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.

## Math Problems and Answers

## 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)

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).

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

**Step 1:**Choose equation (i) and solve for y2x - 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 - 10y = 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:**

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

## Math Word Problem Answers

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.

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.