solve the linear equations x-2y=8 and 5x+3y=1 by algebraic method
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Answered by
147
x-2y=8 is given
multiply it with 5
then it will be 5x-10y=40 --------------- 1
and 2nd equation is 5x+3y=1 ----------------2
by 1 and 2 we get
5x - 10y = 40
(-)5x + 3y= 1
-------------------------
13y = 39
---------------------------
then y=39/13 = 3
y = 3
by substituting it in 2
5x+3(3)=1
5x=1-9
5x=8
x=8/5
multiply it with 5
then it will be 5x-10y=40 --------------- 1
and 2nd equation is 5x+3y=1 ----------------2
by 1 and 2 we get
5x - 10y = 40
(-)5x + 3y= 1
-------------------------
13y = 39
---------------------------
then y=39/13 = 3
y = 3
by substituting it in 2
5x+3(3)=1
5x=1-9
5x=8
x=8/5
Answered by
118
x-2y=8 is given
multiply it with 5
then it will be 5x-10y=40..........(i)
and 2nd equation is 5x+3y=1.........(ii)
by equation (i)and(ii) we get
3x-10y=40
(-)5x+3y=1
...................................
13y=39
then y=39/13=3
y=3
by subsiting it in 2
5x+3(3)=1
5x=1-9
5x=8
x=8/5
multiply it with 5
then it will be 5x-10y=40..........(i)
and 2nd equation is 5x+3y=1.........(ii)
by equation (i)and(ii) we get
3x-10y=40
(-)5x+3y=1
...................................
13y=39
then y=39/13=3
y=3
by subsiting it in 2
5x+3(3)=1
5x=1-9
5x=8
x=8/5
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