To solve the simultaneous equations using the elimination method, we have the following equations:
- \( 2u + 7v = 10 \) (Equation 1)
- \( 3u + v = 6 \) (Equation 2)
We can eliminate one of the variables by making the coefficients of \(v\) in both equations equal. Let's eliminate \(v\).
First, we can multiply Equation 2 by 7 to match the coefficient of \(v\) in Equation 1:
\[ 7(3u + v) = 7(6) \]
This gives us:
\[ 21u + 7v = 42 \quad \text{(Equation 3)} \]
Now we have:
- \( 2u + 7v = 10 \) (Equation 1)
- \( 21u + 7v = 42 \) (Equation 3)
Next, we can subtract Equation 1 from Equation 3:
\[ (21u + 7v) - (2u + 7v) = 42 - 10 \]
This simplifies to:
\[ 21u - 2u + 7v - 7v = 32 \] \[ 19u = 32 \]
Now, solve for \(u\):
\[ u = \frac{32}{19} \]
Next, we substitute \(u\) back into one of the original equations to find \(v\). We can use Equation 2:
\[ 3u + v = 6 \]
Substituting \(u\):
\[ 3\left(\frac{32}{19}\right) + v = 6 \] \[ \frac{96}{19} + v = 6 \]
To isolate \(v\), we subtract \(\frac{96}{19}\) from both sides:
\[ v = 6 - \frac{96}{19} \]
Converting 6 into a fraction with a denominator of 19:
\[ v = \frac{114}{19} - \frac{96}{19} \] \[ v = \frac{114 - 96}{19} \] \[ v = \frac{18}{19} \]
Thus, the solution to the simultaneous equations is:
\[ u = \frac{32}{19}, \quad v = \frac{18}{19} \]
Final Answer:
\((u, v) = \left( \frac{32}{19}, \frac{18}{19} \right)\)