1. ## simultaneous equation

$\frac{3x+y+1}{8}=\frac{x-y}{5}=\frac{x^2-y^2}{5}$

$\frac{3x+y+1}{8}=\frac{x-y}{5}$

$7x+13y=-5$ --- 1
$

\frac{x-y}{5}=\frac{x^2-y^2}{5}
$

$x-y=(x+y)(x-y)$

x+y=1 (I am not sure if i can do this) ----2

So solving 1 and 2 , i gt x=3 and y=-2 .

It looks to me that there is another pair of values x and y but just couldn't find them . I am confusing my self at times . How can i find the other pair ?

2. Originally Posted by thereddevils
$\frac{3x+y+1}{8}=\frac{x-y}{5}=\frac{x^2-y^2}{5}$

$\frac{3x+y+1}{8}=\frac{x-y}{5}$

$7x+13y=-5$ --- 1
$

\frac{x-y}{5}=\frac{x^2-y^2}{5}
$

$x-y=(x+y)(x-y)$

x+y=1 (I am not sure if i can do this) ----2

So solving 1 and 2 , i gt x=3 and y=-2 .

It looks to me that there is another pair of values x and y but just couldn't find them . I am confusing my self at times . How can i find the other pair ?
Hi

Better factor : $x-y=(x+y)(x-y)$

$(x-y)(x+y-1)=0$

2 possibilities :
or x+y-1 = 0 (you have solved this case)
or x-y = 0 (you have forgotten this one)