Can anyone thoroughly explain how to solve these problems as well as how to graph them?
1. Linda works at a pharmacy for $15 an hour. She also baby-sits for $10 an hour. Linda needs to earn at least $90 per week, but she does not want to work more than 20 hours per week. Show and describe the number of hours Linda could work at each job to meet her goals. List two possible solutions.
2. Theo's mother has given him at most $150 to buy clothes for school. The pants cost $30 each and shirts cost $15 each. How many of each can he buy? Write a linear inequality to describe the situations. Graph the linear inequality and give three possible combinations of pants and shirts Theo could buy.
3. A grocer sells mango for $4/lb and apples for $3/1b. The grocer starts with 45 lb of mangos and 50 lb of apples each day. The grocer's goal is to make at least $300 by selling mangos and apples each day. Show and describe all possible combinations of mangos and apples that could be sold to meet the goal. List two possible combinations.
to graph inequalities, draw the lines that represent the equations (as if not inequalities). Then shade the area above or below the line, depending on whether y is greater or less than the values on the line.
For example, on #1, if Linda works p hours at the pharmacy and b hours babysitting,
p+b <= 20
15p+10b >= 90
To see the solution, check out the shaded region here:
http://www.wolframalpha.com/input/?i=plot+p%2Bb+%3C%3D+20+and+15p%2B10b+%3E%3D+90+and+p%3E%3D0+and+b%3E%3D0
Do the others in like wise -- devise your inequalities and then discover the shaded region.
hours worked in pharmacy --- x
hours worked baby-sitting ---- y
15x + 10y ≥ 90 , or after simplifying
3x + 2y ≥ 18
also x+y ≤ 20
sketching the region below y = -x + 20
and the region above the line y = (-3/2)x + 9
in the first quadrant yields many ordered pairs that satisfy both inequalities.
e.g.
10, 10
11, 8
11, 7
try #2 and #3 using the same method I used for the first one.
I still don't know how to graph #2.
I have absolutely no idea of how to solve #3, either.
#2.
30p+15c <= 150
#3.
0 <= m <= 45
0 <= a <= 50
4m + 3a >= 300
see the solution at
http://www.wolframalpha.com/input/?i=solve+0+%3C%3D+m+%3C%3D+45%2C+0+%3C%3D+a+%3C%3D+50%2C+4m+%2B+3a+%3E%3D+300
1. To solve this problem and graph it, we can set up a system of inequalities. Let's define the number of hours Linda works at the pharmacy as x and the number of hours she babysits as y.
Since Linda works at the pharmacy for $15 an hour and babysits for $10 an hour, the amount of money she earns from the pharmacy job is 15x, and the amount she earns from babysitting is 10y.
Linda needs to earn at least $90 per week, so we can write the first inequality as:
15x + 10y ≥ 90
Linda does not want to work more than 20 hours per week, so we can write the second inequality as:
x + y ≤ 20
To graph these inequalities, we can start by rewriting them in slope-intercept form:
First inequality:
10y ≥ -15x + 90
y ≥ -1.5x + 9 (dividing by 10)
Second inequality:
y ≤ -x + 20
Now, let's plot these inequalities on a graph:
For the first inequality, start by plotting the point (0,9) on the y-axis (since y = 9 when x = 0), then use the slope of -1.5 (rise -1.5, run 1) to plot more points and draw a line. Since the inequality is "greater than or equal to," shade the region above the line.
For the second inequality, start by plotting the point (0,20) on the y-axis (since y = 20 when x = 0), then use the slope of -1 (rise -1, run 1) to plot more points and draw a line. Since the inequality is "less than or equal to," shade the region below the line.
The shaded regions where the two inequalities overlap represent the valid solutions.
Two possible solutions for this problem could be:
1. x = 6 (working 6 hours at the pharmacy) and y = 14 (babysitting for 14 hours).
2. x = 10 (working 10 hours at the pharmacy) and y = 8 (babysitting for 8 hours).
2. To solve this problem and graph it, let's define the number of pants Theo buys as x and the number of shirts as y.
Since pants cost $30 each and shirts cost $15 each, the cost of the pants is 30x, and the cost of the shirts is 15y.
Theo's mother has given him at most $150, so we can write the inequality as:
30x + 15y ≤ 150
To graph this inequality, let's rewrite it in slope-intercept form:
30x + 15y ≤ 150
Divide both sides by 15 to simplify:
2x + y ≤ 10
Now, let's plot this inequality on a graph:
Start by plotting the point (0,10) on the y-axis (since y = 10 when x = 0), then use the slope of -2 (rise -2, run 1) to plot more points and draw a line. Since the inequality is "less than or equal to," shade the region below the line.
Possible combinations of pants and shirts that satisfy this inequality could be:
1. x = 5 (buying 5 pants) and y = 0 (not buying any shirts).
2. x = 0 (not buying any pants) and y = 10 (buying 10 shirts).
3. x = 2 (buying 2 pants) and y = 6 (buying 6 shirts).
3. To solve this problem and find all possible combinations, let's define the weight of mangoes sold as x and the weight of apples sold as y.
Since mangoes are sold for $4/lb and apples for $3/lb, the revenue from mangoes sold is $4x, and the revenue from apples sold is $3y.
The grocer's goal is to make at least $300, so we can write the inequality as:
4x + 3y ≥ 300
To simplify this inequality, let's divide both sides by 1 to keep the coefficients whole numbers:
x + 0.75y ≥ 75
To find possible combinations, we need to find integer values for x and y.
Let's list some possible combinations:
1. x = 15 (selling 15 lb of mangoes) and y = 0 (not selling any apples).
2. x = 0 (not selling any mangoes) and y = 100 (selling 100 lb of apples).
3. x = 10 (selling 10 lb of mangoes) and y = 40 (selling 40 lb of apples).
These are two possible combinations that could meet the goal.