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

Given equations are

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

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

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:

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:

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)**.

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

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

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:

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:

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

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

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:

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:

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:

Now

$\frac{1}{x + y}$ = 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:

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:

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:

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

=>

Step 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.