Chapter Test
Sign In
at mostrefers to
less than or equal to.
at leastrefers to
greater than or equal to.
Graph:
Example Solution: 1 trophy and 4 medals
Graph:
Example Solution: 2 trophies and 8 medals
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 |
Therefore, we have an inequality that represents the situation. 12 x+ 3 y≤ 60 Now, to graph the above inequality, we will determine its boundary line by replacing the inequality sign with the equals sign. Inequality &Boundary Line 12x+3y ≤ 60 &12x+3y = 60 Then, we will draw the boundary line. Since the equation of the line is in standard form, we can draw it by finding its intercepts. To find the x-intercept, we substitute y=0 in the equation and solve it for x. We proceed in the same way to find the y-intercept, as well.
| 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.
x= 0, y= 0
Zero Property of Multiplication
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.
x+ y ≥ 6 With the inequality in Part A, we can write the system of inequalities. 12x+3y≤ 60 x+y≥ 6 Proceeding in the same way as we did in Part A, we can graph the system.
Since the overlapping area is the solution, one possible solution can be 2 trophies and 8 medals.