Question

solve the following pair of equation by the method of elimination7(y+3)-2(x+2)=14 ; 4(y-2)+3(x-3)=2
please answer....please

Answer 1

Hi !

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

7y + 21 - 2x -4 = 14

7y - 2x + 17 = 14

7y - 2x = -3 ------> [1]


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

4y - 8 + 3x - 9 = 2

4y + 3x - 17 = 2

4y + 3x = 19 -----> [2]




7y - 2x = -3

4y + 3x = 19

lets make the coefficients of the variable x equal .

3(7y - 2x = -3 )
2(4y + 3x = 19 )

the equation becomes :

21y - 6x = -9 ----> [3]
8y + 6x = 38 ----> [4]

adding equations 3 and 4 ,


21y - 6x = -9
+ 8y + 6x = 38
------------------------
29 y = 29

y = 29/29 = 1

put this value in equation [2]


4y + 3x = 19

4 + 3x = 19

3x = 15
x = 15/3
x = 5


hence ,

x = 5 , y = 1