Simultaneous Equation--why have they got two sets of answers? please explain

3x-5y=16
xy=7
x=7/y
3x=21/y
21/y-5y=16
21-5y=16y
21=21y
y=1, substituting in x=7/y, x=7

but the answers at the book of the back says
x=7. y=1
x=-5/8. y=-21/5

what am I missing here?

thank you

Ya there's one mistake..
after substituting the value of 3x we ll get
21/y - 5y = 16
21/y = 16 + 5y (Shifting -5y to RHS)
21 = 16y + 5y^2 (Taking LCM)
5y^2 + 16y - 21 = 0 (Quadratic Equation)
5y^2 - 5y + 21y - 21 = 0
5y (y-1) + 21 (y-1) = 0
(y-1) (5y+21) = 0
Either y-1 = 0 or 5y+21 = 0
y = 1 y = -21/5
Substituting the value of y, u ll get the values of x as well.
If still needs help then pls do tell..
Cheers!!!!!

you need to solve a quadratic but you were on the correct lines then made a little slip.

3x - 5y = 16 ..........................( 1 )
xy = 7 ..............................( 2 )
so from ( 2 ) x = 7/y
and 3x = 21/y substitute in ( 1 )
21/y -5y = 16 as before and so far so good. Now multiply thro' by y
so 21 - 5y^2 = 16y this is where you made a little mistake
giving 5y^2 +16y - 21 = 0
using the formula to solve ( easiest ) gives y = -21/5 and y = 1 from which you can easily calculate the x values

When you times '21/y - 5y = 16' by y to get 21-5y = 16y, you haven't multiplied the -5y by y.
It should read there:
21/y-5y=16
21-5y^2=16y

You know have a quadratic which when you solve comes up with two values for y. You can then substitute these into one of the original equations, giving you two sets of values.

Well i don't have time to solve this ...but i can tell u ... that when u are substituting x or y ... in any other equation .... the equation will become quadratic mean giving a variable power 2 ..... so it is impossible to get only 1 answer ... when u will solve it ..it would give u 2 answers of both variables.

21-5y=16y should be 21 - 5y^2 = 16y

This is a quadratic equation, not a linear equation. Book is correct!

Be careful.