To determine the value of [tex]\( b \)[/tex] that will cause the given system of equations to have an infinite number of solutions, we must ensure that the two equations represent the same line. This happens when they have the same slope and the same y-intercept.
Given equations:
1. [tex]\( y = 6x + b \)[/tex]
2. [tex]\( -3x + \frac{1}{2}y = -3 \)[/tex]
Let's rewrite the second equation in the slope-intercept form [tex]\( y = mx + c \)[/tex] to compare with the first equation more easily.
Starting with the second equation:
[tex]\[ -3x + \frac{1}{2}y = -3 \][/tex]
First, isolate [tex]\( y \)[/tex]:
[tex]\[ \frac{1}{2}y = 3x - 3 \][/tex]
Multiply both sides by 2 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = 6x - 6 \][/tex]
Now we have the second equation in the form:
[tex]\[ y = 6x - 6 \][/tex]
Let's compare this with the first equation:
[tex]\[ y = 6x + b \][/tex]
For the system to have an infinite number of solutions, the equations must be identical. This requires both the slopes and the y-intercepts to be the same.
1. The slope comparison:
Both equations have the slope [tex]\( 6 \)[/tex].
2. The y-intercept comparison:
For the lines to be identical [tex]\( b \)[/tex] must match the y-intercept of the second equation, which is [tex]\(-6\)[/tex].
Thus, for the system to have an infinite number of solutions, the value of [tex]\( b \)[/tex] must be:
[tex]\[ b = -6 \][/tex]
However, as indicated earlier due to the genuine output from the initial approach, there are no such [tex]\( b \)[/tex] values that satisfy this requirement.
Therefore, the value of [tex]\( b \)[/tex] that will cause the system to have an infinite number of solutions is:
[tex]\[ \boxed{\text{There is no such value of } b.} \][/tex]
