Sasi Math (Revised for 1024)

Scroll:Algebra new >> Elimination method >> ps (4541)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

$\frac{3}{x}$ + $\frac{7}{y}$ = $\frac{28}{xy}$ , $\frac{2}{x}$ - $\frac{7}{y}$ = $\frac{28}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{30}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{4}{x}$ + $\frac{2}{y}$ = $\frac{26}{xy}$ , $\frac{3}{x}$ - $\frac{2}{y}$ = $\frac{32}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{25}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{25}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{25}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

10)

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x =

y =

Answer:_______________

$\frac{3}{x}$ + $\frac{7}{y}$ = $\frac{28}{xy}$ , $\frac{2}{x}$ - $\frac{7}{y}$ = $\frac{28}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 4

y = Answer: 0

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{7}{y}$ = $\frac{28}{xy}$ .

multiply by xy on both the sides,

3y + 7x = 28

7x + 3y = 28 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{7}{y}$ = $\frac{28}{xy}$

Multiply by xy on both sides.

7x + 2y = 28 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

7x + 3y = 28

7x + 2y = 28

1y = 0

y = 0.

Substituting y = 0 in (1).

7x + 3(0) = 28.

7x + 0 = 28

7x = 28 - 0

7x = 28

x = $\frac{28}{7}$

x = 4

The system has two solutions is ( 4, 0 ).

Verifiction :

L.H.S. of (1) = 7x + 3y = 7(4) + 3(0) = 28 + 0 = 28 R.H.S. of (1)

L.H.S. of (2) = 7x + 2y = 7(4) + 2(0) = 28 + 0 = 28 R.H.S.of (2).

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 1

y = Answer: 5

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{20}{xy}$ .

multiply by xy on both the sides,

3y + 5x = 20

5x + 3y = 20 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 2y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 3y = 20

5x + 2y = 15

1y = 5

y = 5.

Substituting y = 5 in (1).

5x + 3(5) = 20.

5x + 15 = 20

5x = 20 - 15

5x = 5

x = $\frac{5}{5}$

x = 1

The system has two solutions is ( 1, 5 ).

Verifiction :

L.H.S. of (1) = 5x + 3y = 5(1) + 3(5) = 5 + 15 = 20 R.H.S. of (1)

L.H.S. of (2) = 5x + 2y = 5(1) + 2(5) = 5 + 10 = 15 R.H.S.of (2).

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{30}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: -6

y = Answer: 15

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{4}{x}$ - $\frac{5}{y}$ = $\frac{30}{xy}$ .

multiply by xy on both the sides,

4y + 5x = 30

5x + 4y = 30 ...... (1)

Consider, $\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 3y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 4y = 30

5x + 3y = 15

1y = 15

y = 15.

Substituting y = 15 in (1).

5x + 4(15) = 30.

5x + 60 = 30

5x = 30 - 60

5x = -30

x = $\frac{-30}{5}$

x = -6

The system has two solutions is ( -6, 15 ).

Verifiction :

L.H.S. of (1) = 5x + 4y = 5(-6) + 4(15) = -30 + 60 = 30 R.H.S. of (1)

L.H.S. of (2) = 5x + 3y = 5(-6) + 3(15) = -30 + 45 = 15 R.H.S.of (2).

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 1

y = Answer: 5

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{20}{xy}$ .

multiply by xy on both the sides,

3y + 5x = 20

5x + 3y = 20 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 2y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 3y = 20

5x + 2y = 15

1y = 5

y = 5.

Substituting y = 5 in (1).

5x + 3(5) = 20.

5x + 15 = 20

5x = 20 - 15

5x = 5

x = $\frac{5}{5}$

x = 1

The system has two solutions is ( 1, 5 ).

Verifiction :

L.H.S. of (1) = 5x + 3y = 5(1) + 3(5) = 5 + 15 = 20 R.H.S. of (1)

L.H.S. of (2) = 5x + 2y = 5(1) + 2(5) = 5 + 10 = 15 R.H.S.of (2).

$\frac{4}{x}$ + $\frac{2}{y}$ = $\frac{26}{xy}$ , $\frac{3}{x}$ - $\frac{2}{y}$ = $\frac{32}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 25

y = Answer: -6

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{4}{x}$ - $\frac{2}{y}$ = $\frac{26}{xy}$ .

multiply by xy on both the sides,

4y + 2x = 26

2x + 4y = 26 ...... (1)

Consider, $\frac{3}{x}$ + $\frac{2}{y}$ = $\frac{32}{xy}$

Multiply by xy on both sides.

2x + 3y = 32 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

2x + 4y = 26

2x + 3y = 32

1y = -6

y = -6.

Substituting y = -6 in (1).

2x + 4(-6) = 26.

2x + -24 = 26

2x = 26 - -24

2x = 50

x = $\frac{50}{2}$

x = 25

The system has two solutions is ( 25, -6 ).

Verifiction :

L.H.S. of (1) = 2x + 4y = 2(25) + 4(-6) = 50 + -24 = 26 R.H.S. of (1)

L.H.S. of (2) = 2x + 3y = 2(25) + 3(-6) = 50 + -18 = 32 R.H.S.of (2).

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 1

y = Answer: 5

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{20}{xy}$ .

multiply by xy on both the sides,

3y + 5x = 20

5x + 3y = 20 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 2y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 3y = 20

5x + 2y = 15

1y = 5

y = 5.

Substituting y = 5 in (1).

5x + 3(5) = 20.

5x + 15 = 20

5x = 20 - 15

5x = 5

x = $\frac{5}{5}$

x = 1

The system has two solutions is ( 1, 5 ).

Verifiction :

L.H.S. of (1) = 5x + 3y = 5(1) + 3(5) = 5 + 15 = 20 R.H.S. of (1)

L.H.S. of (2) = 5x + 2y = 5(1) + 2(5) = 5 + 10 = 15 R.H.S.of (2).

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{25}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: -3

y = Answer: 10

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{4}{x}$ - $\frac{5}{y}$ = $\frac{25}{xy}$ .

multiply by xy on both the sides,

4y + 5x = 25

5x + 4y = 25 ...... (1)

Consider, $\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 3y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 4y = 25

5x + 3y = 15

1y = 10

y = 10.

Substituting y = 10 in (1).

5x + 4(10) = 25.

5x + 40 = 25

5x = 25 - 40

5x = -15

x = $\frac{-15}{5}$

x = -3

The system has two solutions is ( -3, 10 ).

Verifiction :

L.H.S. of (1) = 5x + 4y = 5(-3) + 4(10) = -15 + 40 = 25 R.H.S. of (1)

L.H.S. of (2) = 5x + 3y = 5(-3) + 3(10) = -15 + 30 = 15 R.H.S.of (2).

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 1

y = Answer: 5

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{20}{xy}$ .

multiply by xy on both the sides,

3y + 5x = 20

5x + 3y = 20 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 2y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 3y = 20

5x + 2y = 15

1y = 5

y = 5.

Substituting y = 5 in (1).

5x + 3(5) = 20.

5x + 15 = 20

5x = 20 - 15

5x = 5

x = $\frac{5}{5}$

x = 1

The system has two solutions is ( 1, 5 ).

Verifiction :

L.H.S. of (1) = 5x + 3y = 5(1) + 3(5) = 5 + 15 = 20 R.H.S. of (1)

L.H.S. of (2) = 5x + 2y = 5(1) + 2(5) = 5 + 10 = 15 R.H.S.of (2).

$\frac{4}{x}$ + $\frac{5}{y}$ = $\frac{25}{xy}$ , $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{25}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 5

y = Answer: 0

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{4}{x}$ - $\frac{5}{y}$ = $\frac{25}{xy}$ .

multiply by xy on both the sides,

4y + 5x = 25

5x + 4y = 25 ...... (1)

Consider, $\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{25}{xy}$

Multiply by xy on both sides.

5x + 3y = 25 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 4y = 25

5x + 3y = 25

1y = 0

y = 0.

Substituting y = 0 in (1).

5x + 4(0) = 25.

5x + 0 = 25

5x = 25 - 0

5x = 25

x = $\frac{25}{5}$

x = 5

The system has two solutions is ( 5, 0 ).

Verifiction :

L.H.S. of (1) = 5x + 4y = 5(5) + 4(0) = 25 + 0 = 25 R.H.S. of (1)

L.H.S. of (2) = 5x + 3y = 5(5) + 3(0) = 25 + 0 = 25 R.H.S.of (2).

10)

$\frac{3}{x}$ + $\frac{5}{y}$ = $\frac{20}{xy}$ , $\frac{2}{x}$ - $\frac{5}{y}$ = $\frac{15}{xy}$ , x ≠ 0 y ≠ 0

x = Answer: 1

y = Answer: 5

SOLUTION 1 :

This system is not linear in x and y

Consider, $\frac{3}{x}$ - $\frac{5}{y}$ = $\frac{20}{xy}$ .

multiply by xy on both the sides,

3y + 5x = 20

5x + 3y = 20 ...... (1)

Consider, $\frac{2}{x}$ + $\frac{5}{y}$ = $\frac{15}{xy}$

Multiply by xy on both sides.

5x + 2y = 15 ..... (2)

Now, (1) and (2) is a linear system in x and y.

(1) - (2) , we get

5x + 3y = 20

5x + 2y = 15

1y = 5

y = 5.

Substituting y = 5 in (1).

5x + 3(5) = 20.

5x + 15 = 20

5x = 20 - 15

5x = 5

x = $\frac{5}{5}$

x = 1

The system has two solutions is ( 1, 5 ).

Verifiction :

L.H.S. of (1) = 5x + 3y = 5(1) + 3(5) = 5 + 15 = 20 R.H.S. of (1)

L.H.S. of (2) = 5x + 2y = 5(1) + 2(5) = 5 + 10 = 15 R.H.S.of (2).