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Consider the given equation:
3x + 2y = 8

The equation y = _____ x + _____ represents the line parallel to the given equation and passes through the point (-2, 5).



Answer :

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]

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