Consider the original equation:
[tex]\[ 3x + 2y = 8 \][/tex]
First, we need to rearrange this equation into slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Start by solving for y:
[tex]\[ 2y = -3x + 8 \][/tex]
[tex]\[ y = -\frac{3}{2}x + 4 \][/tex]
So, the slope (m) of this equation is [tex]\(-\frac{3}{2}\)[/tex].
A line parallel to this will have the same slope. Therefore, the slope of the new line is also [tex]\(-\frac{3}{2}\)[/tex].
Next, we use the point-slope form of the equation of a line to find the equation of the line that is parallel and passes through the point [tex]\((-2, 5)\)[/tex]. The point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the given point [tex]\((-2, 5)\)[/tex] and [tex]\( m \)[/tex] is the slope.
Substitute the given point and the slope into the equation:
[tex]\[ y - 5 = -\frac{3}{2}(x + 2) \][/tex]
Next, distribute the slope on the right side:
[tex]\[ y - 5 = -\frac{3}{2}x - 3 \][/tex]
Add 5 to both sides to put the equation into the slope-intercept form (y = mx + b):
[tex]\[ y = -\frac{3}{2}x - 3 + 5 \][/tex]
[tex]\[ y = -\frac{3}{2}x + 2 \][/tex]
Therefore, the equation of the line parallel to [tex]\(3x + 2y = 8\)[/tex] and passing through the point [tex]\((-2, 5)\)[/tex] is:
[tex]\[ y = -\frac{3}{2}x + 2 \][/tex]
So, the correctly filled blanks are:
[tex]\[ y = -\frac{3}{2} x + 2 \][/tex]
