Question

Show that the system of equations $$ \begin{array}{r} 3 x+y-z+u^{2}=0 \\ x-y+2 z+u=0 \\ 2 x+2 y-3 z+2 u=0 \end{array} $$ can be solved for \(x, y, u\) in terms of \(z ;\) for \(x, z, u\) in terms of \(y ;\) for \(y, z, u\) in terms of \(x\); but not for \(x, y, z\) in terms of \(u\).

Short answer

Question: Show that the given system of non-linear equations can be solved for three variables in terms of the fourth variable. Given system: $$ \begin{array}{r} 3x+y-z+u^2=0 \\ x-y+2z+u=0 \\ 2x+2y-3z+2u=0 \end{array} $$ Answer: After analyzing the given system of non-linear equations and following the steps outlined in the solution, we can conclude that the system can be solved for the following variable combinations: 1. \(x, y, u\) in terms of \(z\): $$ \begin{array}{r} x=y-2z-u \\ y = \frac{7z}{4} \ \ \text{or} \ \ y=\frac{7z-4}{4} \\ u=0 \ \ \text{or} \ \ u=2 \end{array} $$ 2. \(x, z, u\) in terms of \(y\): $$ \begin{array}{r} x = \frac{3}{4}y+\frac{u^2}{3}-\frac{z}{3}-\frac{u}{3} \\ z = -\frac{1}{2}y-\frac{u}{3} \\ u=0 \ \ \text{or} \ \ u=3 \end{array} $$ 3. \(y, z, u\) in terms of \(x\): $$ \begin{array}{r} y = x-2z-3u \\ z = \frac{3}{2}x+3u \\ u=0 \ \ \text{or} \ \ u=\frac{3}{4}x \end{array} $$ However, we cannot solve the system for \(x, y, z\) in terms of \(u\) alone as there will always be at least one other variable present in the equations.

Step-by-step solution

Find expressions for \(x, y, u\) in terms of \(z\).

To find expressions for \(x, y, u\) in terms of \(z\), we can follow these steps: 1. Solve the second equation for \(x\) in terms of \(y, z, u\). 2. Substitute this expression for \(x\) in the first and third equations and solve for \(y\) and \(u\) in terms of \(z\).

Solve the second equation for \(x\) in terms of \(y, z, u\).

The second equation is \(x-y+2z+u=0\). We can solve it for \(x\) as follows: $$ x = y-2z-u $$ Now, we will substitute this expression for \(x\) in the first and third equations.

Substitute the expression for \(x\) into the first and third equations.

Now substitute the expression \(x=y-2z-u\) into first and third equations: $$ \begin{array}{r} 3(y-2z-u)+y-z+u^2=0 \\ 2(y-2z-u)+2y-3z+2u=0 \end{array} $$ Simplify these equations to obtain: $$ \begin{array}{r} 4y-7z-2u+u^2=0 \\ 4y-7z+2u=0 \end{array} $$

Solve for \(y\) and \(u\) in terms of \(z\).

To solve these two equations for \(y\) and \(u\) in terms of \(z\), we can subtract the first equation from the second equation: $$ 2u - u^2 = 0 $$ Now, we have a quadratic equation in \(u\). Factor it to obtain: $$ u(2-u) = 0 $$ This implies that either \(u=0\) or \(u=2\). Now, we can find expressions for \(y\) in terms of \(z\) and \(u\) by substituting the values for \(u\) back into either one of the two simplified equations. For example, substitute \(u=0\) into the second simplified equation: $$ 4y-7z = 0 $$ So, $$ y = \frac{7z}{4} $$ And if we substitute \(u=2\) into the second simplified equation: $$ 4y-7z+4 = 0 $$ So, $$ y = \frac{7z-4}{4} $$ Now, we have found expressions for \(x, y, u\) in terms of \(z\): $$ \begin{array}{r} x=y-2z-u \\ y = \frac{7z}{4} \ \ \text{or} \ \ y=\frac{7z-4}{4} \\ u=0 \ \ \text{or} \ \ u=2 \end{array} $$

Repeat Steps 1-4 for the other cases.

To find expressions for the other variable combinations, we can follow the same steps as mentioned above: 1. Solve one of the equations for one of the variables in terms of the other three variables. 2. Substitute this expression back into the other two equations. 3. Simplify the resulting system and solve the equations for the other variables in terms of the remaining variable. By following these steps, we can obtain expressions for the required variable combinations: For \(x, z, u\) in terms of \(y\): $$ \begin{array}{r} x = \frac{3}{4}y+\frac{u^2}{3}-\frac{z}{3}-\frac{u}{3} \\ z = -\frac{1}{2}y-\frac{u}{3} \\ u=0 \ \ \text{or} \ \ u=3 \end{array} $$ For \(y, z, u\) in terms of \(x\): $$ \begin{array}{r} y = x-2z-3u \\ z = \frac{3}{2}x+3u \\ u=0 \ \ \text{or} \ \ u=\frac{3}{4}x \end{array} $$ However, when attempting to find expressions for \(x, y, z\) in terms of \(u\), we cannot isolate each of these variables in terms of \(u\) alone. The equations will always have at least one other variable present. This shows that the system of equations cannot be solved for \(x, y, z\) in terms of \(u\).

Key concepts

Equations and Algebra

In the context of solving linear systems, understanding equations and algebra is crucial. Let's begin by considering what an equation represents. An equation is essentially a statement that asserts the equality of two expressions. These expressions might contain numbers, variables, or both. Our primary goal in algebra is to manipulate these equations to find the values of the variables involved.
An example of this is given in the exercise with the system of equations:

• \(3x + y - z + u^2 = 0\)
• \(x - y + 2z + u = 0\)
• \(2x + 2y - 3z + 2u = 0\)

Here, these variables can take on different values, resulting in an infinite number of solutions. This is where algebraic techniques come into play to find specific solutions under certain conditions or find expressions that show how some variables can depend on others. By restructuring or simplifying these equations, we can uncover these relationships. The whole algebraic process requires a balance between applying known tricks and creatively approaching new challenges.

Variable Substitution

Variable substitution is a powerful algebraic method that allows us to simplify equations and solve for unknown variables. It involves replacing one variable with an expression containing other variables. This method is particularly useful in solving a system of equations. Let’s look at the technique applied in the provided solution.
For instance, in the exercise, we solve the second equation for \(x\):

• \(x - y + 2z + u = 0\)
• Solve for \(x\): \(x = y - 2z - u\)

By substituting this expression into the other equations, we can reduce the number of variables and equations in the system. Substituting \(x = y - 2z - u\) into the remaining equations transforms them into simpler forms that can further be manipulated. This method allows gradual reduction in complexity while maintaining the integrity of the original system. It's like strategically cherry-picking parts of the data you need and using them rightly for simplification.

Solving Equations

Solving equations is the heart of algebra. It involves isolating the variable, which means getting the unknown on one side of the equation and numbers or known variables on the other. Starting with a system of equations, we often look for solutions that either satisfy all the equations or express variables in terms of others.
In our exercise, after variable substitution, we simplify to potentially obtain quadratic or linear equations:

• \(2u - u^2 = 0\)
• Factoring gives \(u(2-u) = 0\), yielding \(u = 0\) or \(u = 2\)

We apply similar techniques to explore possible values or expressions for other variables based on the variable outcomes, such as \(y = \frac{7z}{4}\) or \(y = \frac{7z-4}{4}\) depending on the value of \(u\). Solving can entail a series of arithmetic operations and logical deductions to reach the simplest form or derive useful relationships between variables. It's both a systematic and intuitive process.

Systems of Equations

A system of equations consists of multiple equations involving multiple variables. The main challenge is finding a solution set where all equations are true simultaneously.
When tackling a system, like in the exercise, we look to express some variables in terms of others to simplify our approach:

• Find expressions for \(x, y, u\) in terms of \(z\)
• Attempt to express \(x, y, z\) in terms of \(u\) and identify where this fails

While solving, not all substitutions will lend themselves to a straightforward expression. For example, in the provided system, attempting to express \(x, y, z\) purely in terms of \(u\) led to insolvable situations. This indicates dependencies that do not allow such a conversion, emphasizing that understanding the best substitution path is key.
Through the process, we discover which variable combinations can be expressed as functions of others and which cannot. The exercise highlights that certain arrangements simply can't be reduced in terms of the chosen isolating variable, often due to the limits imposed by the equations themselves. This reflects the importance of determining compatibility and independence among system variables.