solve equation: xy=80, x-y=11?

xy = 80 (solve by using substitution)
x - y = 11

x - y = 11
x = y + 11

xy = 80
(y + 11)y = 80
y^2 + 11y = 80
y^2 + 11y - 80 = 0
y^2 + 16y - 5y - 80 = 0
(y^2 + 16y) - (5y + 80) = 0
y(y + 16) - 5(y + 16) = 0
(y + 16)(y - 5) = 0

y + 16 = 0
y = -16

y - 5 = 0
y = 5

x - y = 11
x - (-16) = 11
x = 11 - 16
x = -5

x - y = 11
x - 5 = 11
x = 11 + 5
x = 16

∴ (x = -5, y = -16), (x = 16, y = 5)

x-y =11

so x = y +11

substitute this in first equation

xy = 80

(y+11)y = 80

y 2 +11y = 80

y 2 + 11 y - 80 = 0

solving for y,
y = -16 or 5

substitute in 2nd equation, x - y = 11
if y = -16 then,
x +16 = 11 or x = -5

or
if y = 5 then
x -5 = 11 or x = 16

so, there are two sets of answers.

ie x = 16 and y = 5

or x = -5 and y = -16

x= 16 y= 5
x= -5 y= -16

x - 80/x = 11

x ² - 80 = 11 x

x ² - 11 x - 80 = 0

( x - 16 ) ( x + 5 ) = 0

x = 16 , x = - 5
y = 5 , y = - 16

(16,5) , (- 5,- 16)

x-y=11 so...
y=x-11 pluy this into the other equation in place of y....
x(x-11)=80
x^2-11x=80
x^2-11x-80=0
(x-16) (x+5)=0
so x is either 16 or 5
i checked 5 its not the correct one

so plug x=16 in either equation to solve for y
xy=80
16y=80
y=5

hence x=16 and y=5

xy=80 gives x=80/y
put it in the 2nd equation which will give
80/y -y =11
or 80 -y^2 =11y
or y^2 +11y -80 =0
or (y+16)(y-5)=0 giving
y= -16 and 5.
Go back to x=80/y and put value of x obtained above.
x will be = 80/-16 =-5 for y=-16 and
x=80/5 =16 for y=5.
so the solution set is {(-5, -16), (16,5)}

x = 80/y

Thus, 80/y - y = 11

80 - y^2 = 11y

y^2 + 11y - 80 =0

Solve using the quadratic formula.

x=16 y=5

To solve this, you can either begin with the equation on the left, or the equation on the right.

To start with the equation on the right, add y to both sides, giving x = y + 11. Then substitute that for x in the left equation, giving (y + 11)y = 80. Then expand, subtract 80 from both sides, and solve the quadratic.

Good luck!

x=80/y, so 80/y -y=11

Solve for y, then plug it in and solve for x.