Solve the following pair of linear (simultaneous) equation using method of substitution.
`{:((2x+1)/(7)+(5y-3)/(3)=12),((3x+2)/(2)-(4y+3)/(9)=13)):}`

Answer

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

**Step 1: Write down the equations.**

The equations given are:
1. \(\frac{2x + 1}{7} + \frac{5y - 3}{3} = 12\) (Equation 1)
2. \(\frac{3x + 2}{2} - \frac{4y + 3}{9} = 13\) (Equation 2)

**Step 2: Simplify Equation 1 to express \(x\) in terms of \(y\).**

Starting with Equation 1:
\[ \frac{2x + 1}{7} + \frac{5y - 3}{3} = 12 \]

First, we can eliminate the fractions by multiplying through by the least common multiple (LCM) of the denominators, which is 21:
\[ 21 \left(\frac{2x + 1}{7}\right) + 21 \left(\frac{5y - 3}{3}\right) = 21 \cdot 12 \]
This simplifies to:
\[ 3(2x + 1) + 7(5y - 3) = 252 \]
Expanding this gives:
\[ 6x + 3 + 35y - 21 = 252 \]
Combining like terms:
\[ 6x + 35y - 18 = 252 \]
Adding 18 to both sides:
\[ 6x + 35y = 270 \]
Now, we can express \(x\) in terms of \(y\):
\[ 6x = 270 - 35y \]
\[ x = \frac{270 - 35y}{6} \]

**Step 3: Substitute \(x\) in Equation 2.**

Now, we substitute this expression for \(x\) into Equation 2:
\[ \frac{3\left(\frac{270 - 35y}{6}\right) + 2}{2} - \frac{4y + 3}{9} = 13 \]

First, simplify the left side:
\[ \frac{\frac{810 - 105y}{6} + 2}{2} - \frac{4y + 3}{9} = 13 \]
Multiply the numerator by 2:
\[ \frac{810 - 105y + 12}{12} - \frac{4y + 3}{9} = 13 \]
This simplifies to:
\[ \frac{822 - 105y}{12} - \frac{4y + 3}{9} = 13 \]

Now, eliminate the fractions by multiplying through by the LCM of 12 and 9, which is 36:
\[ 36 \left(\frac{822 - 105y}{12}\right) - 36 \left(\frac{4y + 3}{9}\right) = 36 \cdot 13 \]
This simplifies to:
\[ 3(822 - 105y) - 4(4y + 3) = 468 \]
Expanding gives:
\[ 2466 - 315y - 16y - 12 = 468 \]
Combining like terms:
\[ 2454 - 331y = 468 \]
Subtracting 2454 from both sides:
\[ -331y = 468 - 2454 \]
\[ -331y = -1986 \]
Dividing by -331:
\[ y = \frac{1986}{331} = 6 \]

**Step 4: Substitute \(y\) back to find \(x\).**

Now substitute \(y = 6\) back into the expression for \(x\):
\[ x = \frac{270 - 35(6)}{6} \]
Calculating this gives:
\[ x = \frac{270 - 210}{6} = \frac{60}{6} = 10 \]

**Final Solution:**
Thus, the solution to the system of equations is:
\[ x = 10, \quad y = 6 \]

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