What is the value of u in the system of equations 3 (2u + v) = 7uv, 3 (u + 3v) = 11uv?
1
We are given a system of two equations with two variables, u and v:
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.
Let's assume \(u \neq 0\) and \(v \neq 0\). We can divide both sides of each equation by \(uv\).
Start with the first equation:
\[3 (2u + v) = 7uv\]Distribute the 3 on the left side:
\[6u + 3v = 7uv\]Divide both sides by \(uv\):
\[\frac{6u + 3v}{uv} = \frac{7uv}{uv}\] \[\frac{6u}{uv} + \frac{3v}{uv} = 7\]Cancel out common terms:
\[\frac{6}{v} + \frac{3}{u} = 7\]Now, transform the second equation:
\[3 (u + 3v) = 11uv\]Distribute the 3 on the left side:
\[3u + 9v = 11uv\]Divide both sides by \(uv\):
\[\frac{3u + 9v}{uv} = \frac{11uv}{uv}\] \[\frac{3u}{uv} + \frac{9v}{uv} = 11\]Cancel out common terms:
\[\frac{3}{v} + \frac{9}{u} = 11\]We now have a new system of equations in terms of \(1/u\) and \(1/v\):
Let \(x = \frac{1}{u}\) and \(y = \frac{1}{v}\). The system becomes a linear system in variables \(x\) and \(y\):
We can solve this linear system using methods like substitution or elimination. Let's use the elimination method.
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:
\[2 \times (9x + 3y) = 2 \times 11\] \[18x + 6y = 22\]Let's call this new equation Equation 3:
Now we subtract the first equation (\(3x + 6y = 7\)) from Equation 3:
\[(18x + 6y) - (3x + 6y) = 22 - 7\] \[18x + 6y - 3x - 6y = 15\] \[15x = 15\]Now, solve for \(x\):
\[x = \frac{15}{15}\] \[x = 1\]We defined \(x = \frac{1}{u}\). We found that \(x = 1\). So:
\[\frac{1}{u} = 1\]Solving for \(u\):
\[u = \frac{1}{1}\] \[u = 1\]We can also find \(y\) and consequently \(v\) by substituting \(x=1\) into either of the linear equations. Using \(3x + 6y = 7\):
\[3(1) + 6y = 7\] \[3 + 6y = 7\] \[6y = 7 - 3\] \[6y = 4\] \[y = \frac{4}{6}\] \[y = \frac{2}{3}\]Since \(y = \frac{1}{v}\), we have:
\[\frac{1}{v} = \frac{2}{3}\] \[v = \frac{3}{2}\]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.
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.
| Original Equation | Transformed Equation (\(x=1/u, y=1/v\)) |
|---|---|
| \(3(2u+v) = 7uv\) | \(3x + 6y = 7\) |
| \(3(u+3v) = 11uv\) | \(9x + 3y = 11\) |
| Step | Description | Detail |
|---|---|---|
| 1 | Analyze the system | Identify as non-linear due to the \(uv\) term. |
| 2 | Transform equations | Assuming \(u,v \neq 0\), divide by \(uv\) to get equations in terms of \(1/u\) and \(1/v\). |
| 3 | Substitute variables | Let \(x = 1/u\) and \(y = 1/v\) to form a linear system. |
| 4 | Solve linear system | Use elimination or substitution to find values for \(x\) and \(y\). |
| 5 | Convert back | Use \(u = 1/x\) and \(v = 1/y\) to find the original variables. |
Solving systems of non-linear equations can be more complex than linear systems. Different techniques are used depending on the form of the equations.
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).
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