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Question 1:
4x-8≤-5 or x-7≥1
We solve each side separately
4x-8≤-5 add 8 to both sides
4x≤3 divide both sides by 4
x≤3/4
x-7≥1 add 7 to both sides
x≥6
So we have x≤3/4 or x≥6
This means we’d have a solid dot at 6 with the line shaded to the right, and a solid dot at 3/4 with the line
shaded to the left (or in your case, brackets instead of parentheses, which would be graph C).
Question 2:
Determine the important information.
- 16 commercial spots needed
- commercials are 30 or 60 seconds
- commercial time is 13 minutes
Set up an equation, with x being 30 second commercials and y being 60 second commercials
X+y=16
.5x+y=13 (we use .5 for 30 seconds and 1 for 60 seconds to make the math use smaller numbers)
Now we solve by substitution
Y=16-x (from the first equation solved for y)
.5x+16-x=13 (replace y with 16-x)
16-.5x=13 (combine like terms)
-.5x=-3 (subtract 16 from both sides)
x=6 (multiply both sides by –2)
So there were 6 30-second commercials.
6+y=16
y=10 (10 60-second commercials)
Now we substitute the values into the second equation to check
.5*6 + 10 = 3+10=13
Second equation works; answer is 6 30-second commercials and 10 60-second commercials
Question 3:
These lines are parallel, which means there is no solution (no point at which they will cross)
If it has no solutions, that means it is inconsistent.
If it has no solutions, that also means it is neither dependent (infinitely many solutions) or independent
(only one solution) but if you have to pick one, I would pick independent, since dependent means that they
are the same line.
We can use the intercepts to identify the equations for the lines. The red has a y-intercept of 6 and a slope
of –1 (since it goes down one for every one it goes right). The equation is y=-x+6. The blue has a yintercept of 3 and a slope of –1. The equation is y=-x+3. y=-x+6 matches the second option. We can write
the equation of the blue line as x+y=3 and then it also matches the second option, so that is the correct
choice.
Question 4:
3x+9y=12
-9x+y=76
To use substitution, we solve one of the equations for one of the variables. Because 76 is not easily
divisible by 9, I’ll solve the first equation for x.
3x+9y=12
x+3y=4 (divide both sides by 3)
x=4-3y (subtract 3y from both sides)
Now I put in the value for x in the second equation
-9(4-3y)+y=76
-36+27y+y=76 (using the distributive property)
-36+28y=76 (combine like terms)
28y=112 (add 36 to both sides)
y=4 (divide both sides by 28)
In the first equation, put 4 in for y.
3x+9*4=12
3x+36=12
3x=-24 (subtract 36 from both sides)
x=-8 (divide both sides by 3)
x=-8, y=4
Check using second equation
-9*-8+4=72+4=76.
Equation works out, answer is correct: -8, 4
Question 5:
Here we can just check each given number.
y-8>2y-4
Check 6:
6-8=-2
2*6-4=12-4=8
-2 is not greater than 8; 6 is not a solution
Check –15:
-15-8=-22
2*-15-4=-30-4=-34
-22 is greater than –34; -15 is a solution
Check –17:
-17-8=-25
2*-17-4=-34-4=-38
-25 is greater than –38; -17 is a solution
Check –2:
-2-8=-10
2*-2-4=-4-4=-8
-10 is not greater than –8; -2 is not a solution
We could also solve the inequality to check:
y-8>2y-4
y-4>2y (add 4 to both sides)
-4>y (add y to both sides)
y<-4 so this means any numbers less than –4 will work, which means –15 and –17 work but 6 and –2 do
not.
Question 6:
-1/2 x ≥ -3/4
Multiply both sides by –2
2*(-1/2)x≥2*(-3/4)
x≥-3/2
Question 7:
Identify important information:
- 2 points for win and 1 point for tie
- 58 points total
- 11 more wins than ties
Let w=wins and t=ties
2w+t=58
w-t=11
Solve second equation for w by adding t to both sides
W=11+t
Substitute in first equation
2(11+t) +t=58
22+2t+t=58
22+3t=58
3t=36 (subtract 22 from both sides)
t=12 (divide both sides by 3)
2w+12=58 (substitute 12 for t)
2w=46 (subtract 12 from both sides)
w=23 (divide both sides by 2)
t=12, w=23
Use second equation to check
23-12=11
Answer: 23 wins, 12 ties
Question 8:
If x<9 and x≥-6, this means x is between –6 and 9
Because the symbol for 6 is “or equal to” we use a bracket instead of parenthesis.
Graph C is correct.
Writing in interval notation, we use the same parenthesis/bracket format.
[-6, 9)
We use brackets for greater than/equal to and less than/equal to; we use parentheses for greater than and
less than.
Question 9:
X≥0 and y≥0 mean that the shaded region will only be in the first quadrant (upper right)
Then we can choose values to make points for the other two equations.
X+3y≤9
Set y=0, this gives us x=9. So our first point is 9, 0
Set x=0, this gives us y=3. So our second point is 0, 3.
We put a dot at 0,3 and 9,0 and connect them.
3x+y≤9
Set y=0, this gives us x=3. So our first point is 3,0
Set x=0, this gives us y=9. So our second point is 0, 9.
We put a dot at 3,0 and 0,9 and connect them.
Since all signs are “or equal to” we use solid lines.
The shaded section will be in the first quadrant underneath both lines.
If the written instruction is unclear, you can use this tool to help you visualize the graphs:
http://www.ronblond.com/M11/LinIne/index.html
(this website will help you with all of the graphing ones)
Question 10:
2x+3y≤6
Set x=0, then y=2 (3*2=6)
Set y=0, then x=3 (2*3=6)
So the points are 0, 2 and 3, 0; connect these with a solid line and shade underneath
Question 11:
3x-(4x+5)≤2(x+2)+x
First get rid of parentheses
3x-4x-5≤2x+4+x
Combine like terms
-x-5≤3x+4
Add x to both sides
-5≤4x+4
Subtract 4 from both sides
-9≤4x
Divide both sides by 4
-9/4≤x
x≥-9/4
Using set builder notation, we do
(x|x≥-9/4)
Question 12:
4x-12y=1
4x-12y=2
This system has no solutions, since the same variables equation different numbers. I noticed the second
equation is indented, so if there is a – sign that didn’t copy properly, this is the method.
4x-12y=1
-4x-12y=2
Add equations together
0x-24y=3
y=-3/24=-1/8
4x-12(-1/8)=1
4x+12/8=1
4x=1-(12/8)=1-3/2=-1/2
4x=-1/2
x=-1/8
Check in second equation:
-4(-1/8)-12(-1/8)=-4/8+12/8=1/2+3/2=2
So if the second equation should have a negative sign in front, the solution is x=-1/8, y=-1/8
Question 13:
5x+y<-4
Test numbers given
5*-5+3=-25+3=-22
-22<-4, so yes
Question 14:
If we graph the two equations, we find that they cross at 6, -4 (you can confirm this by using the graphing
website I showed you above). If you were to use the method explained before, this is the procedure.
6x-y=40
Using 0 for x gives us 40 for y, which is quite large, so we’ll use larger numbers for x.
X=5. Then 30-y=40 so y=-10
5, -10 is our first point
x=8. Then 48-y=40 so y=8
8, 8 is our second point.
6x+7y=8
x=-1. Then –6+7y=8; 7y=14; y=2
-1, 2 is our first point
x=6. Then 36+7y=8; 7y=-28; y=-4
6, -4 is our second point.
Since there is a solution, this system is consistent and independent. Graphs crossing at 6, -4 matches graph
B.
Question 15:
Determine important information:
- replacement means repair >80%
- repair cost 6400
- car not replaced
So we have 6400<80%of the value
6400<.8x
Divide both sides by .8
8000<x
So x>8000, meaning the car’s value is greater than 8000(B)
Question 16:
3/4(3x-1/2)-1/3<2/3
Add 1/3 to both sides
3/4(3x-1/2)<1
Get rid of parentheses
(9/4)x-3/8<1
Add 3/8 to both sides
(9/4)x<11/8
Divide both sides by 9/4 (this is the same as multiplying 11/8 * 4/9)
X<44/72
Simplify fraction
X<11/18
Set-builder notation:
(x|x<11/18)
Question 17:
T+19≥6
Subtract 19 from both sides
T≥-13
Bracket on –13 and line shaded right (graph C)
Question 18:
X+y=90
X+.25y=57.75
X=90-y
90-y+.25y=57.75
90-.75y=57.75
-.75y=-32.25
Divide both sides by -.75
Y=43
X+43=90; x=47
Check with second equation
47+.25*43=47+10.75=57.75
Angles are 43, 47
Question 19:
C=.25(t-400)+46.95
Need 88.45<c<110.95
88.45=.25(t-400)+46.95
41.5=.25(t-400) (subtract 46.95 from both sides)
166=t-400 (divide both sides by .25)
566=t (add 400 to both sides)
110.45=.25(t-400)+46.95
63.5=.25(t-400)
254=t-400
654=t
t must be between 566 and 654 minutes
Question 20:
X=6 is a vertical line at 6
Y=-8 is a horizontal line at –8
They will cross at 6, -8
One solution, so it is consistent and independent
Question 21:
-11≤2x+5
Subtract 5 from both sides
-16≤2x
Divide both sides by 2
-8≤x
2x+5<13
Subtract 5 from both sides
2x<8
Divide both sides by 2
X<4
So –8≤x<4
Bracket at –8 and parentheses and 4, shaded between (graph A)
Question 22:
Interval notation means we need a bracket at 6 (since the number line has a bracket) and then it goes to
infinity since there is no stopping point (graph A, because we use parentheses for infinity)
Question 23:
X+9y=17
-x+6y=-2
Add equations together
15y=15
y=1
Substitute y in first equation
X+9=17
X=8
Check using second equation
-8+6*1=-8+6=-2
Solution is 8, 1
Question 24:
Graph 1 has a slope of –1 and a y-intercept of –3 and is shaded below with a solid line.
Y≤-x-3
Y+x≤3 (equation b)
Graph 2 has a slope of 3/4 and a y-intercept of 3 and is shaded below with a solid line.
Y=(3/4x)+3
4y=3x+12
4y-3x=12
We can rearrange this to match equation b
4y-3x=12
-3x=12-4y
3x=-12+4y
3x=4y-12
Graph 3’s options are not visible but I will give you the equation in several forms and hopefully you will be
able to determine the correct one (you can also leave a note and I’ll help you out)
Slope is 3/2 and y-intercept is –2; shaded above with dashed line
Y=(3/2)x-2
2y=3x-4
3x+2y-4
3x=-2y-4