Sasi Math (Revised for 1024)

Scroll:Algebra new >> Elimination method >> ps (4539)

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.

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y =

x =

Answer:_______________

x + $\frac{y}{3}$ = 5, $\frac{x}{3}$ + 3y = 7

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y =

x =

Answer:_______________

x + $\frac{y}{3}$ = 5, $\frac{x}{4}$ + 4y = 7

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 5, $\frac{x}{3}$ + 2y = 7

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 5, $\frac{x}{4}$ + 3y = 7

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y =

x =

Answer:_______________

10)

x + $\frac{y}{3}$ = 6, $\frac{x}{3}$ + 2y = 9

y =

x =

Answer:_______________

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y = Answer: 2

x = Answer: 3

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 4 ....... (1)

$\frac{(2}{x+y)}$ /2 = 4x2

2x + y = 8 ...... (2)

$\frac{x}{3}$ + 2y = 5 ...... (3)

(x + $\frac{6y)}{3}$ = 5x3

x + 6y = 15 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 8

(4) x (2) ⇒ 2x + 12y = 30

-11y = -22

y = $\frac{22}{11}$ = 2

y = 2

Substituting y = 2 in (2)

2x + 2 = 8

2x = 8 - 2

2x = 6

x = $\frac{6}{2}$

x = 3.

The solution is (2, 3).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3 + $\frac{2}{2}$ = 3 + 1 = 4 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3}{3}$ + 2(2) = 1 + 4 = 5 = R.H.S . of (2).

x + $\frac{y}{3}$ = 5, $\frac{x}{3}$ + 3y = 7

y = Answer: 1.85

x = Answer: 4.38

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{3}$ = 5 ....... (1)

$\frac{(3}{x+y)}$ /3 = 5x3

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

$\frac{x}{3}$ + 3y = 7 ...... (3)

(x + $\frac{9y)}{3}$ = 7x3

x + 9y = 21 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 3x + y = 15

(4) x (3) ⇒ 3x + 27y = 63

-26y = -48

y = $\frac{48}{26}$ = 1.85

y = 1.85

Substituting y = 1.85 in (2)

3x + 1.85 = 15

3x = 15 - 1.85

3x = 13.15

x = $\frac{13.15}{3}$

x = 4.38.

The solution is (1.85, 4.38).

Verification :

L.H.S of (1) = x + $\frac{y}{3}$ = 4.38 + $\frac{1.85}{3}$ = 4.38 + 0.62 = 5 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 3y = $\frac{4.38}{3}$ + 3(1.85) = 1.46 + 5.55 = 7.01 = R.H.S . of (2).

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y = Answer: 2

x = Answer: 3

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 4 ....... (1)

$\frac{(2}{x+y)}$ /2 = 4x2

2x + y = 8 ...... (2)

$\frac{x}{3}$ + 2y = 5 ...... (3)

(x + $\frac{6y)}{3}$ = 5x3

x + 6y = 15 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 8

(4) x (2) ⇒ 2x + 12y = 30

-11y = -22

y = $\frac{22}{11}$ = 2

y = 2

Substituting y = 2 in (2)

2x + 2 = 8

2x = 8 - 2

2x = 6

x = $\frac{6}{2}$

x = 3.

The solution is (2, 3).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3 + $\frac{2}{2}$ = 3 + 1 = 4 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3}{3}$ + 2(2) = 1 + 4 = 5 = R.H.S . of (2).

x + $\frac{y}{3}$ = 5, $\frac{x}{4}$ + 4y = 7

y = Answer: 1.47

x = Answer: 4.51

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{3}$ = 5 ....... (1)

$\frac{(3}{x+y)}$ /3 = 5x3

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

$\frac{x}{4}$ + 4y = 7 ...... (3)

(x + $\frac{16y)}{4}$ = 7x4

x + 16y = 28 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 3x + y = 15

(4) x (3) ⇒ 3x + 48y = 84

-47y = -69

y = $\frac{69}{47}$ = 1.47

y = 1.47

Substituting y = 1.47 in (2)

3x + 1.47 = 15

3x = 15 - 1.47

3x = 13.53

x = $\frac{13.53}{3}$

x = 4.51.

The solution is (1.47, 4.51).

Verification :

L.H.S of (1) = x + $\frac{y}{3}$ = 4.51 + $\frac{1.47}{3}$ = 4.51 + 0.49 = 5 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{4}$ + 4y = $\frac{4.51}{4}$ + 4(1.47) = 1.13 + 5.88 = 7.01 = R.H.S . of (2).

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y = Answer: 2

x = Answer: 3

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 4 ....... (1)

$\frac{(2}{x+y)}$ /2 = 4x2

2x + y = 8 ...... (2)

$\frac{x}{3}$ + 2y = 5 ...... (3)

(x + $\frac{6y)}{3}$ = 5x3

x + 6y = 15 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 8

(4) x (2) ⇒ 2x + 12y = 30

-11y = -22

y = $\frac{22}{11}$ = 2

y = 2

Substituting y = 2 in (2)

2x + 2 = 8

2x = 8 - 2

2x = 6

x = $\frac{6}{2}$

x = 3.

The solution is (2, 3).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3 + $\frac{2}{2}$ = 3 + 1 = 4 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3}{3}$ + 2(2) = 1 + 4 = 5 = R.H.S . of (2).

x + $\frac{y}{2}$ = 5, $\frac{x}{3}$ + 2y = 7

y = Answer: 2.91

x = Answer: 3.55

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 5 ....... (1)

$\frac{(2}{x+y)}$ /2 = 5x2

2x + y = 10 ...... (2)

$\frac{x}{3}$ + 2y = 7 ...... (3)

(x + $\frac{6y)}{3}$ = 7x3

x + 6y = 21 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 10

(4) x (2) ⇒ 2x + 12y = 42

-11y = -32

y = $\frac{32}{11}$ = 2.91

y = 2.91

Substituting y = 2.91 in (2)

2x + 2.91 = 10

2x = 10 - 2.91

2x = 7.09

x = $\frac{7.09}{2}$

x = 3.55.

The solution is (2.91, 3.55).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3.55 + $\frac{2.91}{2}$ = 3.55 + 1.46 = 5.01 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3.55}{3}$ + 2(2.91) = 1.18 + 5.82 = 7 = R.H.S . of (2).

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y = Answer: 2

x = Answer: 3

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 4 ....... (1)

$\frac{(2}{x+y)}$ /2 = 4x2

2x + y = 8 ...... (2)

$\frac{x}{3}$ + 2y = 5 ...... (3)

(x + $\frac{6y)}{3}$ = 5x3

x + 6y = 15 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 8

(4) x (2) ⇒ 2x + 12y = 30

-11y = -22

y = $\frac{22}{11}$ = 2

y = 2

Substituting y = 2 in (2)

2x + 2 = 8

2x = 8 - 2

2x = 6

x = $\frac{6}{2}$

x = 3.

The solution is (2, 3).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3 + $\frac{2}{2}$ = 3 + 1 = 4 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3}{3}$ + 2(2) = 1 + 4 = 5 = R.H.S . of (2).

x + $\frac{y}{2}$ = 5, $\frac{x}{4}$ + 3y = 7

y = Answer: 2.00

x = Answer: 4.00

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 5 ....... (1)

$\frac{(2}{x+y)}$ /2 = 5x2

2x + y = 10 ...... (2)

$\frac{x}{4}$ + 3y = 7 ...... (3)

(x + $\frac{12y)}{4}$ = 7x4

x + 12y = 28 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 10

(4) x (2) ⇒ 2x + 24y = 56

-23y = -46

y = $\frac{46}{23}$ = 2.00

y = 2.00

Substituting y = 2.00 in (2)

2x + 2.00 = 10

2x = 10 - 2.00

2x = 8

x = $\frac{8}{2}$

x = 4.00.

The solution is (2.00, 4.00).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 4.00 + $\frac{2.00}{2}$ = 4.00 + 1.00 = 5 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{4}$ + 3y = $\frac{4.00}{4}$ + 3(2.00) = 1.00 + 6.00 = 7 = R.H.S . of (2).

x + $\frac{y}{2}$ = 4, $\frac{x}{3}$ + 2y = 5

y = Answer: 2

x = Answer: 3

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{2}$ = 4 ....... (1)

$\frac{(2}{x+y)}$ /2 = 4x2

2x + y = 8 ...... (2)

$\frac{x}{3}$ + 2y = 5 ...... (3)

(x + $\frac{6y)}{3}$ = 5x3

x + 6y = 15 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 2x + y = 8

(4) x (2) ⇒ 2x + 12y = 30

-11y = -22

y = $\frac{22}{11}$ = 2

y = 2

Substituting y = 2 in (2)

2x + 2 = 8

2x = 8 - 2

2x = 6

x = $\frac{6}{2}$

x = 3.

The solution is (2, 3).

Verification :

L.H.S of (1) = x + $\frac{y}{2}$ = 3 + $\frac{2}{2}$ = 3 + 1 = 4 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{3}{3}$ + 2(2) = 1 + 4 = 5 = R.H.S . of (2).

10)

x + $\frac{y}{3}$ = 6, $\frac{x}{3}$ + 2y = 9

y = Answer: 3.71

x = Answer: 4.76

SOLUTION 1 :

This system is linear in x and y.

x + $\frac{y}{3}$ = 6 ....... (1)

$\frac{(3}{x+y)}$ /3 = 6x3

3x + y = 18 ...... (2)

$\frac{x}{3}$ + 2y = 9 ...... (3)

(x + $\frac{6y)}{3}$ = 9x3

x + 6y = 27 ...... (4)

Solving (2) and (4), we get

(2) ⇒ 3x + y = 18

(4) x (3) ⇒ 3x + 18y = 81

-17y = -63

y = $\frac{63}{17}$ = 3.71

y = 3.71

Substituting y = 3.71 in (2)

3x + 3.71 = 18

3x = 18 - 3.71

3x = 14.29

x = $\frac{14.29}{3}$

x = 4.76.

The solution is (3.71, 4.76).

Verification :

L.H.S of (1) = x + $\frac{y}{3}$ = 4.76 + $\frac{3.71}{3}$ = 4.76 + 1.24 = 6 = R.H.S. of (1)

L.H.S of (3) = $\frac{x}{3}$ + 2y = $\frac{4.76}{3}$ + 2(3.71) = 1.59 + 7.42 = 9.01 = R.H.S . of (2).