Solve any math problem online and get quick solutions to all your math difficulties. Solving math problems online is easy, quick and effective. Online math helpers have a lot of experience in solving math problems for all grades and levels. Whether it is algebra, geometry, quadratic equations or trigonometry that you find tough, you can be assured that you will find help online. Online math help is becoming very popular among students who find its easy accessibility and 24x7 availability a huge advantage.

## Solve Math Problem Online

## Solved Examples

**Question 1:**Solve

2x + 3y = 5 and 5x - 2y = 3

**Solution:**

Given equations are

2x + 3y = 5 ........................(1)

5x - 2y = 3 ........................(2)

To eliminate x:

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

=> 5(2x + 3y = 5)

=> 10x + 15y = 25 ........................(3)

and 2(5x - 2y = 3)

=> 10x - 4y = 6 ............................(4)

Subtract equation (4) from equation (3)

=> 10x + 15y - (10x - 4y) = 25 - 6

=> 10x + 15y - 10x + 4y = 19

=> 19x = 19

Divide each side by 19

=> x = 1

Put x = 1 in equation (1)

=> 2 * 1 + 3y = 5

=> 2 + 3y = 5

=> 3y = 5 - 2 = 3

Divide each side by 3

=> y = 1

Hence the solution to the system is

2x + 3y = 5 ........................(1)

5x - 2y = 3 ........................(2)

Step 1:Step 1:

To eliminate x:

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

=> 5(2x + 3y = 5)

=> 10x + 15y = 25 ........................(3)

and 2(5x - 2y = 3)

=> 10x - 4y = 6 ............................(4)

Step 2:Step 2:

Subtract equation (4) from equation (3)

=> 10x + 15y - (10x - 4y) = 25 - 6

=> 10x + 15y - 10x + 4y = 19

=> 19x = 19

Divide each side by 19

=> x = 1

Step 3:Step 3:

Put x = 1 in equation (1)

=> 2 * 1 + 3y = 5

=> 2 + 3y = 5

=> 3y = 5 - 2 = 3

Divide each side by 3

=> y = 1

Hence the solution to the system is

**(x, y) = (1, 1)**.**Question 2:**A boat takes 20 hours to go 40 km downstream and 30 km upstream. Again the same boat takes 30 hours to go 20 km downstream and 25 km upstream. Find the speed of the boat and the current.

**Solution:**

Let speeds of the boat and the current be x km/hr and y km/hr.

Speed of the boat in downstream = (x + y) km/hr

Speed of the boat in upstream = (x - y) km/hr

According to problem:

$\frac{40}{x + y}$ + $\frac{30}{x - y}$ = 20 .....................(1)

and

$\frac{20}{x + y}$ + $\frac{25}{x - y}$ = 30 ..................... (2)

Put $\frac{1}{x + y}$ = u and $\frac{1}{x - y}$ = v

(1) => 40u + 30v = 20

or 4u + 3v = 2 ..........................(3)

and

(2) => 20u + 25v = 30

or 4u + 5v = 6 ............................(4)

Subtract (4) from (3)

=> 4u + 3v - (4u + 5v) = 2 - 6

=> 4u + 3v - 4u - 5v = -4

=> - 2v = - 4

=>

=> 4u + 3 * 2 = 2

=> 4u + 6 = 2

=> 4u = - 4

=>

$\frac{1}{x + y}$ = u = -1

=> x + y = -1

or y = -1 - x ......................(5)

and $\frac{1}{x - y}$ = v = 2

=> 2x - 2y = 1, Put (5) in this equation

=> 2x - 2(-1 - x) = 1

=> 2x + 2 + 2x = 1

=> 4x = -1

=> x = $\frac{-1}{4}$, put in (5)

=> y = -1 - $\frac{-1}{4}$

=> y = -1 + $\frac{-1}{4}$

=> y = $\frac{-3}{4}$

[Since distance and speed can't be negative.]

Hence, speeds of the boat and the current be $\frac{1}{4}$ km/hr and $\frac{3}{4}$ km/hr respectively.

Speed of the boat in downstream = (x + y) km/hr

Speed of the boat in upstream = (x - y) km/hr

Step 1:Step 1:

According to problem:

$\frac{40}{x + y}$ + $\frac{30}{x - y}$ = 20 .....................(1)

and

$\frac{20}{x + y}$ + $\frac{25}{x - y}$ = 30 ..................... (2)

Step 2:Step 2:

Put $\frac{1}{x + y}$ = u and $\frac{1}{x - y}$ = v

(1) => 40u + 30v = 20

or 4u + 3v = 2 ..........................(3)

and

(2) => 20u + 25v = 30

or 4u + 5v = 6 ............................(4)

Step 3:Step 3:

Subtract (4) from (3)

=> 4u + 3v - (4u + 5v) = 2 - 6

=> 4u + 3v - 4u - 5v = -4

=> - 2v = - 4

=>

**v = 2**, put in equation (3)=> 4u + 3 * 2 = 2

=> 4u + 6 = 2

=> 4u = - 4

=>

**u = -1**

Step 4:

NowStep 4:

$\frac{1}{x + y}$ = u = -1

=> x + y = -1

or y = -1 - x ......................(5)

and $\frac{1}{x - y}$ = v = 2

=> 2x - 2y = 1, Put (5) in this equation

=> 2x - 2(-1 - x) = 1

=> 2x + 2 + 2x = 1

=> 4x = -1

=> x = $\frac{-1}{4}$, put in (5)

=> y = -1 - $\frac{-1}{4}$

=> y = -1 + $\frac{-1}{4}$

=> y = $\frac{-3}{4}$

[Since distance and speed can't be negative.]

Hence, speeds of the boat and the current be $\frac{1}{4}$ km/hr and $\frac{3}{4}$ km/hr respectively.