The graphs of the equations 3x - 20y - 2 = 0 and 11x - 5y + 61 = 0 intersect at P(a, b). What is the value of (a 2+ b 2- ab)/(a 2- b 2+ ab)?
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.
We are given the two equations:
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:
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 \(b = -1\).
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 \(ab\):
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 \(\frac{31}{41}\).
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 \(\frac{31}{41}\).
| Concept | Description |
|---|---|
| Intersection Point | The single point where two lines cross on a graph. It is the solution that satisfies both equations simultaneously. |
| System of Linear Equations | A set of two or more linear equations with the same variables. |
| Elimination Method | A method for solving a system of equations by adding or subtracting equations to eliminate one variable. |
| Substitution | Replacing a variable with its calculated value (or equivalent expression) in another equation or expression. |
| Evaluating Expression | Finding the numerical value of an algebraic expression by substituting known values for the variables. |
There are multiple methods to solve a system of two linear equations in two variables (like x and y):
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.
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