(X-30) * [600 - (x-40) * 10] = 10000 bivariate linear equation (x-30)*[600-(x-40)*10]=10000 linear equation in two unknowns

(X-30) * [600 - (x-40) * 10] = 10000 bivariate linear equation (x-30)*[600-(x-40)*10]=10000 linear equation in two unknowns

600*（x-30）-10*（x-40）*（x-30）=10000
60x-1800-x²+70x-1200=1000
x²-130x+4000=0
（x-80）*（x-50）=0
So x = 50 or 80

The sum of X and y can be obtained from the quadratic equation x + y = 450 x (1-60%) = y (1-40%) - 30

x+y=450
Then y = 450-x
0.4x=0.6y-30
Substituting y = 450-x into
0.4x=270-0.6x-30
x=240
y=450-x=210

Solution equation: (X-30) [600-10 (X-400)] = 10000

（x-30)[600-10(x-400)]=10000
（X-30）(600-10X+4000）=10000
（X-30）（4600-10X）=10000
-10X^2+4900X=14800
X^2-490X=-14800
X^2-490X+245^2-245^2=-14800
(X-245)^2=45225
X-245 = positive and negative 15 lower 201
X = 245 plus or minus 15
After half an hour, I should think about this problem myself

How about [600-10 × (x-40)] (X-30) = 10000

X-30) [600 - (x-40) × 10] = 10000,10 (X-30) (100-x) = 10000, divide both sides by 10 to get (X-30) (100-x) = 1000, multiplication of polynomial 100x-x ^ 2-3000 + 30x = 1000, - x ^ 2 + 130x-3000 = 1000x ^ 2-130x + 4000 = 0 (x-50) (x-80) = 0x1 = 50, X2 = 80