Solving System of Equations: Finding the Value of u

We are given a system of two equations with two variables, u and v:

1. \(3 (2u + v) = 7uv\)
2. \(3 (u + 3v) = 11uv\)

These are non-linear equations because of the \(uv\) term. To solve them, we can try to transform them into a simpler form, possibly a system of linear equations. A common technique when variables appear in a product like this is to divide by the product of the variables, assuming the variables are not zero.

Transforming the System of Equations

Let's assume \(u \neq 0\) and \(v \neq 0\). We can divide both sides of each equation by \(uv\).

Equation 1 Transformation:

Start with the first equation:

Distribute the 3 on the left side:

Divide both sides by \(uv\):

Cancel out common terms:

Equation 2 Transformation:

Now, transform the second equation:

Distribute the 3 on the left side:

Divide both sides by \(uv\):

Cancel out common terms:

Forming a Linear System

We now have a new system of equations in terms of \(1/u\) and \(1/v\):

1. \(\frac{3}{u} + \frac{6}{v} = 7\)
2. \(\frac{9}{u} + \frac{3}{v} = 11\)

Let \(x = \frac{1}{u}\) and \(y = \frac{1}{v}\). The system becomes a linear system in variables \(x\) and \(y\):

1. \(3x + 6y = 7\)
2. \(9x + 3y = 11\)

We can solve this linear system using methods like substitution or elimination. Let's use the elimination method.

Solving the Linear System using Elimination

We want to eliminate one of the variables, say \(y\). We can multiply the second equation (\(9x + 3y = 11\)) by 2 so that the coefficient of \(y\) becomes 6, matching the first equation:

Let's call this new equation Equation 3:

3. \(18x + 6y = 22\)

Now we subtract the first equation (\(3x + 6y = 7\)) from Equation 3:

Now, solve for \(x\):

Finding the Value of u

We defined \(x = \frac{1}{u}\). We found that \(x = 1\). So:

Solving for \(u\):

We can also find \(y\) and consequently \(v\) by substituting \(x=1\) into either of the linear equations. Using \(3x + 6y = 7\):

Since \(y = \frac{1}{v}\), we have:

So, the solution we found is \(u=1\) and \(v=3/2\).

Note: The division by \(uv\) assumed \(u \neq 0\) and \(v \neq 0\). If \(u=0\), the original equations become \(3(v) = 0 \implies v=0\) and \(3(3v)=0 \implies 9v=0 \implies v=0\). So, \((0,0)\) is also a solution to the original system. However, the options provided for the value of u are non-zero, indicating the question seeks the non-zero solution.

Conclusion on the Value of u

Based on the method of transforming the system by dividing by \(uv\), we found a non-zero solution where the value of u is 1.

The value of u is 1.

Revision Table: Key Steps to Solve the System

Additional Information on Solving Non-Linear Systems

Solving systems of non-linear equations can be more complex than linear systems. Different techniques are used depending on the form of the equations.

• Substitution: If one equation can be easily solved for one variable in terms of the other, substitute this expression into the other equation.
• Elimination: Sometimes, equations can be added or subtracted to eliminate a variable, similar to linear systems, but this often requires transforming the equations first, as done in this problem.
• Graphical Methods: For simple systems, plotting the graphs of the equations can show intersection points which are the solutions. However, this is not always precise.
• Special Transformations: As seen here, specific forms (like terms being products of variables) might suggest specific transformations (like dividing by the product) to simplify the system.

It's important to check all potential solutions in the original equations, especially when transformations (like division) are used, which might exclude certain cases (like variables being zero).