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Chapter 1: Problem 114

Solve this system of two equations with two unknowns using EES: $$ \begin{aligned} x^{3}-y^{2} &=10.5 \\ 3 x y+y &=4.6 \end{aligned} $$

Short Answer

Expert verified
Based on the step-by-step solution above, provide a short answer to the problem: To solve the system of equations using the Elimination by Expansion System (EES) method, first rearrange both equations with one variable expressed in terms of the other. Then expand one of the equations to eliminate one of the variables and substitute the remaining equation to find the other variable's value. Finally, solve for the other variable, obtaining the values of x and y as the solutions to the system of equations. Note that a numerical method may be required to solve for x in this particular problem, and once x is found, simply substitute it back into the equation for y to find its value.

Step by step solution

01

Rearrange both equations

We can rearrange the first equation to express x in terms of y and the second equation to express y in terms of x: First equation: $$ x^3 = y^2 + 10.5 \Rightarrow x = \sqrt[3]{y^2 + 10.5} $$ Second equation: $$ 3xy + y = 4.6 \Rightarrow y(3x + 1) = 4.6 \Rightarrow y = \frac{4.6}{3x + 1} $$
02

Expand one of the equations

We will expand the first equation by squaring both sides in order to eliminate y: $$ (\sqrt[3]{y^2 + 10.5})^2 = y^{\frac{4}{3}} + 2y^{\frac{2}{3}}(10.5)^\frac{1}{3} + 10.5^{\frac{2}{3}} $$
03

Substitute the remaining equation

Now, we will substitute y in the expanded equation with the expression for y in terms of x from the second equation: $$ (\sqrt[3]{(\frac{4.6}{3x + 1})^2 + 10.5})^2 = (\frac{4.6}{3x + 1})^{\frac{4}{3}} + 2(\frac{4.6}{3x + 1})^{\frac{2}{3}}(10.5)^\frac{1}{3} + 10.5^{\frac{2}{3}} $$
04

Solve for x and y

At this point, we have a single equation with one unknown, x. To solve for x, use a numerical method such as the bisection method or Newton-Raphson method. Once the value of x is found, substitute it back into the expression for y in terms of x in the second equation: $$ y = \frac{4.6}{3x + 1} $$ Finally, we get the values of x and y, which are the solutions to the given system of equations.

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Most popular questions from this chapter

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

Consider a person whose exposed surface area is \(1.7 \mathrm{~m}^{2}\), emissivity is \(0.5\), and surface temperature is \(32^{\circ} \mathrm{C}\). Determine the rate of heat loss from that person by radiation in a large room having walls at a temperature of \((a) 300 \mathrm{~K}\) and (b) \(280 \mathrm{~K}\). Answers: (a) \(26.7 \mathrm{~W}\), (b) \(121 \mathrm{~W}\)

An aluminum pan whose thermal conductivity is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has a flat bottom with diameter \(15 \mathrm{~cm}\) and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(1400 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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