To fill in the missing information in the chart for the linear equation \( y = 4x - 2 \), we can derive several points by plugging in values for \( x \).
- Find \( y \) when \( x = 3 \): \[ y = 4(3) - 2 = 12 - 2 = 10 \] Thus, the point when \( x = 3 \) is \( (3, 10) \).
Now, we have one completed entry in the chart:
- For \( x = 3 \), \( y = 10 \), which gives us the point \( (3, 10) \).
- Find another \( x \) value. Let's plug in \( x = 0 \): \[ y = 4(0) - 2 = -2 \] Thus, the point when \( x = 0 \) is already given as \( (0, -2) \).
Now let's fill in the chart:
| \( x \) | \( y \) | |---------|---------------| | _____ | _____ | | 3 | 10 | | 0 | -2 |
- Now, we need to determine another \( x \) value and find its corresponding \( y \). We can choose \( x = 1 \) for this purpose: \[ y = 4(1) - 2 = 4 - 2 = 2 \] Thus, we can add the point \( (1, 2) \).
Now we fill the completed chart as follows:
| \( x \) | \( y \) | |---------|------------| | 1 | 2 | | 3 | 10 | | 0 | -2 |
So, the missing numbers you can fill in are \( x = 1 \) and \( y = 2 \) for the first row.