Finding the Intersection Point of Linear Equations

The question asks us to find the intersection point of two linear equations and then evaluate an expression based on the coordinates of that point. The given linear equations are:

1. \(3x - 2y = 8\)
2. \(4x + 3y = 5\)

The intersection point is where both equations are satisfied simultaneously. We can find this point by solving the system of linear equations. Let's use the elimination method.

Our goal is to eliminate one variable, either \(x\) or \(y\), so we can solve for the other. We can multiply the first equation by 3 and the second equation by 2. This will make the coefficients of \(y\) opposites (-6 and +6):

• Multiply equation (1) by 3: \(3 \times (3x - 2y) = 3 \times 8 \implies 9x - 6y = 24\) (Equation 3)
• Multiply equation (2) by 2: \(2 \times (4x + 3y) = 2 \times 5 \implies 8x + 6y = 10\) (Equation 4)

Now, we add Equation 3 and Equation 4 to eliminate \(y\):

\((9x - 6y) + (8x + 6y) = 24 + 10\)

\(9x + 8x - 6y + 6y = 34\)

\(17x = 34\)

Now, solve for \(x\):

\(x = \frac{34}{17}\)

\(x = 2\)

So, the x-coordinate of the intersection point is 2.

Next, substitute the value of \(x=2\) into either of the original equations to find the value of \(y\). Let's use Equation 1:

\(3x - 2y = 8\)

\(3(2) - 2y = 8\)

\(6 - 2y = 8\)

Subtract 6 from both sides:

\(-2y = 8 - 6\)

\(-2y = 2\)

Divide by -2:

\(y = \frac{2}{-2}\)

\(y = -1\)

So, the y-coordinate of the intersection point is -1.

The intersection point is given as P(α, β). From our calculations, we found \(x=2\) and \(y=-1\). Therefore, \(\alpha = 2\) and \(\beta = -1\).

The question asks for the value of the expression \((2\alpha - \beta)\). Now, we substitute the values of \(\alpha\) and \(\beta\) into this expression:

\(2\alpha - \beta = 2(2) - (-1)\)

\( = 4 - (-1)\)

\( = 4 + 1\)

\( = 5\)

Thus, the value of \((2\alpha - \beta)\) is 5.

Understanding Linear Equations Intersection

When the graphs of two linear equations intersect, they meet at a single point. This point represents the unique solution (if one exists) that satisfies both equations simultaneously. Solving a system of linear equations helps us find these coordinates.

Since the slopes are different, the lines are not parallel and will intersect at exactly one point, which we found to be (2, -1).

Revision Table: Key Concepts

Additional Information on Solving Systems

Besides the elimination method used above, another common method to solve a system of linear equations is the substitution method. In this method, you solve one equation for one variable in terms of the other, and then substitute that expression into the other equation.

For example, from \(3x - 2y = 8\), you could solve for \(x\):

\(3x = 2y + 8\)

\(x = \frac{2y + 8}{3}\)

Then substitute this expression for \(x\) into the second equation \(4x + 3y = 5\):

\(4\left(\frac{2y + 8}{3}\right) + 3y = 5\)

\(\frac{8y + 32}{3} + 3y = 5\)

Multiply the entire equation by 3 to clear the denominator:

\(8y + 32 + 9y = 15\)

\(17y + 32 = 15\)

\(17y = 15 - 32\)

\(17y = -17\)

\(y = -1\)

Once \(y = -1\), substitute it back into the expression for \(x\):

\(x = \frac{2(-1) + 8}{3}\)

\(x = \frac{-2 + 8}{3}\)

\(x = \frac{6}{3}\)

\(x = 2\)

Both methods yield the same intersection point (2, -1), confirming our result.