Question - System of Linear Equations Problem
Solution:
The given system of equations is:
\[\begin{cases} 3(2x - 3y) - 4(x + y) = 7x - 3y - 5 \\ 7(x - 2y) - 5(2x - y) = -9x + 9y + 7 \end{cases}\]
First, expand and simplify both equations:
Equation 1: \(6x - 9y - 4x - 4y = 7x - 3y - 5\)
Rearrange and combine like terms:
\(2x - 13y = 7x - 3y - 5 \Rightarrow -5x + 10y = -5\)
Divide by -5 to simplify the equation:
\(x - 2y = 1\) ... (i)
Equation 2: \(7x - 14y - 10x + 5y = -9x + 9y + 7\)
Rearrange and combine like terms:
\(-3x - 9y = -9x + 9y + 7 \Rightarrow 6x - 18y = 7\)
Divide by 6 to simplify the equation:
\(x - 3y = \dfrac{7}{6}\) ... (ii)
Now we can solve the system using equations (i) and (ii):
From equation (i):
\(x = 2y + 1\)
Substitute \(x\) from equation (i) into equation (ii):
\(2y + 1 - 3y = \dfrac{7}{6}\)
\(-y + 1 = \dfrac{7}{6}\)
\(-y = \dfrac{7}{6} - 1\)
\(-y = \dfrac{1}{6}\)
\(y = -\dfrac{1}{6}\)
Now substitute \(y\) into equation (i):
\(x = 2(-\dfrac{1}{6}) + 1\)
\(x = -\dfrac{1}{3} + 1\)
\(x = \dfrac{2}{3}\)
The solution to the system of equations is \(x = \dfrac{2}{3}\), \(y = -\dfrac{1}{6}\).
Tutor Solution:
The system of equations from the image is:
\[ \begin{align*} 3(2x - 3y) - 4(x + y) &= 1x - 3y - 5 \\ 7(x - 2y) - 5(2x - y) &= -9x + 19y + 7 \end{align*} \]Let's solve each equation separately.
For the first equation:
\[ \begin{align*} 3(2x - 3y) - 4(x + y) &= x - 3y - 5 \\ 6x - 9y - 4x - 4y &= x - 3y - 5 \\ (6x - 4x - x) + (-9y - 4y + 3y) &= -5 \\ 2x - 10y &= -5 \quad \text{(Equation 1)} \end{align*} \]Now, let's tackle the second equation:
\[ \begin{align*} 7(x - 2y) - 5(2x - y) &= -9x + 19y + 7 \\ 7x - 14y - 10x + 5y &= -9x + 19y + 7 \\ (-10x + 7x + 9x) + (-14y - 5y - 19y) &= 7 \\ 6x - 28y &= 7 \quad \text{(Equation 2)} \]We now have a simplified system of equations:
\[ \begin{align*} 2x - 10y &= -5 \quad \text{(Equation 1)} \\ 6x - 28y &= 7 \quad \text{(Equation 2)} \end{align*} \]To solve this system, we can use either the substitution method or the elimination method. I will use the elimination method. First, let's make the coefficient of \( x \) in both equations the same. We can multiply Equation 1 by 3 and Equation 2 by 1 and get:
\[ \begin{align*} 3(2x - 10y) &= 3(-5) \\ 6x - 30y &= -15 \quad \text{(Equation 3)} \end{align*} \]Now we subtract Equation 2 from Equation 3:
\[ \begin{align*} (6x - 30y) - (6x - 28y) &= -15 - 7 \\ 6x - 6x - 30y + 28y &= -22 \\ -2y &= -22 \\ y &= \frac{-22}{-2} \\ y &= 11 \end{align*} \]Now that we have the value of \( y \), we can substitute it back into either Equation 1 or Equation 2 to find \( x \). Let's use Equation 1:
\[ \begin{align*} 2x - 10y &= -5 \\ 2x - 10(11) &= -5 \\ 2x - 110 &= -5 \\ 2x &= -5 + 110 \\ 2x &= 105 \\ x &= \frac{105}{2} \\ x &= 52.5 \end{align*} \]Therefore, the solution to the system of equations is \( x = 52.5 \) and \( y = 11 \).
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