To solve the system of equations [tex]\( y = 4x - 1 \)[/tex] and [tex]\( y = -3x + 5 \)[/tex] using the table of values provided, let's follow Gina's approach to see if the y-values from both equations match for any x-value within the specified range, which is between 0.5 and 1.0.
First, we consider the table provided:
[tex]\[
\begin{array}{|c|c|c|}
\hline
x & y=4x-1 & y=-3x+5 \\
\hline
0.5 & 1 & 3.5 \\
\hline
0.6 & 1.4 & 3.2 \\
\hline
0.7 & 1.8 & 2.9 \\
\hline
0.8 & 2.2 & 2.6 \\
\hline
0.9 & 2.6 & 2.3 \\
\hline
1.0 & 3 & 2 \\
\hline
\end{array}
\][/tex]
Now, let's check each row to see if and where the y-values of both equations are equal:
1. For [tex]\( x = 0.5 \)[/tex]:
- [tex]\( y = 4(0.5) - 1 = 1 \)[/tex]
- [tex]\( y = -3(0.5) + 5 = 3.5 \)[/tex]
- The y-values are not equal.
2. For [tex]\( x = 0.6 \)[/tex]:
- [tex]\( y = 4(0.6) - 1 = 1.4 \)[/tex]
- [tex]\( y = -3(0.6) + 5 = 3.2 \)[/tex]
- The y-values are not equal.
3. For [tex]\( x = 0.7 \)[/tex]:
- [tex]\( y = 4(0.7) - 1 = 1.8 \)[/tex]
- [tex]\( y = -3(0.7) + 5 = 2.9 \)[/tex]
- The y-values are not equal.
4. For [tex]\( x = 0.8 \)[/tex]:
- [tex]\( y = 4(0.8) - 1 = 2.2 \)[/tex]
- [tex]\( y = -3(0.8) + 5 = 2.6 \)[/tex]
- The y-values are not equal.
5. For [tex]\( x = 0.9 \)[/tex]:
- [tex]\( y = 4(0.9) - 1 = 2.6 \)[/tex]
- [tex]\( y = -3(0.9) + 5 = 2.3 \)[/tex]
- The y-values are not equal.
6. For [tex]\( x = 1.0 \)[/tex]:
- [tex]\( y = 4(1.0) - 1 = 3 \)[/tex]
- [tex]\( y = -3(1.0) + 5 = 2 \)[/tex]
- The y-values are not equal.
Conclusively, by checking all the provided x-values and corresponding y-values, we see that there is no single x-value where the y-values from [tex]\( y = 4x - 1 \)[/tex] and [tex]\( y = -3x + 5 \)[/tex] are equal.
Therefore, there is no solution to the system of equations within the specified range of x-values between 0.5 and 1.0.
