Anybody knows how to do this??

Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit

coldgirl [10]

Changing the equation into slope form: $y = mx + c$ , where $m$ is the slope [gradient] and $c$ is the y-intercept.

$2x+3y=1470$
$3y = -2x+1470$
$y= - \frac{2}{3}x+ \frac{1470}{3}$
$y=- \frac{2}{3}x+490$

The gradient is $- \frac{2}{3}$ and y-intercept is at $y=490$

Graphing $y=- \frac{2}{3}x+490$ using slope-intercept method:
a) The slope is a negative slope. The line will go 'down hill'
b) The line must pass the point (0, 490)
c) The line will intercept the x-axis at y = 0
   $0 = - \frac{2}{3}x+490$
   $\frac{2}{3}x = 490$
   $2x = 1470$
   $x = \frac{1470}{2}$
   $x = 735$
So, x-intercept is at (735, 0)

The graph of this function is shown below. The intercepts are labelled at:
y = 490
x = 735

-----------------------------------------------------------------------------------------------------------

Next month's profit equation

$2x+3y=1593$

Rewriting this into slope-equation form

$3y = -2x+1593$
$y=- \frac{2}{3}+ \frac{1593}{3}$
$y= - \frac{2}{3}+531$

The gradient, $m$ , equals to $- \frac{2}{3}$
The y-intercept, $c$ , equals to 531

The equation still has the same gradient with last month's profit equation but different y-intercept.

-------------------------------------------------------------------------------------------------------------

A linear graph show points of (0, 300) and (450, 0)

We work out the slope:
$\frac{300-0}{0-450} = \frac{300}{-450}=- \frac{20}{30} =- \frac{2}{3}$

Y-intercept at x = 0, so it's at y = 300

Equation $y = - \frac{2}{3}+300$