Chapter Test
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Note that the phrase at most
refers to less than or equal to.
Note that the phrase at least
refers to greater than or equal to.
Inequality: 12x+3y≤ 60
Graph:
Example Solution: 1 trophy and 4 medals
System:12x+3y≤ 60 x+y≥ 6
Graph:
Example Solution: 2 trophies and 8 medals
We want to write and graph an inequality for the given situation. Let's recall the given information.
Cost Of One Trophy:& $ 12 Cost Of One Medal:& $3 Spending Limit:& $60 Let x be the number of trophies and let y be the number of medals. By using the given information, we will first write an inequality in an organized table.
| Verbal Expression | Algebraic Expression |
|---|---|
| Cost of x trophies ($) | 12 x |
| Cost of y medals ($) | 3 y |
| Total cost ($) | 12 x+ 3 y |
| Total cost is less than or equal to $60. | 12 x+ 3 y≤ 60 |
| x-intercept | y-intercept | ||
|---|---|---|---|
| Substitution | Point | Substitution | Point |
| 12x+3( 0)=60 | (5,0) | 12( 0)+3y = 60 | (0,20) |
Next, we will plot the intercepts and draw the boundary line that passes through these points. Note that the number of boxes cannot be negative, so the line will be restricted by the axes. Additionally, since the inequality is non-strict, the line will be solid.
Finally, we will decide which side of the line we should shade. We will choose a point either side of the line and substitute it into the inequality. If it satisfies the inequality, we shade the region that contains the point. Otherwise, we shade the other region. Let (0,0) be our test point!
x= 0, y= 0
Zero Property of Multiplication
Therefore, we will shade below the line.
The points with whole number coordinates in the shaded region are the solutions to the inequality.
Therefore, one possible solution can be 1 trophy and 4 medals.
Since we want to buy at least 6 items, the total number of trophies and medals must be greater than or equal to 6. Let's use this information to write the inequality.
Since the overlapping area is the solution, one possible solution can be 2 trophies and 8 medals.