Larry's Sandwich Shop sells subs and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit on every sub is $3. Larry made a profit of $1,800 from lunch specials last month. The equation 2x + 3y = 1,800 represents Larry's profits last month, where x is the number of sandwich lunch specials sold and y is the number of sub lunch specials sold.
Change the equation to slope-intercept form. Identify the slope and y-intercept of the equation. Be sure to show all your work.
Describe how you would graph this line using the slope-intercept method. Be sure to write using complete sentences.
Write the equation in function notation. Explain what the graph of the function represents. Be sure to use complete sentences.
Graph the function. On the graph, be sure to label your axes and indicate the increment markings, and make sure to label the intercepts. You may graph your equation by hand on a piece of paper and scan your work or you may use graphing technology.
Suppose Larry's total profit on lunch specials for the next month is $2,040. The profit amounts are the same: $2 for each sandwich and $3 for each sub. In a paragraph of at least three complete sentences, explain how the graphs of the functions for the two months are similar and how they are different.
Below is a graph that represents the total profits for a third month. Write the equation of the line that represents this graph. Show your work or explain how you determined the equations.

graph of line going through ordered pairs (0, 300) and (450, 0)

Place the y-intercept ( Point P ) in the coordinates system
The slope - (2/3) is -2/3. We move from point P, 2 units down which means 200 in the y-axis, and from that, we move 3 units to the right 300 in the coordinates; this last point Q is another point on the line with two points any straight line is defined Q( 400, 300).

We can check our job calculating the x-intercept

y = 0 = (- 2/3)×x + 600 ⇒ (- 2/3)×x = -600

x = (3×600)/2 x = 900

Intercept point R ( 900 ; 0)

The line pass throug it

In function notation the equation becomes:

f(x) = ( -2/3) ×x + 600

Next Month profit amounts are the same, we call y₂ the next line then

2040 = 2×x + 3×y₂ or y₂ = - (2/3)×x + 680

slope m = -2/3 y-intercept P´ ( 0 ; 680 ) x-intercept R´( 1020;0)

In attached file in green

The two lines have the same slope ( they are parallel)
As the prices are the same the situation is like the first straight line was moved (slide) diagonal keeping the slope

For the third month, we have to point for the graph the intercepts

y-axis intercept P´´ ( 0 ; 300 )

x-axis intercept R´´ ( 450 ; 0 )

Again the line keeps the slope 300/450 = 0.6666

(red line in attached file)

With the two points the line equation is:

y₃ = -06666 × x + 450

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