Sasi Math (Revised for 1024)

Scroll:Algebra new >> Elimination method >> ps (4538)

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.

Solve each of the folloeing system of equations by elimination method.

x + 4y = 9, x - 4y = 3

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 4y = 7, x - 4y = 1

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 5y = 9, x - 5y = 3

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 4y = 9, x - 4y = 1

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 5y = 9, x - 5y = 1

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 2y = 9, x - 2y = 1

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 3y = 9, x - 3y = 3

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 6y = 9, x - 6y = 3

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 3y = 7, x - 3y = 3

X =

Y =

Answer:_______________

10)

Solve each of the folloeing system of equations by elimination method.

x + 2y = 7, x - 2y = 3

X =

Y =

Answer:_______________

Solve each of the folloeing system of equations by elimination method.

x + 4y = 9, x - 4y = 3

X = Answer: 6

Y = Answer: $\frac{3}{4}$

SOLUTION 1 :

x + 4y = 9 ..... (1)

x - 4y = 3 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 4y = 9

x - 4y = 3

2x = 12

x = $\frac{12}{2}$ = 6.

Substituing x = 6 in (1), We get

6 + 4y = 9

4y = 9 - 6

4y = 3

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

Hence, ( 6 , $\frac{3}{4}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 6 + 4 x $\frac{3}{4}$ = 6 + 3 = 9.

R.H.S. of (2) = x - 2y = 6 - 4 x $\frac{3}{4}$ = 6 - 3 = 3.

Solve each of the folloeing system of equations by elimination method.

x + 4y = 7, x - 4y = 1

X = Answer: 4

Y = Answer: $\frac{3}{4}$

SOLUTION 1 :

x + 4y = 7 ..... (1)

x - 4y = 1 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 4y = 7

x - 4y = 1

2x = 8

x = $\frac{8}{2}$ = 4.

Substituing x = 4 in (1), We get

4 + 4y = 7

4y = 7 - 4

4y = 3

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

Hence, ( 4 , $\frac{3}{4}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 4 + 4 x $\frac{3}{4}$ = 4 + 3 = 7.

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

Solve each of the folloeing system of equations by elimination method.

x + 5y = 9, x - 5y = 3

X = Answer: 6

Y = Answer: $\frac{3}{5}$

SOLUTION 1 :

x + 5y = 9 ..... (1)

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

This system is linear in x and y,

Adding (1) and (2)

x + 5y = 9

x - 5y = 3

2x = 12

x = $\frac{12}{2}$ = 6.

Substituing x = 6 in (1), We get

6 + 5y = 9

5y = 9 - 6

5y = 3

y = $\frac{3}{5}$

Hence, ( 6 , $\frac{3}{5}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 6 + 5 x $\frac{3}{5}$ = 6 + 3 = 9.

R.H.S. of (2) = x - 2y = 6 - 5 x $\frac{3}{5}$ = 6 - 3 = 3.

Solve each of the folloeing system of equations by elimination method.

x + 4y = 9, x - 4y = 1

X = Answer: 5

Y = Answer: $\frac{4}{4}$

SOLUTION 1 :

x + 4y = 9 ..... (1)

x - 4y = 1 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 4y = 9

x - 4y = 1

2x = 10

x = $\frac{10}{2}$ = 5.

Substituing x = 5 in (1), We get

5 + 4y = 9

4y = 9 - 5

4y = 4

y = $\frac{4}{4}$

Hence, ( 5 , $\frac{4}{4}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 5 + 4 x $\frac{4}{4}$ = 5 + 4 = 9.

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

Solve each of the folloeing system of equations by elimination method.

x + 5y = 9, x - 5y = 1

X = Answer: 5

Y = Answer: $\frac{4}{5}$

SOLUTION 1 :

x + 5y = 9 ..... (1)

x - 5y = 1 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 5y = 9

x - 5y = 1

2x = 10

x = $\frac{10}{2}$ = 5.

Substituing x = 5 in (1), We get

5 + 5y = 9

5y = 9 - 5

5y = 4

y = $\frac{4}{5}$

Hence, ( 5 , $\frac{4}{5}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 5 + 5 x $\frac{4}{5}$ = 5 + 4 = 9.

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

Solve each of the folloeing system of equations by elimination method.

x + 2y = 9, x - 2y = 1

X = Answer: 5

Y = Answer: $\frac{4}{2}$

SOLUTION 1 :

x + 2y = 9 ..... (1)

x - 2y = 1 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 2y = 9

x - 2y = 1

2x = 10

x = $\frac{10}{2}$ = 5.

Substituing x = 5 in (1), We get

5 + 2y = 9

2y = 9 - 5

2y = 4

y = $\frac{4}{2}$

Hence, ( 5 , $\frac{4}{2}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 5 + 2 x $\frac{4}{2}$ = 5 + 4 = 9.

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

Solve each of the folloeing system of equations by elimination method.

x + 3y = 9, x - 3y = 3

X = Answer: 6

Y = Answer: $\frac{3}{3}$

SOLUTION 1 :

x + 3y = 9 ..... (1)

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

This system is linear in x and y,

Adding (1) and (2)

x + 3y = 9

x - 3y = 3

2x = 12

x = $\frac{12}{2}$ = 6.

Substituing x = 6 in (1), We get

6 + 3y = 9

3y = 9 - 6

3y = 3

y = $\frac{3}{3}$

Hence, ( 6 , $\frac{3}{3}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 6 + 3 x $\frac{3}{3}$ = 6 + 3 = 9.

R.H.S. of (2) = x - 2y = 6 - 3 x $\frac{3}{3}$ = 6 - 3 = 3.

Solve each of the folloeing system of equations by elimination method.

x + 6y = 9, x - 6y = 3

X = Answer: 6

Y = Answer: $\frac{3}{6}$

SOLUTION 1 :

x + 6y = 9 ..... (1)

x - 6y = 3 .... (2)

This system is linear in x and y,

Adding (1) and (2)

x + 6y = 9

x - 6y = 3

2x = 12

x = $\frac{12}{2}$ = 6.

Substituing x = 6 in (1), We get

6 + 6y = 9

6y = 9 - 6

6y = 3

y = $\frac{3}{6}$

Hence, ( 6 , $\frac{3}{6}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 6 + 6 x $\frac{3}{6}$ = 6 + 3 = 9.

R.H.S. of (2) = x - 2y = 6 - 6 x $\frac{3}{6}$ = 6 - 3 = 3.

Solve each of the folloeing system of equations by elimination method.

x + 3y = 7, x - 3y = 3

X = Answer: 5

Y = Answer: $\frac{2}{3}$

SOLUTION 1 :

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

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

This system is linear in x and y,

Adding (1) and (2)

x + 3y = 7

x - 3y = 3

2x = 10

x = $\frac{10}{2}$ = 5.

Substituing x = 5 in (1), We get

5 + 3y = 7

3y = 7 - 5

3y = 2

y = $\frac{2}{3}$

Hence, ( 5 , $\frac{2}{3}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 5 + 3 x $\frac{2}{3}$ = 5 + 2 = 7.

R.H.S. of (2) = x - 2y = 5 - 3 x $\frac{2}{3}$ = 5 - 2 = 3.

10)

Solve each of the folloeing system of equations by elimination method.

x + 2y = 7, x - 2y = 3

X = Answer: 5

Y = Answer: $\frac{2}{2}$

SOLUTION 1 :

x + 2y = 7 ..... (1)

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

This system is linear in x and y,

Adding (1) and (2)

x + 2y = 7

x - 2y = 3

2x = 10

x = $\frac{10}{2}$ = 5.

Substituing x = 5 in (1), We get

5 + 2y = 7

2y = 7 - 5

2y = 2

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

Hence, ( 5 , $\frac{2}{2}$ ) is the solution to given system of equations.

Verification:

L.H.S. of (1) = x + 2y = 5 + 2 x $\frac{2}{2}$ = 5 + 2 = 7.

R.H.S. of (2) = x - 2y = 5 - 2 x $\frac{2}{2}$ = 5 - 2 = 3.