Question

3 x-5y+4
2(x+1)+3(y-3)​

Answer 1

Answer:

linear equation on two variable

Step-by-step explanation:

3x - 5y + 2(x+1) + 3(y-3)

Answer 2

Answer:

thank you

Step-by-step explanation:

(i) We have, 3x−5y−4=0

⇒3x−5y=4...(i)

Again 9x=2y+7

⇒9x−2y=7...(ii)

By Elimination Method:

Multiplying equation (i) by 3, we get

9x−15y=12...(iii)

Subtracting (ii) from (iii), we get

   9x−15y=12

   9x−2y=7

     

  −13y=5      

​  

 

⇒y=−  

13

5

​  

 

Putting the value of equation (ii), we get

9x−2(−  

13

5

​  

)=7

⇒9x+  

13

10

​  

=7

⇒9x=7−  

13

10

​  

 

⇒9x=  

13

91−10

​  

 

⇒9x=  

13

81

​  

 

⇒x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

,y=−  

13

5

​  

 

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

4+5y

​  

 

Substituting the value of x in equation (ii), we get

9×(  

3

4+5y

​  

)−2y=7

⇒3×(4+5y)−2y=7

⇒12+15y−2y=7

⇒13y=7−12

∴y=−  

13

5

​  

 

Putting the value of y in equation (i), we have

3x−5×(−  

13

5

​  

)=4

⇒3x+  

13

25

​  

=4

⇒3x=4−  

13

25

​  

 

⇒3x=  

13

27

​  

 

∴x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

 , y=−  

13

5

​  

.

(ii) We have,  

2

x

​  

+  

2

2y

​  

=−1

⇒  

6

3x+4y

​  

=−1

∴3x+4y=−6...(i)

And x−  

2

y

​  

=3⇒  

3

3x−y

​  

=3

∴3x−y=9...(ii)

By Elimination Method:

Subtracting (ii) from (i), we get

5y=−15 ⇒y=−  

5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+r×(−3)=−6

⇒3x−12=−6

⇒3x=−6+12⇒3x=6

Hence, Solution is x=2 , y=−3

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

−6−4y

​  

 

Substituting the value of x in equation (ii) from equation (i), we get

3×(  

3

−6−4y

​  

)−y=9

⇒−6−4y−y=9

⇒−6−5y=9

⇒−5y=9+6=15

∴y=  

−5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+×(−3)=−6

⇒3x−12=−6

⇒3x=12−6=6

∴x=  

3

6

​  

=2

Hence, the required solution is x=2,y=−3(i) We have, 3x−5y−4=0

⇒3x−5y=4...(i)

Again 9x=2y+7

⇒9x−2y=7...(ii)

By Elimination Method:

Multiplying equation (i) by 3, we get

9x−15y=12...(iii)

Subtracting (ii) from (iii), we get

   9x−15y=12

   9x−2y=7

     

  −13y=5      

​  

 

⇒y=−  

13

5

​  

 

Putting the value of equation (ii), we get

9x−2(−  

13

5

​  

)=7

⇒9x+  

13

10

​  

=7

⇒9x=7−  

13

10

​  

 

⇒9x=  

13

91−10

​  

 

⇒9x=  

13

81

​  

 

⇒x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

,y=−  

13

5

​  

 

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

4+5y

​  

 

Substituting the value of x in equation (ii), we get

9×(  

3

4+5y

​  

)−2y=7

⇒3×(4+5y)−2y=7

⇒12+15y−2y=7

⇒13y=7−12

∴y=−  

13

5

​  

 

Putting the value of y in equation (i), we have

3x−5×(−  

13

5

​  

)=4

⇒3x+  

13

25

​  

=4

⇒3x=4−  

13

25

​  

 

⇒3x=  

13

27

​  

 

∴x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

 , y=−  

13

5

​  

.

(ii) We have,  

2

x

​  

+  

2

2y

​  

=−1

⇒  

6

3x+4y

​  

=−1

∴3x+4y=−6...(i)

And x−  

2

y

​  

=3⇒  

3

3x−y

​  

=3

∴3x−y=9...(ii)

By Elimination Method:

Subtracting (ii) from (i), we get

5y=−15 ⇒y=−  

5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+r×(−3)=−6

⇒3x−12=−6

⇒3x=−6+12⇒3x=6

Hence, Solution is x=2 , y=−3

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

−6−4y

​  

 

Substituting the value of x in equation (ii) from equation (i), we get

3×(  

3

−6−4y

​  

)−y=9

⇒−6−4y−y=9

⇒−6−5y=9

⇒−5y=9+6=15

∴y=  

−5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+×(−3)=−6

⇒3x−12=−6

⇒3x=12−6=6

∴x=  

3

6

​  

=2

Hence, the required solution is x=2,y=−3(i) We have, 3x−5y−4=0

⇒3x−5y=4...(i)

Again 9x=2y+7

⇒9x−2y=7...(ii)

By Elimination Method:

Multiplying equation (i) by 3, we get

9x−15y=12...(iii)

Subtracting (ii) from (iii), we get

   9x−15y=12

   9x−2y=7

     

  −13y=5      

​  

 

⇒y=−  

13

5

​  

 

Putting the value of equation (ii), we get

9x−2(−  

13

5

​  

)=7

⇒9x+  

13

10

​  

=7

⇒9x=7−  

13

10

​  

 

⇒9x=  

13

91−10

​  

 

⇒9x=  

13

81

​  

 

⇒x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

,y=−  

13

5

​  

 

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

4+5y

​  

 

Substituting the value of x in equation (ii), we get

9×(  

3

4+5y

​  

)−2y=7

⇒3×(4+5y)−2y=7

⇒12+15y−2y=7

⇒13y=7−12

∴y=−  

13

5

​  

 

Putting the value of y in equation (i), we have

3x−5×(−  

13

5

​  

)=4

⇒3x+  

13

25

​  

=4

⇒3x=4−  

13

25

​  

 

⇒3x=  

13

27

​  

 

∴x=  

13

9

​  

 

Hence, the required solution is x=  

13

9

​  

 , y=−  

13

5

​  

.

(ii) We have,  

2

x

​  

+  

2

2y

​  

=−1

⇒  

6

3x+4y

​  

=−1

∴3x+4y=−6...(i)

And x−  

2

y

​  

=3⇒  

3

3x−y

​  

=3

∴3x−y=9...(ii)

By Elimination Method:

Subtracting (ii) from (i), we get

5y=−15 ⇒y=−  

5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+r×(−3)=−6

⇒3x−12=−6

⇒3x=−6+12⇒3x=6

Hence, Solution is x=2 , y=−3

By Substitution Method:

Expressing x in terms of y from equation (i), we have

x=  

3

−6−4y

​  

 

Substituting the value of x in equation (ii) from equation (i), we get

3×(  

3

−6−4y

​  

)−y=9

⇒−6−4y−y=9

⇒−6−5y=9

⇒−5y=9+6=15

∴y=  

−5

15

​  

=−3

Putting the value of y in equation (i), we get

3x+×(−3)=−6

⇒3x−12=−6

⇒3x=12−6=6

∴x=  

3

6

​  

=2

Hence, the required solution is x=2,y=−3hgujhmn