Questions: Nevin made tables of values to approximate the solution to a system of equations. First he found that the x-value of the solution was between 1 and 2. and then he found that it was between 1 and 1.5. Next, he made this table. x y=4x-3 y=-5x+9 --------- 1 1 4 1.1 1.4 3.5 1.2 1.8 3 1.3 2.2 2.5 1.4 2.6 2 1.5 3 1.5 Which ordered pair is the best approximation of the exact solution? A. (1.3,1.8) B. (1.5,3) C. (1.3,2.3) D. (1.1,21)

Solution Steps

To find the best approximation of the exact solution to the system of equations, we need to identify the point where the two equations intersect. This occurs when the values of \( y \) from both equations are equal. By examining the table, we look for the \( x \) value where the \( y \) values from both equations are closest to each other.

The system of equations given is:

1. \( y = 4x - 3 \)
2. \( y = -5x + 9 \)

We have a table of values for \( x \) ranging from 1 to 1.5. We will calculate the corresponding \( y \) values for both equations at these \( x \) values and look for the point where the \( y \) values are closest to each other.

Using the equations, we calculate the \( y \) values for each \( x \):

• For \( x = 1 \): \( y_1 = 1 \), \( y_2 = 4 \)
• For \( x = 1.1 \): \( y_1 = 1.4 \), \( y_2 = 3.5 \)
• For \( x = 1.2 \): \( y_1 = 1.8 \), \( y_2 = 3 \)
• For \( x = 1.3 \): \( y_1 = 2.2 \), \( y_2 = 2.5 \)
• For \( x = 1.4 \): \( y_1 = 2.6 \), \( y_2 = 2 \)
• For \( x = 1.5 \): \( y_1 = 3 \), \( y_2 = 1.5 \)

We calculate the absolute differences between the \( y \) values from both equations:

• For \( x = 1.3 \): \( |2.2 - 2.5| = 0.3 \)

This is the smallest difference found, indicating that the best approximation of the solution occurs at \( (1.3, 2.2) \).

Final Answer