Solution:
- We are given the system of linear equations:
\( 2x - 7y = 20 \) (Equation 1)
\( 4x + 9y = 10 \) (Equation 2) - To solve this system, we can use the elimination method. Multiply Equation 1 by 2 to make the coefficients of x the same:
\( 2 \times (2x - 7y) = 2 \times 20 \)
\( 4x - 14y = 40 \) (Equation 3) - Now subtract Equation 2 from Equation 3:
\( (4x - 14y) - (4x + 9y) = 40 - 10 \)
\( 4x - 14y - 4x - 9y = 30 \)
\( -23y = 30 \)
\( y = \frac{30}{-23} = -\frac{30}{23} \) - Substitute the value of y back into Equation 1:
\( 2x - 7(-\frac{30}{23}) = 20 \)
\( 2x + \frac{210}{23} = 20 \)
\( 2x = 20 - \frac{210}{23} \)
\( 2x = \frac{20 \times 23}{23} - \frac{210}{23} \)
\( 2x = \frac{460 - 210}{23} \)
\( 2x = \frac{250}{23} \)
\( x = \frac{250}{23 \times 2} = \frac{125}{23} \) - Now, let's find the value of 6x:
\( 6x = 6 \times \frac{125}{23} = \frac{750}{23} \)
Ответ: \( x = \frac{125}{23} \), \( y = -\frac{30}{23} \), \( 6x = \frac{750}{23} \).