Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit on every wrap is $3. Sal made a profit of $1,470 from lunch specials last month. The equation 2x + 3y = 1,470 represents Sal's profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold

4.Graph the function. On the graph, 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.

5.Suppose Sal's total profit on lunch specials for the next month is $1,593. The profit amounts are the same: $2 for each sandwich and $3 for each wrap. 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.

2x + 3y = 1470

3y = -2x + 1470
y = -2/3 + 490

slope = -2/3, y int is 490

To graph the function, we can start by finding the x and y-intercepts.

To find the x-intercept, we set y = 0 in the equation 2x + 3y = 1,470.
2x + 3(0) = 1,470
2x + 0 = 1,470
2x = 1,470
x = 1,470/2
x = 735

So the x-intercept is (735, 0).

To find the y-intercept, we set x = 0 in the equation 2x + 3y = 1,470.
2(0) + 3y = 1,470
0 + 3y = 1,470
3y = 1,470
y = 1,470/3
y = 490

So the y-intercept is (0, 490).

We can plot these intercepts on a graph and connect them with a straight line.

Next, let's consider the profit for the next month, which is $1,593. The profit equation remains the same: 2x + 3y = 1,593.

Since the profit amounts are the same for each lunch special, the shape of the graph will be the same as the previous month's graph. However, the intercepts will likely be different.

We can expect the x-intercept to move to a new point, and the y-intercept to also shift. The slope between these two intercepts will remain constant since the profit amounts are constant.

In summary, the graphs for the two months will be similar in shape since the profit equation remains constant. However, the intercepts will vary depending on the profit amount for each month.

To graph the equation 2x + 3y = 1,470, we can start by finding the intercepts.

To find the x-intercept, we set y = 0 and solve for x:
2x + 3(0) = 1,470
2x = 1,470
x = 735

So the x-intercept is (735, 0).

To find the y-intercept, we set x = 0 and solve for y:
2(0) + 3y = 1,470
3y = 1,470
y = 490

So the y-intercept is (0, 490).

Now we can plot these points on a graph and draw a line through them to represent the equation.

For the second part of the question, suppose the total profit for the next month is $1,593. We can use the same equation 2x + 3y = 1,593, where x represents the number of sandwich lunch specials and y represents the number of wrap lunch specials.

We can calculate the x-intercept and y-intercept for this equation in the same way as before.

The x-intercept is found by setting y = 0:
2x + 3(0) = 1,593
2x = 1,593
x = 796.5

So the x-intercept is (796.5, 0).

The y-intercept is found by setting x = 0:
2(0) + 3y = 1,593
3y = 1,593
y = 531

So the y-intercept is (0, 531).

When we plot these intercepts on a graph and draw a line through them, we can compare it to the graph for the previous month.

The graphs for the two months are similar in that they both represent the equation 2x + 3y, and they both have a positive slope. This means that as the number of sandwich lunch specials increases, the number of wrap lunch specials must decrease in order to maintain the same total profit.

The graphs are different in terms of the intercepts and the slope. The intercepts represent the number of sandwich and wrap lunch specials needed to earn a profit of $0. In the second month, the intercepts are higher, indicating that more lunch specials need to be sold to achieve the same profit. The slope represents the ratio of sandwiches to wraps needed to maintain the profit. A steeper slope means a higher ratio of sandwiches to wraps. So, the second month's graph may have a steeper slope, indicating a higher ratio of sandwiches to wraps compared to the first month.