To determine the best approximation for the solution to the system of equations using the table provided, we need to find the [tex]\(x\)[/tex]-value in the table where the [tex]\(y\)[/tex]-values of both equations are closest to each other.
Let's analyze the table:
[tex]\[
\begin{array}{|c|c|c|}
\hline
x & y=2x-3 & y=-5x+2 \\
\hline
0.5 & -2 & -0.5 \\
0.6 & -1.8 & -1 \\
0.7 & -1.6 & -1.5 \\
0.8 & -1.4 & -2 \\
0.9 & -1.2 & -2.5 \\
1.0 & -1 & -3 \\
\hline
\end{array}
\][/tex]
We compare the pairs of [tex]\(y\)[/tex]-values corresponding to each [tex]\(x\)[/tex] value:
- For [tex]\(x = 0.5\)[/tex]:
[tex]\[
y = 2(0.5)-3 = -2 \\
y = -5(0.5)+2 = -0.5 \\
\text{Difference} = | -2 - (-0.5)| = 1.5
\][/tex]
- For [tex]\(x = 0.6\)[/tex]:
[tex]\[
y = 2(0.6)-3 = -1.8 \\
y = -5(0.6)+2 = -1 \\
\text{Difference} = | -1.8 - (-1)| = 0.8
\][/tex]
- For [tex]\(x = 0.7\)[/tex]:
[tex]\[
y = 2(0.7)-3 = -1.6 \\
y = -5(0.7)+2 = -1.5 \\
\text{Difference} = | -1.6 - (-1.5)| = 0.1
\][/tex]
- For [tex]\(x = 0.8\)[/tex]:
[tex]\[
y = 2(0.8)-3 = -1.4 \\
y = -5(0.8)+2 = -2 \\
\text{Difference} = | -1.4 - (-2)| = 0.6
\][/tex]
- For [tex]\(x = 0.9\)[/tex]:
[tex]\[
y = 2(0.9)-3 = -1.2 \\
y = -5(0.9)+2 = -2.5 \\
\text{Difference} = | -1.2 - (-2.5)| = 1.3
\][/tex]
- For [tex]\(x = 1.0\)[/tex]:
[tex]\[
y = 2(1)-3 = -1 \\
y = -5(1)+2 = -3 \\
\text{Difference} = | -1 - (-3)| = 2
\][/tex]
The closest [tex]\(y\)[/tex]-values occur when [tex]\(x = 0.7\)[/tex] with [tex]\(y\)[/tex]-values [tex]\(-1.6\)[/tex] and [tex]\(-1.5\)[/tex] giving a difference of [tex]\(0.1\)[/tex].
To approximate the corresponding ordered pair for [tex]\(x = 0.7\)[/tex]:
The approximate [tex]\(y\)[/tex]-value is around the middle of [tex]\(-1.6\)[/tex] and [tex]\(-1.5\)[/tex]:
[tex]\[
y \approx \left(\frac{-1.6 + (-1.5)}{2}\right) = \left(\frac{-3.1}{2}\right) = -1.55
\][/tex]
Thus, the best approximation of the solution in the table is:
[tex]\[ (0.7, -1.55) \][/tex]
Among the given answer choices, the closest one is:
C. [tex]\((0.7, -1.5)\)[/tex]
