Question

Find the value of x so that the line passing through (x, 10) and (-4, 8) has a slope of 2/3

Answer 1

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut point with ele axis and

$m = \frac {y2-y1} {x2-x1}$

How we have:

$\frac {2} {3} = \frac {8-10} {- 4-x}\\\frac {8-10} {- 4-x} = \frac {2} {3}\\\frac {-2} {- 4-x} = \frac {2} {3}$

We clear the value of "x"

$2 (-4-x) = - 6\\-8-2x = -6\\-2x = -6 + 8\\-2x = 2\\x = \frac {2} {- 2}\\x = -1$

Answer:

$x = -1$

Answer 2

Answer:

x = -1

Step-by-step explanation:

We are given the following two points from which the line passes and has a slope of $\frac{2}{3}$.

We are to find the value of x.

Slope = $\frac { y _ 2 - y _ 1 } { x _ 2 - x _ 1 }$

$\frac{2}{3}$ = $\frac{10-8}{x-(-4)}$

$\frac{2}{3}$ = $\frac{2}{x+4}$

By cross multiplication:

$2(x+4)=3 \times 2$

$2x+8=6$

$2x=-2$

x = -1