Understanding the Problem of Intersecting Lines

The question asks us to find the value of a specific algebraic expression involving the coordinates of the intersection point of two linear equations. First, we need to find the point P(a, b) where the graphs of the two given equations intersect. This point (a, b) is the solution (x, y) to the system of these two linear equations. Once we find the values of 'a' and 'b', we substitute them into the given expression and calculate its value.

Solving the System of Linear Equations

We are given the two equations:

• Equation 1: \(3x - 20y - 2 = 0 \implies 3x - 20y = 2\)
• Equation 2: \(11x - 5y + 61 = 0 \implies 11x - 5y = -61\)

We can solve this system using the elimination method. Our goal is to make the coefficient of either 'x' or 'y' the same (or opposite) in both equations so that we can eliminate one variable by adding or subtracting the equations.

Let's eliminate 'y'. The coefficient of 'y' in Equation 1 is -20, and in Equation 2 is -5. We can multiply Equation 2 by 4 to make the coefficient of 'y' -20.

Multiply Equation 2 by 4:

\(4 \times (11x - 5y) = 4 \times (-61)\)

\(44x - 20y = -244\) (Equation 3)

Now we have the system:

• Equation 1: \(3x - 20y = 2\)
• Equation 3: \(44x - 20y = -244\)

Subtract Equation 1 from Equation 3:

\((44x - 20y) - (3x - 20y) = -244 - 2\)

\(44x - 20y - 3x + 20y = -246\)

\(41x = -246\)

Now, solve for 'x':

\(x = \frac{-246}{41}\)

\(x = -6\)

Now substitute the value of 'x' (\(-6\)) into either Equation 1 or Equation 2 to find 'y'. Let's use Equation 1:

\(3x - 20y = 2\)

\(3(-6) - 20y = 2\)

\(-18 - 20y = 2\)

Add 18 to both sides:

\(-20y = 2 + 18\)

\(-20y = 20\)

Solve for 'y':

\(y = \frac{20}{-20}\)

\(y = -1\)

The intersection point P(a, b) is (-6, -1). So, \(a = -6\) and \(b = -1\).

Evaluating the Given Expression

The expression we need to evaluate is \(\frac{a^2 + b^2 - ab}{a^2 - b^2 + ab}\). We have \(a = -6\) and \(b = -1\).

First, calculate the terms \(a^2\), \(b^2\), and \(ab\):

• \(a^2 = (-6)^2 = 36\)
• \(b^2 = (-1)^2 = 1\)
• \(ab = (-6)(-1) = 6\)

Now substitute these values into the numerator and the denominator of the expression.

Numerator: \(a^2 + b^2 - ab = 36 + 1 - 6 = 37 - 6 = 31\)

Denominator: \(a^2 - b^2 + ab = 36 - 1 + 6 = 35 + 6 = 41\)

So, the value of the expression is \(\frac{31}{41}\).

Conclusion

The intersection point of the given lines is (-6, -1). Substituting these values into the expression \(\frac{a^2 + b^2 - ab}{a^2 - b^2 + ab}\) gives the value \(\frac{31}{41}\).

Revision Table: Key Concepts

Additional Information: Solving Linear Systems

There are multiple methods to solve a system of two linear equations in two variables (like x and y):

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method (or Addition Method): Multiply one or both equations by constants so that the coefficients of one variable are opposites or identical. Then add or subtract the equations to eliminate that variable.
• Graphical Method: Graph both equations on the same coordinate plane. The coordinates of the point where the lines intersect are the solution. This method can be less precise if the intersection point has non-integer coordinates.
• Matrix Method (Cramer's Rule or Matrix Inverse): More advanced methods using matrices, suitable for larger systems or when using computational tools.

The choice of method often depends on the specific equations. The elimination method is usually efficient when coefficients can be easily made equal or opposite.