To solve this system of equations, we can use substitution.
From the first equation, we have x*y = 320.
Now, let's solve for x in terms of y from the first equation:x*y = 320x = 320/y
Now, substitute x = 320/y into the second equation:(x+4)(y+1) = 420(320/y + 4)(y+1) = 420(320 + 4y)(y+1) = 420y320y + 320 + 4y^2 + 4y = 420y4y^2 + 4y - 100y + 320 = 04y^2 - 96y + 320 = 0y^2 - 24y + 80 = 0
Now, we can factor the quadratic equation:(y-20)(y-4) = 0
Therefore, y = 20 or y = 4
Now, substitute these values back into x = 320/y to find the corresponding values of x:If y = 20, then x = 320/20 = 16If y = 4, then x = 320/4 = 80
Therefore, the solutions to the system of equations are x = 16, y = 20 or x = 80, y = 4.
To solve this system of equations, we can use substitution.
From the first equation, we have x*y = 320.
Now, let's solve for x in terms of y from the first equation:
x*y = 320
x = 320/y
Now, substitute x = 320/y into the second equation:
(x+4)(y+1) = 420
(320/y + 4)(y+1) = 420
(320 + 4y)(y+1) = 420y
320y + 320 + 4y^2 + 4y = 420y
4y^2 + 4y - 100y + 320 = 0
4y^2 - 96y + 320 = 0
y^2 - 24y + 80 = 0
Now, we can factor the quadratic equation:
(y-20)(y-4) = 0
Therefore, y = 20 or y = 4
Now, substitute these values back into x = 320/y to find the corresponding values of x:
If y = 20, then x = 320/20 = 16
If y = 4, then x = 320/4 = 80
Therefore, the solutions to the system of equations are x = 16, y = 20 or x = 80, y = 4.