If x+y=40,y+z=58,z+x=52 then find the value of z?
Answers
Answered by
3
x+y=40 then ....
x=40-y
if z+x =52 THEN
Z+40-y=52
so z = 52-40+y = 12+y then
y+z=58
y+12+y=58
2y=58-12
=y=46\2
y=23
we known that
y+z =58 then
23+z=58
z= 58-23
z=35 ...................................
x=40-y
if z+x =52 THEN
Z+40-y=52
so z = 52-40+y = 12+y then
y+z=58
y+12+y=58
2y=58-12
=y=46\2
y=23
we known that
y+z =58 then
23+z=58
z= 58-23
z=35 ...................................
Answered by
1
x + y = 40..........(1)
y + z = 58⇒ y = 58 - z
z + x = 52...........(3)
sub (2) in (1)
we get
x + 58-z = 40
x-z = -18
z-x = 18...........(4)
solve (3) and (4)
we get ,
2z = 70
z = 35
y + z = 58⇒ y = 58 - z
z + x = 52...........(3)
sub (2) in (1)
we get
x + 58-z = 40
x-z = -18
z-x = 18...........(4)
solve (3) and (4)
we get ,
2z = 70
z = 35
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