Let x and y = the numbers
$x + y = 2$ → Equation (1)
$1 + y' = 0$
$y' = -1$
$z = x^3 + y^2$ → Equation (2)
$dz/dx = 3x^2 + 2y \, y' = 0$
$3x^2 + 2y(-1) = 0$
$y = \frac{3}{2}x^2$
From Equation (1)
$x + \frac{3}{2}x^2 = 2$
$2x + 3x^2 = 4$
$3x^2 + 2x - 4 = 0$
$x = 0.8685 \, \text{&} \, -1.5352$
Use
$x = 0.8685$
$y = \frac{3}{2}(0.8685^2)$
$y = 1.1315$
$z = 0.8685^3 + 1.1315^2$
$z = 1.9354$ answer