Find the equation of the line that is perpendicular to y=1/6x+3 and contains the point (-3,23).
y = [?]x + [?]

y = -6x + 6

Step-by-step explanation:

The general equation of a line is

y = mx + b

where m is the slope and b is the y-intercept

The slope in the first equation is 1/6

Mathematically, when two lines are perpendicular, the product of their slopes is -1

1/6 * m2 = -1

m2 = -6

So we have the equation as;

y-y1 = m(x-x1)

y-23 = -6(x + 3)

y = -6x -18 +23

y = -6x + 6

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3