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Solve this system of three equations with three unknowns using EES: $$\begin{aligned} &2 x-y+z=7\\\ &\begin{array}{c} 3 x^{2}+3 y=z+3 \\ x y+2 z=4 \end{array} \end{aligned}$$

Short Answer

Expert verified
System of equations: $$\begin{aligned} &2 x-y+z=7 \\\\ &3 x^{2}+3 y=z+3 \\\\ &x y+2 z=4 \end{aligned}$$ Solution: x = -1, y = 2, and z = 5.

Step by step solution

01

Analyze the equations

First, let's take a look at the system of equations: $$\begin{aligned} &2 x-y+z=7 \hspace{10mm}(1)\\\ &3 x^{2}+3 y=z+3 \hspace{10mm}(2)\\\ &x y+2 z=4 \hspace{10mm}(3) \end{aligned}$$
02

Eliminate one variable from the first equation

To eliminate one variable from the first equation, we will express the z variable from equation (1) and substitute it into the other equations: $$\begin{aligned} z = 7 - 2x + y \hspace{10mm}(4) \end{aligned}$$ Now we substitute equation (4) into equations (2) and (3).
03

Substitute z and create a new system of equations

Substituting z from equation (4) into equations (2) and (3), we get: $$\begin{aligned} &3 x^{2}+3 y = 7 - 2x + y + 3 \hspace{10mm}(5)\\\ &x y + 2(7 - 2x + y) = 4 \hspace{10mm}(6) \end{aligned}$$ Simplify the equations: $$\begin{aligned} &3 x^{2}- 2x + 2y = 10 \hspace{10mm}(7)\\\ &x y - 4x + 2y = -10 \hspace{10mm}(8) \end{aligned}$$
04

Solve for one variable

Now, we will solve equation (8) for one of the variables. We choose to solve for x: $$x(y-4) = -2(y-5) \implies x = \frac{-2(y-5)}{y-4}$$
05

Substitute the variable into the second equation

Now, we will substitute x into equation (7): $$3\left(\frac{-2(y-5)}{y-4}\right)^{2} - 2\left(\frac{-2(y-5)}{y-4}\right) + 2y = 10$$
06

Solve for y

Simplify and solve the above equation for y: $$3\left(\frac{4(y-5)^{2}}{(y-4)^{2}}\right) - 2\left(\frac{-2(y-5)}{y-4}\right) + 2y = 10 \implies y = 2$$
07

Solve for x

Substitute y back into the equation for x: $$x = \frac{-2(2-5)}{2-4} \implies x = -1$$
08

Solve for z

Finally, substitute x and y back into the equation for z: $$z = 7 - 2(-1) + 2 \implies z = 5$$
09

Write the solution

The solution of the system of equations is x = -1, y = 2, and z = 5.

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