Question

3.5.3 quiz: two - variable systems: graphing
gina made tables of values to solve a system of equations. first, she found that the x - value of the solution was between 0 and 1, and then she found that it was between 0.5 and 1. next, she made this table.

x	y = 4x - 1	y = - 3x + 5
0.6	1.4	3.2
0.7	1.8	2.9
0.8	2.2	2.6
0.9	2.6	2.3
1.0	3	2

which ordered pair is the best approximation of the exact solution?
a. (0.6, 2.5)
b. (0.9, 2.1)
c. (0.7, 1.8)
d. (0.9, 2.4)

Explanation:

Step1: Understand the solution of a system of equations

The solution to a system of equations \( y = 4x - 1 \) and \( y=-3x + 5 \) is the point where the two lines intersect, i.e., where \( 4x-1=-3x + 5 \) and the \( y \)-values from both equations are equal (or as close as possible in a table). We need to find the \( x \) and \( y \) where the \( y \)-values of both equations are closest to each other.

Step2: Analyze the table values

Let's list the \( y \)-values for each \( x \) and find the difference between the two \( y \)-values:
- For \( x = 0.5 \): \( y_1=1 \), \( y_2 = 3.5 \), difference \( |1 - 3.5|=2.5 \)
- For \( x = 0.6 \): \( y_1 = 1.4 \), \( y_2=3.2 \), difference \( |1.4 - 3.2| = 1.8 \)
- For \( x = 0.7 \): \( y_1=1.8 \), \( y_2 = 2.9 \), difference \( |1.8 - 2.9|=1.1 \)
- For \( x = 0.8 \): \( y_1=2.2 \), \( y_2 = 2.6 \), difference \( |2.2 - 2.6| = 0.4 \)
- For \( x = 0.9 \): \( y_1=2.6 \), \( y_2=2.3 \), difference \( |2.6 - 2.3|=0.3 \)
- For \( x = 1.0 \): \( y_1=3 \), \( y_2 = 2 \), difference \( |3 - 2| = 1 \)

Wait, maybe I made a mistake. Wait, the solution is where \( 4x-1=-3x + 5 \). Let's solve the equation: \( 4x+3x=5 + 1 \), \( 7x=6 \), \( x=\frac{6}{7}\approx0.857 \), and \( y = 4\times\frac{6}{7}-1=\frac{24}{7}-1=\frac{17}{7}\approx2.428 \) or \( y=-3\times\frac{6}{7}+5=\frac{-18 + 35}{7}=\frac{17}{7}\approx2.428 \). Now let's check the options:

- Option A: \( (0.6,2.5) \). For \( x = 0.6 \), \( y_1=1.4 \), \( y_2 = 3.2 \), 2.5 is not close to either.
- Option B: \( (0.9,2.1) \). For \( x = 0.9 \), \( y_1=2.6 \), \( y_2=2.3 \). 2.1 is not close to 2.6 or 2.3.
- Option C: \( (0.7,1.8) \). For \( x = 0.7 \), \( y_2=2.9 \), 1.8 is only the \( y \) of the first equation, not close to the second.
- Option D: Wait, maybe I misread the options. Wait, the original options: Wait, maybe I miscalculated. Wait, when \( x = 0.9 \), \( y_1=2.6 \), \( y_2=2.3 \). The average or the value where they are closest. Wait, the exact solution is around \( x\approx0.857 \), \( y\approx2.428 \). Let's check the options again. Wait, maybe the option D is \( (0.9,2.4) \)? Wait, the user's option D: "D. (0.9,2.4)". Let's recalculate:

For \( x = 0.9 \), \( y = 4x-1=4\times0.9 - 1=3.6 - 1 = 2.6 \); \( y=-3x + 5=-2.7 + 5 = 2.3 \). The midpoint or the value closest. Wait, maybe the question is about the best approximation. Let's see the \( y \)-values getting closer as \( x \) increases from 0.5 to 0.9, then at \( x = 0.9 \), \( y_1=2.6 \), \( y_2=2.3 \). The exact \( y \) is around 2.4. Let's check the options:

- Option A: (0.6,2.5): \( y_1=1.4 \), \( y_2=3.2 \), 2.5 is not related.
- Option B: (0.9,2.1): \( y_1=2.6 \), \( y_2=2.3 \), 2.1 is far.
- Option C: (0.7,1.8): \( y_2=2.9 \), 1.8 is only for first equation.
- Option D: (0.9,2.4): Let's see, for \( x = 0.9 \), the \( y \)-values are 2.6 and 2.3. The average of 2.6 and 2.3 is 2.45, close to 2.4. And \( x = 0.9 \) is close to the exact \( x\approx0.857 \). Also, when \( x = 0.8 \), \( y_1=2.2 \), \( y_2=2.6 \), midpoint is 2.4, and \( x = 0.8 \) is close to 0.857. Wait, maybe I made a mistake in option labels. Wait, the user's options:

Wait the user's options:

A. (0.6, 2.5)

B. (0.9, 2.1)

C. (0.7, 1.8)

D. (0.9, 2.4)

Let's re-express:

We need to find the ordered pair where the \( y \)-values of both equations are closest. Let's check the difference between \( y = 4x - 1 \) and \( y=-3x + 5 \):

For each option's \( x \):

- Option A: \( x = 0.6 \), \( y = 2.5 \). Let's see what \( y \) should be for \( x = 0.6 \): \( y_1=1.4 \), \( y_2=3.2 \). 2.5 is between 1.4 and 3.2, but the difference between 1.4 and 3.2 is 1.8, so 2.5 is the midpoint? No, not helpful.

- Option B: \( x = 0.9 \), \( y = 2.1 \). \( y_1=2.6 \), \( y_2=2.3 \). 2.1 is below both, not close.

- Option C: \( x = 0.7 \), \( y = 1.8 \). \( y_1=1.8 \) (which is \( 4\times0.7 - 1=2.8 - 1=1.8 \)), but \( y_2=-3\times0.7 + 5=-2.1 + 5 = 2.9 \). So \( y = 1.8 \) is only for the first equation, not the second.

- Option D: \( x = 0.9 \), \( y = 2.4 \). Let's see \( y_1=2.6 \), \( y_2=2.3 \). The average of 2.6 and 2.3 is \( \frac{2.6 + 2.3}{2}=2.45 \), close to 2.4. And \( x = 0.9 \) is close to the exact \( x=\frac{6}{7}\approx0.857 \). Also, as \( x \) increases from 0.8 to 0.9, \( y_1 \) goes from 2.2 to 2.6, \( y_2 \) goes from 2.6 to 2.3. At \( x = 0.8 \), \( y_1=2.2 \), \( y_2=2.6 \); at \( x = 0.9 \), \( y_1=2.6 \), \( y_2=2.3 \). The exact \( y \) is around 2.4, which is between 2.2 (x=0.8) and 2.6 (x=0.9) for \( y_1 \), and between 2.6 (x=0.8) and 2.3 (x=0.9) for \( y_2 \). So the best approximation is (0.9, 2.4) because at \( x = 0.9 \), the \( y \)-values are 2.6 and 2.3, and 2.4 is close to the average, and \( x = 0.9 \) is close to the exact \( x \approx 0.857 \). Also, looking at the differences: at \( x = 0.9 \), the difference between \( y_1 \) and \( y_2 \) is \( |2.6 - 2.3| = 0.3 \), which is the smallest difference among the given \( x \) values (except maybe miscalculating earlier). Wait, at \( x = 0.8 \), difference is \( |2.2 - 2.6| = 0.4 \), at \( x = 0.9 \), difference is 0.3. So \( x = 0.9 \) has a smaller difference. So the ordered pair (0.9, 2.4) is the best approximation.

Answer:

D. (0.9, 2.4)