Question

7(y+3)-2 (x+2)=14,4(y-2)+3(x-3)=2

Answer 1

7(y+3)-2(x+2)=14

=>7y+21-2x-4=14

=>7y+17-2x=14

=>7y-2x=14-17

=>7y-2x=-3 ------1st equation

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

=>4y-8+3x-9=2

=>4y+3x-17=2

=>4y+3x=2+17

=>4y+3x=19 ------2nd equation

3[7y-2x=-3]

2[4y+3x=19]

21y-6x=-9 -------3rd equation

8y+6x=38  -------4th equation

by elimination 

adding eqation 3 and 4

we get 29y=29

=>y=1

put 'y' is equal to 1 in equation 3

7-2x=-3

-2x=-10

=>x=5

therefore y=1,x=5

Answer 2

Simplify given equations

7(y+3)−2(x+2)=14

or 7y−2x=−3

and 4(y−2)+3(x−3)=2

or 4y+3x=19

new set of equations is :

7y−2x=−3...(1)

4y+3x=19...(2)

Let us use the elimination method to solve the given system of equation.

Multiply (1) by 3 and (2) by 2. And add both the

equations.

Adding  

29y=29⇒y=

29

29

​

 =1

21y−6x=−98y+6x=38

​

 

Substitute the value of x in equation (1), we have

7(1)−2x=−3

x=5

Answer: x=5 and y=1

Answer 3

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