Question

m = ½ p(2, 4)
2x - 4y = 3
-4y = -2x + 3
y = ½x - ¾
2x - 4y = 3
y + 4 = ½(x - 2)
y + 4 = ½x - 1
y = ½x - 5

Explanation:

Step1: Identify the problem type

This is a problem about linear equations and their manipulation, likely involving finding the equation of a line or solving for variables. Let's assume we need to find the intersection point of two lines, say \( y = \frac{1}{2}x - \frac{3}{4} \) and \( y = \frac{1}{2}x - 5 \) (or another pair, but let's check the first two visible equations: \( y = -2x - 1 \) and \( 2x - 4y = 3 \)). Wait, maybe we need to solve the system of equations \(

$$\begin{cases} y = -2x - 1 \\ 2x - 4y = 3 \end{cases}$$

\).

Step2: Substitute \( y \) from the first equation into the second

Substitute \( y = -2x - 1 \) into \( 2x - 4y = 3 \):
\( 2x - 4(-2x - 1) = 3 \)
Simplify: \( 2x + 8x + 4 = 3 \)
Combine like terms: \( 10x + 4 = 3 \)
Subtract 4: \( 10x = -1 \)
Divide by 10: \( x = -\frac{1}{10} \)

Step3: Find \( y \) using \( x = -\frac{1}{10} \) in \( y = -2x - 1 \)

\( y = -2(-\frac{1}{10}) - 1 = \frac{2}{10} - 1 = \frac{1}{5} - 1 = -\frac{4}{5} \)

Wait, maybe the problem is about the point-slope form. The first cell has \( m = \frac{1}{2} \), \( P(2, 4) \)? Wait, the handwriting is a bit unclear. Let's re-examine. The first cell: \( m = \frac{1}{2} \), \( P(\underline{2}, \underline{4}) \)? Then the next cell has \( 2x - 4y = 3 \), solving for \( y \): \( -4y = -2x + 3 \), so \( y = \frac{1}{2}x - \frac{3}{4} \). The last cell: \( y + 4 = \frac{1}{2}(x - 2) \), expanding: \( y + 4 = \frac{1}{2}x - 1 \), so \( y = \frac{1}{2}x - 5 \).

Maybe the problem is to find the equation of a line with slope \( \frac{1}{2} \) passing through \( (2, 4) \). Let's do that.

Step1: Use point-slope form

Point-slope form: \( y - y_1 = m(x - x_1) \), where \( m = \frac{1}{2} \), \( (x_1, y_1) = (2, 4) \)
So \( y - 4 = \frac{1}{2}(x - 2) \)

Step2: Simplify to slope-intercept form

\( y - 4 = \frac{1}{2}x - 1 \)
Add 4: \( y = \frac{1}{2}x + 3 \)

Wait, but the last cell has \( y = \frac{1}{2}x - 5 \), maybe a different point. Alternatively, maybe the first cell is to fill in the slope and point. Given the slope \( m = \frac{1}{2} \), and the point \( P(2, 4) \) (since the underlines are for 2 and 4). Then the equation using point-slope is \( y - 4 = \frac{1}{2}(x - 2) \), which simplifies to \( y = \frac{1}{2}x + 3 \), but the last cell has \( y = \frac{1}{2}x - 5 \), so maybe the point is \( (2, -4) \)? Let's check: \( y - (-4) = \frac{1}{2}(x - 2) \) → \( y + 4 = \frac{1}{2}x - 1 \) → \( y = \frac{1}{2}x - 5 \), which matches the last cell. So the point is \( (2, -4) \), so the first cell should have \( m = \frac{1}{2} \), \( P(2, -4) \).

Answer:

For the first cell (filling in the point): \( P(2, -4) \) (assuming the slope is \( \frac{1}{2} \) and the point leads to the equation in the last cell). If solving the system \(

$$\begin{cases} y = -2x - 1 \\ 2x - 4y = 3 \end{cases}$$

\), the solution is \( x = -\frac{1}{10} \), \( y = -\frac{4}{5} \).