Find the slope of the line [tex]7y + 3x + 2 = 0[/tex].

(a) [tex]-\frac{3}{7}[/tex]
(b) [tex]\frac{7}{4}[/tex]
(c) [tex]\frac{7}{3}[/tex]
(d) [tex]-\frac{7}{3}[/tex]
(e) [tex]-\frac{7}{4}[/tex]



Answer :

To determine the slope of the line given by the equation [tex]\(7y + 3x + 2 = 0\)[/tex], we need to rearrange the equation into the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept.

Step 1: Begin with the given equation:
[tex]\[ 7y + 3x + 2 = 0 \][/tex]

Step 2: Isolate the term with [tex]\(y\)[/tex] on one side of the equation. To do this, we need to move the terms involving [tex]\(x\)[/tex] and the constant to the other side:
[tex]\[ 7y = -3x - 2 \][/tex]

Step 3: Solve for [tex]\(y\)[/tex] by dividing all terms by 7:
[tex]\[ y = -\frac{3}{7}x - \frac{2}{7} \][/tex]

Now, we have the equation in the slope-intercept form [tex]\(y = mx + b\)[/tex]. From this form, we can see that [tex]\(m = -\frac{3}{7}\)[/tex].

Therefore, the slope of the line is:
[tex]\[ -\frac{3}{7} \][/tex]

The correct answer is:
(a) [tex]\( -\frac{3}{7} \)[/tex]