Solve the following paris of linear (simultaneous) equation using method of elimination by substution:
2(x - 3) + 3(y - 5) = 0
5(x - 1) + 4(y - 4) = 0

Answer

To solve the given pair of linear equations using the method of elimination by substitution, we will follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( 2(x - 3) + 3(y - 5) = 0 \)
2. \( 5(x - 1) + 4(y - 4) = 0 \)

First, we expand both equations:

**For the first equation:**
\[ 2(x - 3) + 3(y - 5) = 0 \\ 2x - 6 + 3y - 15 = 0 \\ 2x + 3y - 21 = 0 \\ \Rightarrow 2x + 3y = 21 \quad \text{(Equation 1)} \]

**For the second equation:**
\[ 5(x - 1) + 4(y - 4) = 0 \\ 5x - 5 + 4y - 16 = 0 \\ 5x + 4y - 21 = 0 \\ \Rightarrow 5x + 4y = 21 \quad \text{(Equation 2)} \]

### Step 2: Express one variable in terms of the other
We will express \( x \) in terms of \( y \) using Equation 1:
\[ 2x + 3y = 21 \\ 2x = 21 - 3y \\ x = \frac{21 - 3y}{2} \]

### Step 3: Substitute the expression for \( x \) into the second equation
Now, we substitute \( x \) in Equation 2:
\[ 5x + 4y = 21 \\ 5\left(\frac{21 - 3y}{2}\right) + 4y = 21 \]

### Step 4: Simplify and solve for \( y \)
Multiply through by 2 to eliminate the fraction:
\[ 5(21 - 3y) + 8y = 42 \\ 105 - 15y + 8y = 42 \\ 105 - 7y = 42 \]
Now, isolate \( y \):
\[ -7y = 42 - 105 \\ -7y = -63 \\ y = 9 \]

### Step 5: Substitute back to find \( x \)
Now that we have \( y \), substitute it back into the expression for \( x \):
\[ x = \frac{21 - 3(9)}{2} \\ x = \frac{21 - 27}{2} \\ x = \frac{-6}{2} \\ x = -3 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = -3, \quad y = 9 \]