Understanding the Problem: Graph Cutting the X-Axis

The question asks us to find a specific value related to a point where the graph of a linear equation intersects the x-axis. When a graph cuts the x-axis, it means the point of intersection lies directly on the x-axis. A key property of any point on the x-axis is that its y-coordinate is always zero.

Identifying the Point of Intersection P(α, β)

We are given that the point where the graph of the equation \(2x = 5 - 3y\) cuts the x-axis is P\((\alpha, \beta)\). Since P is on the x-axis, its y-coordinate must be 0. Therefore, we know that \(\beta = 0\).

The point P\((\alpha, \beta)\) is also on the line represented by the equation \(2x = 5 - 3y\). This means that the coordinates \((\alpha, \beta)\) must satisfy the equation.

Substituting Coordinates into the Equation

Let's substitute the coordinates \((\alpha, \beta)\) into the given equation \(2x = 5 - 3y\):

\(2(\alpha) = 5 - 3(\beta)\)

Now, we know that \(\beta = 0\). Substitute this value into the equation:

\(2\alpha = 5 - 3(0)\)

\(2\alpha = 5 - 0\)

\(2\alpha = 5\)

Solving for α

We now have a simple equation to solve for \(\alpha\):

\(2\alpha = 5\)

To find \(\alpha\), divide both sides by 2:

\(\alpha = \frac{5}{2}\)

Determining the Values of α and β

From our analysis, we have found the values of \(\alpha\) and \(\beta\) for the point P\((\alpha, \beta)\):

• \(\alpha = \frac{5}{2}\)
• \(\beta = 0\)

So, the point where the graph cuts the x-axis is P\((\frac{5}{2}, 0)\).

Calculating the Value of (2α + β)

The question asks for the value of the expression \((2\alpha + \beta)\). Now that we have the values of \(\alpha\) and \(\beta\), we can substitute them into the expression:

\(2\alpha + \beta = 2\left(\frac{5}{2}\right) + 0\)

First, calculate \(2\left(\frac{5}{2}\right)\):

\(2 \times \frac{5}{2} = \frac{10}{2} = 5\)

Now, add \(\beta\):

\(5 + 0 = 5\)

Therefore, the value of \((2\alpha + \beta)\) is 5.

Summary of Steps

• Identify that cutting the x-axis means the y-coordinate is 0.
• Use the point P\((\alpha, \beta)\) and the fact that it's on the x-axis to set \(\beta = 0\).
• Substitute \((\alpha, \beta)\) into the equation \(2x = 5 - 3y\).
• Substitute \(\beta = 0\) into the equation and solve for \(\alpha\).
• Substitute the values of \(\alpha\) and \(\beta\) into the expression \((2\alpha + \beta)\) and calculate the result.

Revision Table: Key Concepts

Additional Information: Linear Equations and Intercepts

The equation \(2x = 5 - 3y\) is a linear equation because the variables x and y are raised to the power of 1. Linear equations graph as straight lines.

The point where the graph cuts the x-axis is also known as the x-intercept. To find the x-intercept of any linear equation, you set \(y=0\) and solve for x. Similarly, to find the y-intercept (where the graph cuts the y-axis), you set \(x=0\) and solve for y.

In this case, setting \(y=0\) gives \(2x = 5 - 3(0) \implies 2x = 5 \implies x = 5/2\). The x-intercept is \((5/2, 0)\), which is our point P\((\alpha, \beta)\) where \(\alpha = 5/2\) and \(\beta = 0\).