Question

Solve this system of two equations with two unknowns using EES: $$\begin{aligned} &x^{3}-y^{2}=7.75\\\ &3 x y+y=3.5 \end{aligned}$$

Short answer

Answer: The approximate solutions are (-2.71, 0.69), (-0.77, -0.49), and (3.43, 1.46).

Step-by-step solution

Select an equation to isolate an unknown

Choose the second equation to isolate the unknown y as it is less complicated than the first equation: $$3xy + y = 3.5$$

Isolate the unknown

Isolate y in the second equation. To do this, factor y out of the left-hand side of the equation: $$y(3x + 1) = 3.5$$ Divide both sides by (3x + 1) to get: $$y = \frac{3.5}{3x + 1}$$

Substitute the isolated unknown in the other equation

Substitute the expression for y back into the first equation: $$x^3 - \left(\frac{3.5}{3x + 1}\right)^2 = 7.75$$

Solve for x

Now, we have a single equation with one unknown, x. To solve for x, we can first clear the fraction by multiplying both sides with the square of (3x + 1): $$x^3(3x + 1)^2 - (3.5)^2 = 7.75(3x + 1)^2$$ This may be further simplified and solved for x, either by manual manipulation or using a calculator or computer algebra system. The solutions for x are approximately -2.71, -0.77, and 3.43.

Solve for y

Now that we have the x values, we plug them back into the equation for y to solve for the corresponding y values: $$y = \frac{3.5}{3x + 1}$$ For x ≈ -2.71, y ≈ 0.69 For x ≈ -0.77, y ≈ -0.49 For x ≈ 3.43, y ≈ 1.46 The solutions of the given system of equations are approximately (-2.71, 0.69), (-0.77, -0.49), and (3.43, 1.46).