1. The answer is Yes
The equation:

represents a <em>straight line </em>with slope
and y-intercept
. To know if the point
is a solution to the equation, let's substitute the x and y coordinates into the equation, therefore:

Since our result is true, this point is a solution to the equation.
2. The answer is No
The equation:

represents a <em>straight line </em>with slope
and no y-intercept, that is, it passes through the origin. By substituting this point into the equation:

Since our result is false, this point is not a solution to the equation.
3. In this problem, we have four points, so let's solve it step by step. Our equation is:

and represents a <em>straight line </em>with slope
and y-intercept 
3.1 <u>For the point</u> (4, 0)
<em>This point is a solution to the equation y = x - 4</em>
By substituting this point into the equation, we have:

Since our result is true, this point is a solution to the equation.
3.2 <u>For the point</u> (3, -1)
<em>This point is a solution to the equation y = x - 4</em>
By substituting this point into the equation, we have:

Since our result is true, this point is a solution to the equation.
3.3 <u>For the point</u> (6, 3)
<em>This point is not a solution to the equation y = x - 4</em>
By substituting this point into the equation, we have:

Since our result is false, this point is not a solution to the equation.
3.4 <u>For the point</u> (2, -4)
<em>This point is not a solution to the equation y = x - 4</em>
By substituting this point into the equation, we have:

Since our result is false, this point is not a solution to the equation.
4. In this problem we have:
- There are 8 crackers in 1 serving.
- There are 16 crackers in 2 servings.
- There are 24 crackers in 3 servings, and so on.
So, there is a <em>linear relationship</em> between crackers and servings. Thus, it is easy to write a linear equation like this:

Therefore, for x = 12 servings, we have:

Finally:
<em>There are 96 crackers in 12 servings</em>
5. An open equation is an equation containing one or more <em>variables </em>such that the <em>truth or falsehood </em>of the equation depends on the values of the variables assumed in a specific instance. An example of an open equation is:


For this particular x-value, the equation is true.

For this particular x-value, the equation is false.
So, we can conclude that this equation is only true when x = 7
6. Equations can be used to model and solve real-life problems. In this way, we can make predictions from an equation that models a real-life situation. For instance, a kitchen an <em>appliance manufacturing company </em>can determine the total cost in dollars of producing units of a blender by using the equation:

Thus, the cost of producing zero units is $2500, found by the y-intercept, that is the fixed cost of production. It includes costs that must be paid regardless of the number of units produced. The slope
means that the cost of producing each unit is $20, called by the Economists <em>the marginal cost </em>(cost per unit).
7. In this exercise, pizza costs $1.50 per slice. This can be modeled by the following equation:

This can be represented by the following Table as well:

So, in this Table we have indicated the total cost up to 5 slices of pizza.