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.

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).

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

2x - y = 10

y = 2x - 10

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

3x + 4x - 20 = 1

7x - 20 = 1

7x = 21

x = 3 (Divide each side by 7)

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

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

Let us start with LHS

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

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

Solved Example

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.

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.