16x}$ = $\frac{1}{15}$

=> $x^2 + 16x = 15(2x + 16)$

=> $x^2 - 14x - 240 = 0$

=> By solving above equation, we have

x = 24  or x = -10

x = -10 is neglected (number of days cannot be negative)

Hence x = 24.