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Let x=number of quarters
Then x+2=number of dimes
And 3(x+2)=number of nickels
Now lets deal in pennies:
25x+10(x+2)+3*5(x+2)=250
25x+10x+20+15x+30=250 subtract 20 and 30 from each side
25x+10x+20-20+15x+30-30=250-20-30 collect like terms
50x=200
x=4-------------------------number of quarters
x+2=4+2=6----------------------number of dimes
3(x+2)=3*6=18-------------------number of nickels
CK
4*25=100
6*10=60
18*5=90
100+60+90=250
250=250
Hope this helps----ptaylor

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OK
x+4=y subtract 4 and also y from each side
x-y=-4----------------------eq1
5y=5x+35 subtract 5y and also 35 from each side
5x-5y=-35 divide each term by 5
x-y=-5-----------------------eq 2
Equations 1 and 2 represent parallel lines, therefore they are independent and inconsistent
Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
also
Average speed =(total distance)/(total time)
Total distance=20+5+2=27 miles
Time traveling at 55 mph=20/55=4/11 hr
Time travelling at 40 mph=5/40=1/8 hr
Time traveling at 25 mph=2/25 hr
Total time=4/11 +1/8 + 2/25=0.364 +0.125+0.08=0.569 hr
Average speed=27/0.569=47.452 mph
Hope this helps----ptaylor

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If 24 men can complete the tower in 15 days, there are 24*15=360 man-days of labor involved in the project
In other words, 1 man can complete the project in 360 days.
9 men can complete the job in 360/9=40 days
And, of course, it takes 24 men 360/24=15 days to complete the job
Another way:
Let x=number of days it takes 9 men to do the job
The men work at the rate of 1/360 of the project per day
so, (9/360)*x=1 (1 tower, that is)
9x=360
x=40 days
Hope this helps---ptaylor

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Let x=amount of water needed
Now we can set up a simple ratio:
50 is to 6.5 as 0.5 is to x and we write this as follows:
50/6.5=0.5/x multiply each term by 6.5x (or simply cross-multiply)
50x=3.25
x= 0.065 qts per 1/2lb----amount of water needed
CK
(100)*(1/2)gal=50gal
100*0.065=6.5 qts
Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let change mi/hr into ft/sec:
1mi=5280ft
1 hr=60*60=3600sec, soooo
1 mi/hr=5280ft/3600sec=1.467 ft/sec, soooo
65 mi/hr=65*(1.467) ft/sec=95 1/3 ft
Hope this helps---ptaylor

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Let x=number of oranges that Leslie had at the beginning
After the first person came and bought a third, she had x-(1/3)x=(2/3)x left
After the second person bought 4, she had (2/3)x-4 left
After the third person bought a fourth of the remaining, she had ((2/3)x-4))-1/4((2/3)x-4) left and we are told this equals 15, sooooo
((2/3)x-4)-1/4((2/3)x-4)=15 lets first multiply each term by 4
4(2/3)x-4)-(2/3)x+4=60
(8/3)x-16-(2/3)x+4=60
(6/3)x-12=60
2x=72
x=36----amount she had at the beginning
Ck
36-(1/3)*36=24
24-4=20
20-(1/4)*20=15
15=15
Hope this helps---ptaylor

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Let x=son's present age
And let y father's present age
x-2=son's age 2 yrs ago; x+3=son's age 3 yrs hence
y-2=fathers age 2 yrs ago; y+3=father's age 3 yrs hence
Now we are told the following:
y-2=8(x-2) simplify
y-2=8x-16
y-8x=-14-------------------eq1
and
y+3=4.5(x+3)simplify
y+3=4.5x+13.5
y-4.5x=10.5----------------eq2
subtract eq1 from eq2
3.5x=24.5
x=7 yrs old-----------------present age of son
from eq1
y-56=-14
y=42---------------present age of father
CK
two years ago:
42-2=8*5
40=40
3 years from present:
42+3=(4.5)*10
45=45
Hope this helps-----ptaylor

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Initial speed of ball=55mph---------------------speed of train
Final speed of ball=initial speed(55mph) + speed that it was pitched(55mph)=110mph for a brief period until it slows down.
Try dropping something eco-friendly out the window of a moving car and observe what it does. It will head out in the direction the car is traveling, bouncing along the highway, at the same speed of the car for a brief instance.
Hope this helps-----ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Now you need to think about this a little. The cars are traveling at different rates, so X can't be the "rate of the cars". But X can be the rate of one of the cars:
Let X=rate of north-bound car
Then X+15=rate of south-bound car
Distance north-bound car travels after 6 hours=X*6
Distance south-bound car travels after 6 hours=(x+15)*6
Now we are told that these two distances add up to 630 mi, soooo
6X+6(X+15)=630 get rid of parens
6X+6X+90=630 subtract 90 from each side
12X=540
X=45 mph---------------------------rate of north-bound car
X+15=45+15=60 mph----------------------rate of south-bound car
CK
45*6+60*6=630
270+360=630
630=630
Another way:
We know the cars are separating at the rate of X+X+15mph=2X+15 mph
in 6 hours they have traveled (2X+15)*6=12X+90 miles and we are told that this equals 630 mi, sooo
12X+90=630----same as before
Hope this helps---ptaylor

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(n/2)-1/3=13/6 First, lets get rid of the fractions. We do this by multiplying each term in the equation by the lowest common denominator which is 6, sooo
6*(n/2)-6*(1/3)=6*(13/6) and this gives us
3n-2=13 add 2 to each side
3n-2+2=13+2 collect like terms
3n=15 divide each side by 3
n=5
CK
(5/2)-1/3=13/6 we could multiply by the lowest common denominator again or we could just add the fractions. Lets add the fractions:
5/2=15/6 (multiply numerator and denominator by 3)
1/3=2/6 (multiply numerator and denominator by 2), sooooo
15/6-2/6=13/6
13/6=13/6
Note that I used parens for some of the fractions (eg (n/2)) just to make sure there is no confusion.
n/2-1/3 could be read as n/(2-1/3) or (n/2)-1/3
Parens helps to underscore that some expressions are one quantity.

Hope this helps----ptaylor

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Let x=amount of water that needs to be boiled off
Now we know that the total amount of salt that exists before before x amount of water is boiled off has to equal the total amount of salt that exists after the water is boiled off.
Total amount of salt before the water is boiled off is (2 liters)*90 g/l=180g
Total amount of salt after x amount of water is boiled off is (2-x)liters*180g/l
Soooo
180g=(2-x)180g/l
180=360-180x
-180x=-180
x=1 liter
Hope this helps ---ptaylor

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OK
Let x=Sammy's age
Three times Sammy's age is 3x
Four years more than three times Sammy's age is 3x+4 and this is Wendy's age, sooo
x+3x+4=60 subtract 4 from each side
4x=56
x=14-----------Sammy's age
3x+4=3*14+4=46---Wendy's age

Hope this helps---ptaylor

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Let x=amount of soybean meal needed
Then 320-x=amount of cornmeal needed
Now we know that the amount of pure protein that exists before the mixture takes place has to equal the amount of pure protein that exists after the mixture takes place, sooooo:
0.16x+0.08(320-x)=0.11*320
0.16x+25.6-0.08x=35.2 subtract 25.6 from each side
0.08x=9.6
x=120 pounds----amount of soybean meal needed
320-120=200 pounds--amount of cornmeal needed
CK
0.16*120+0.08*200=35.2
19.2+16=35.2
35.2=35.2
Hope this helps----ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Ok, lets deal in minutes:
5 meters/sec=5*60 =300 meters/min
2 meters/sec=2*60=120 meters/min
Let d=one-way distance
t1=time required jogging out
t2=time required walking back
t1+t2=35
t1=d/300
t2=d/120
d/300+d/120=35 multiply each term by 600
2d+5d=21000
7d=21000
d=3000 meters
2d=6000 meters total meters travelled
Hope this helps--ptaylor

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60 sec=1 min
in 1 minute she runs 20*60 ft or 1200 ft
in 30 minutes, she runs 30*1200 ft=36000 ft and this equals 36000/5280= 6.8 mi
Hope this helps---ptaylor

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Well, we know that 1/2 times y is y/2 just like we know that 1/2 times 4 is 4/2=2
Now if we divide that by 3, we have (y/2)/3. This is a complex fraction and we can simplify it by multiplying the numerator and denominator by (1/3), like so:
(y/2)(1/3)/3(1/3) and this reduces to (y/6)/1 or y/6
And we are told that this equals 2, sooooo
(y/6)=2 multiply each side by 6
y=12
CK
1/2 times 12=6; 6 divided by 3=2
Hope this helps---ptaylor

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6r+10=4r+22 First, subtract 4r and also 10 from each side and we get:
6r-4r+10-10=4r-4r+22-10 Next, collect like terms
2r=12 Now divide each side by 2
r=6
Now you can review the rules that allow me to do what I did
CK
6*6+10=4*6+22
36+10=24+22
46=46
Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
ok. Looks like you gave it a shot!!! I am assuming that the cars are going in the same direction. If they were going toward each other, we would have to know how far apart they were to begin with.
When car B starts, car A will have already travelled 70*1=70km.
Car B will have overtaken car A when they have both travelled the same distance
Let t =the amount of time needed for them to travel the same distance AFTER car B starts.
Distance travelled by car A=70+70t
Distance travelled by car B=90t
When these distances are the same, the two cars are together, soooo:
70+70t=90t
20t=70
t=3.5 hr
Distance car A travelled =70+70*3.5=315km
Distance car B travelled=90*3.5=315km

Hope this helps----ptaylor

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0.6x+5<1.4x-8 subtract 1.4x and also 5 from each side
0.6x-1.4x+5-5<1.4x-1.4x-8-5 collect like terms
-0.8x<-13 multiply each side by -1 (Must now change the inequality sign)
0.8x>13
x>16.25
Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
So:
d=rt
22=r*47
r=22/47= km/min
t=d/r
t=37/(22/47)=37*(47/22)=1739/22=79.045 min
You can solve it by dealing with decimals instead of fractions if you like
Hope this helps---ptaylor

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Let x=amount of time it takes Abbie to paint the room
Then Abbie paints at the rate of 1/x rooms/min
2x=amount of time it takes Beth
Beth paints at the rate of 1/2x rooms/min
3x=amount of time it takes Cathy
And Cathy paints at the rate of 1/3x rooms/min
Together they paint at the rate of 1/x + 1/2x +1/3x=11/6x rooms/min
Now we are told that (11/6x)*60=1 (1 room, that is), or
x=110 min----------------Amount of time it takes Abbie
and 2*110=220 min--------Amount of time it takes Beth
3*110=330 min---------------Amount of time it takes Cathy
CK
In 60 min Abbie paints 60/110 of the room
Beth paints 60/220 of the room
Cathy paints 60/330 of the room
60/110 + 60/220 + 60/330=1 (1 room, that is)
6/11 + 6/22 + 6/33=1 multiply each term by 66
36+18+12=66
66=66
Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Now we know that the two cars are separating at the rate of 60+72=132kph
So, in 90 min (1.5hr) they are 132*(1.5)=198 km apart.
Another way:
In 1.5hr, car A travels 60*(1.5)=90km
In 1.5hr, car B travels 72*(1.5)=108km
90+108=198km

Hope this helps---ptaylor

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let t1= time to travel from A to B
And let t2= time to travel from B to A
Now we are told that t1 + t2=5 and we know that:
t1=d/81 and
t2=d/54 soooooo:
d/81 + d/54 = 5 multiply each term by 162
2d+3d=810
5d=810
d=162 mi
t1=162/81=2 hours---------------------trip by car
and t2=162/54=3 hours--------------------trip by bus
Trip by car + trip by bus=5 hrs
2+3=5
5=5
Hope this helps---ptaylor

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Let x=rate per hour that Cindy's clock loses time, so:
5*x=4/10
x=4/50=2/25 minute per hour
In 14 hours, then, Cindy's clock will lose (2/25)*14=28/25= 1.12 min
We could also solve this problem by setting up a ratio
5 is to 0.4 as 14 is to ?
Hope this helps---ptaylor

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Let w=width of garden
We are told that the perimeter of the garden is 32 meters and we know that the perimeter equals 2l+2w, so:
2*4+2w=32
2w=32-8
2w=24
w=12 meters
Hope this helps---ptaylor

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Let x= the number
then 60% of x=0.60x
So, we are told that: x+0.60x=192 or
1.6x=192
x=120
CK
0.60*120=72
120+72=192
192=192
Hope this helps---ptaylor

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Distance(d) equals Rate (r) times Time (t) or d=rt; r=d/t and t=d/r
Let d=distance to her appointment
And t=time required to travel to her appointment
Then d=60t
On the return trip:
d=28(t+24/60)=28(t+2/5) (Lets deal in hours for the time being )
So our equation to solve is:
60t=28(t+2/5)
60t=28t+56/5
32t=56/5
160t=56
t=56/160=7/20 hr or 21 min
d=60t=60*7/20 =21 mi
CK
at 28 mph on the return trip, it would take her
21/28 = 3/4 hr or 45 min
45-21=24 min longer

Hope this helps---ptaylor

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Let x=amount of 60% solution needed
Then 20-x=amount of 40% solution needed
Now we know that the amount of pure alcohol that exists before the mixture takes place is equal to the amount of pure alcohol after the mixture takes place, so:
0.60x+0.40(20-x)=amount of pure alcohol before the mixture takes place, and
0.45*20=amount of pure alcohol after the mixture takes place,so
0.60x+0.40(20-x)=0.45*20 simplify
0.60x+8-0.40x=9 subtract 8 from each side
0.20x=1
x=5 liters-------------------------amount of 60% solution needed
20-x=20-5=15 liters----------------amount of 40% solution needed
CK
0.60*5+0.40*15=0.45*20
3+6=9
9=9
Hope this helps---ptaylor

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Let x=the smaller part
Then 55-x=the larger part
Now we are told the following:
2(55-x)=5x+12 simplify
110-2x=5x+12 add 2x to and subtract 12 from each side
110-2x+2x-12=5x+2x+12-12 collect like terms
98=7x
x=14 in--------------------smaller part
55-x=55-14=41 in-----------larger part
CK
2 times larger part=2*41=82 in
5 times smaller part=5*14=70 in
70+12=82
82=82
Hope this helps -----ptaylor

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Let x=boy's age now
And let y=sister's age now
We are told that:
x+2=(3/4)(y+2) ---------------------eq1
We are also told that:
x-2=(2/3)(y-2)----------------------eq2
simplifying eq1, we get:
x+2=(3/4)y+6/4 multiply each term by 4:
4x+8=3y+6
or 4x-3y=-2------------------------------------eq1a
Now we simplify eq2:
x-2=(2/3)y-4/3 multiply each term by 3:
3x-6=2y-4
3x-2y=2--------------------------------------eq2a
Next, we multiply each term in eq1a by 2 and each term in eq2a by 3:
8x-6y=-4----eq1b
9x-6y=6------eq2b
subtract eq1b from eq2b:
x=10--------------------------------boy's age now
substitute x=10 into eq 1a:
40-3y=-2
-3y=-42
y=14----sister's age now
CK
10+2=(3/4)(14+2)
12=12
and
10-2=(2/3)(14-2)
8=8
Hope this helps---ptaylor