linear using substitution method
x + y = 8 ----equation 1
x + x/y = 5 ----equation 2
rearranging equation 1 you'll have
x = 8 - y ---equation 3
substitute equation 3 to equation 2
x + x/y = 5
(8 - y) + ( 8 - y)/y = 5
getting the LCD
[y(8-y) + (8-y)]/y = 5
cross multiplying y
8y - y² + 8 - y = 5y
-y² + 8y - 5y - y + 8 = 0
-y² +2y + 8 = 0
dividing the whole equation with -1 you'll have
y² - 2y - 8 = 0
(y - 4)(y + 2) = 0
y = 4, y = -2
substitute y=4 to equation 3
x = 8 - y
x = 8 - 4
x = 4
substituting y = -2 to equation 3
x = 8 - y
x = 8 - (-2)
x = 10
checking if all the values of x and y are true
from equation 1
x + y = 8
y=4,x=4
4 + 4 =8
8 = 8 ok!
y=-2, x=10
10+ -2 = 8
8 = 8. ok!
from equation 2
x + x/y = 5
y=4, x=4
4 + 4/4 = 5
4 + 1 = 5
5=5. ok!
y=-2, x =10
10 + 10/-2 = 5
10 - 5 = 5
5=5. ok!
therefore the values are
(4,4) and (10,-2)