Solve the following paris of linear (simultaneous) equation using method of elimination by substution:
2x + 7y = 39
3x + 5y = 31

Answer

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

### Step 1: Write down the equations
The given equations are:
1. \( 2x + 7y = 39 \) (Equation 1)
2. \( 3x + 5y = 31 \) (Equation 2)

### Step 2: Solve one equation for one variable
Let's solve Equation 2 for \( x \):
\[ 3x + 5y = 31 \]
Subtract \( 5y \) from both sides:
\[ 3x = 31 - 5y \]
Now, divide by 3:
\[ x = \frac{31 - 5y}{3} \quad (Equation 3) \]

### Step 3: Substitute the expression for \( x \) into the other equation
Now, substitute Equation 3 into Equation 1:
\[ 2\left(\frac{31 - 5y}{3}\right) + 7y = 39 \]
Multiply through by 3 to eliminate the fraction:
\[ 2(31 - 5y) + 21y = 117 \]
Distribute \( 2 \):
\[ 62 - 10y + 21y = 117 \]
Combine like terms:
\[ 62 + 11y = 117 \]

### Step 4: Solve for \( y \)
Subtract 62 from both sides:
\[ 11y = 117 - 62 \]
\[ 11y = 55 \]
Now, divide by 11:
\[ y = \frac{55}{11} = 5 \]

### Step 5: Substitute back to find \( x \)
Now that we have \( y \), substitute \( y = 5 \) back into Equation 3 to find \( x \):
\[ x = \frac{31 - 5(5)}{3} \]
Calculate:
\[ x = \frac{31 - 25}{3} = \frac{6}{3} = 2 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = 5 \]