steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Is there an x,y that solves the equations?
a = 7
b = 4
a = 2*x + y*b
b = x*y + a/4
a = 7
b = 4
a = 2*x + y*b
b = x*y + a/4
Adding, is there an analytical solution?Is there an x,y that solves the equations?
a = 7
b = 4
a = 2*x + y*b
b = x*y + a/4
Y=9/(4x)2*x + y*4 = 7 = a
Xy + 7/4 = 4 = b
2*x + y*4 = 7
4y = 7-2x
Y= 7/4 -x/2.
X=-2y+7/2
Xy + 7/4 = 4
Xy = 4 - 7/4
Y = 4/x - 7/4x (x!=0)...
X = 4/y - 7/4y (y!=0)...
(7/4 - x/2 = 4/x - 7/4x)*4x
7x - 2(x^2) = 16 - 7
- 2x^2 + 7x - 9 = 0 (has no real roots)
(-2y+7/2 = 4/y - 7/4y)*-4y
8y*y+14y=-16+7
8y*y+14y+23=0 (has no real roots)
From here it looks like I have to take -7*9/(y^2)-7*9/23*2=x and get that in terms of Y. Only one of the two quadrants it traverses (y>0) can intersect with the other function, where Y > 0 and X < 0, so the relation can be broken into two functions and the function of negative Y can be discarded.Y=9/(4x)2*x + y*4 = 7 = a
Xy + 7/4 = 4 = b
2*x + y*4 = 7
4y = 7-2x
Y= 7/4 -x/2.
X=-2y+7/2
Xy + 7/4 = 4
Xy = 4 - 7/4
Y = 4/x - 7/4x (x!=0)...
X = 4/y - 7/4y (y!=0)...
(7/4 - x/2 = 4/x - 7/4x)*4x
7x - 2(x^2) = 16 - 7
- 2x^2 + 7x - 9 = 0 (has no real roots)
(-2y+7/2 = 4/y - 7/4y)*-4y
8y*y+14y=-16+7
8y*y+14y+23=0 (has no real roots)
X=9/(4y)
- 2x^2 + 7x - 9 = 0
- 2x^2 + 7(9/(4y)) - 9 = 0
2x^2 + 9 = 7(9/(4y))
2*4x^2/9*7 + 4/7= (1/4*y)
1/(2*4x^2/9*7) + 1/(4/7)= y
9*7/2*4x^2 + 7/4= y
8y*y+23=-14(9/(4x))
2y*y+23*4=-14(9/x)
-(y*y)/7*9-23*2/7*9=(1/x)
-7*9/(y^2)-7*9/23*2=x
...
These two functions appear to intersect in the upper left quadrant, which is where I expect the solution must be.
A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form Ax+By+C=0. Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.