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Let x=second number
Then x+3=first number
Now our equation to solve is:
x(x+3)=89-49=40 simplify
x^2+3x-40=0 quadratic in standard form and it can be factored
(x+8)(x-5)=0
x=-8
and
x=5
CK for x=-8
(-8)(-8+3)=40
40=40
CK for x=5
5(8)=40
40=40
Both answers work
Hope this helps--ptaylor

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Let x=your brother's age
Now the problem tells us that 98 years ago, your were 2 years old and your brother was (1/2)*2=1 year old
So our equation to solve is:
x-98=1 add 98 to each side
x=99 years old----your brother's age
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 walked one way; 2d=total distance walked
Time walking at 5 mi/hr=d/5
Time walking at 4 mi/hr=d/4
And we are told that the above times add up to 3 3/20 hr.
3 3/20 hr=63/20 hr
So our equation to solve is:
d/5 +d/4=63/20 multiply each term by 20
4d+5d=63
9d=63
d=7 mi
2d = total distance walked =2*7=14 mi---ans
Hope this helps---ptaylor

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Area of Rectangle (A)=Length(L)*Width(W) or A(Rectangle)=LW
Area of a Square (A)=Side(S)* Side(S) or A(square)=S^2
Let W=width of rectangle which is also the length of the side of a square
Then the length,L, of the rectangle=W+8
Area, A, of rectangle=W(W+8)
Area of Square=W^2
Now we are told the following:
W^2+160=W(W+8) simplify
W^2+160=W^2+8W collect like terms (W^2's cancel out)
8W=160
W=20 ft------------------width of rectangle
W+8=20+8=28 ft----Length of rectangle
CK
20*28=20*20+160
560=400+160
560=560
Hope this helps----ptaylor

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3/8-1/2+5/6
First, we need to find the Least Common Multiple (LCM) of the three denominators (8, 2, 6). The LCM is the smallest number that is a multiple of each denominator. There are several ways to find the LCM. I will just list some of the multiples of each denominator and see what we have:
8=8, 16, 24, 32, 40----
2=2, 4, 6, 8, 10, 12 , 14, 16, 18, 20, 22, 24-----
6=6, 12, 18, 24,----
Sooo, by inspection, we see that 24 is the LCM
3/8=9/24
1/2=12/24
5/6=20/24
Now we have
9/24-12/24+20/24 Now we can add these fractions and find out how many 24th's the we have:
29/24-12/24=17/24-----------------answer
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 r=Anna's rowing speed in still water
Then r+3=Anna's rowing speed downstream
and
r-3=Anna's rowing speed upstream
Time required to row downstream=216/(r+3)
Time required to row upstream=216/(r-3) and we are told that this took 18 hours longer, sooooooo:
216/(r+3) +18=216/(r-3) multiply each term by (r+3)(r-3)
216(r-3)+18(r+3)(r-3)=216(r+3) next divide each term by 18 and simplify
12r-36+r^2-9=12r+36 collect like terms
r^2-81=0
r=+or-9 ---- -9 is no good; can't have negative rowing speeds
so
r= 9 mph Anna's roving speed in still water
CK
216/12 + 18=216/6
18+18=36
36=36
Hope this helps---ptaylor

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Let x=Linda's age
Then x+11=Steve's age
Now we are told that:
x+x+11=113 subtract 11 from each side
x+x+11-11=113-11 collect like terms
2x=102 divide both sides by 2
x=51 years old------------------Lindas age
x+11=51+11=62 years old----Steve's age
CK
51+62=113
113=113
Hope this helps---ptaylor

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Let x=time it takes Jacqueline to do the job alone
So Jacqueline works at the rate of 1/x of the job per hour
Then x-15 =time it takes Chase to do the job alone
And Chase works at the rate of 1/(x-15) of the job per hour
Together they work at the rate of 1/x + 1/(x-15) of the job per hour
But we are told that they can do the job, working together, in 4 hours
So together they work at the rate of 1/4 of the job per hour
Then our equation to solve is:
1/x + 1/(x-15)=1/4 multiply each term by 4x(x-15)
4(x-15)+4x=x(x-15) simplify
4x-60+4x=x^2-15x collect like terms
x^2-23x+60=0 quadratic in standard form and it can be factored
(x-20)(x-3)=0
x=20 hours---Time it takes Jacqueline to do the job working alone
x-15=20-15=5 hours-----Time it takes Chase to do the job working alone
and
x=3 hours---- IMPOSSIBLE ANSWER---WHY????
CK
1/20+1/5=1/4
1/20+4/20=1/4
5/20=1/4
1/4=1/4
Hope this helps----ptaylor

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Area of a rectangle=Length*Width or A=LW
A=(x+6)(x-9)=x^2-3x-54
Hope this helps--ptaylor

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Let x=amount of time it takes John to do the work
Then John works at the rate of 1/x of the job per day
x+10=amount of time it takes Ganesh to do the work
Then Ganesh works at the rate of 1/(x+10) of the job per day
Together they complete the work in 12 days
So, together they work at the rate of 1/12 of the job per day
So our equation to solve is:
1/x +1/(x+10)=1/12 multiply each term by 12x(x+10)
12(x+10)+12x=x(x+10)
12x+120+12x=x^2+10x collecting like terms, we get
x^2-14x-120=0 quadratic in standard form and it can be factored
(x-20)(x+6)=0
x=20-----------------amount of time it takes John to do the work
x+10=20+10=30 days----ANSWER-- time it takes Ganesh working alone
and
x=-6---------------work is positive---no good
CK
1/20+1/30=1/12
3/60+2/60=1/12
5/60=1/12
1/12=1/12
Hope this helps---ptaylor

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Let x= the positive number; x^2=the square of the positive number
x+4= four more than the number; (x+4)^2=the square of 4 more than the positive number
Now we are told that:
x^2+(x+4)^2=58 simplify
x^2+x^2+8x+16=58
2x^2+8x+16=58
2x^2+8x-42=0 divide each term by 2
x^2+4x-21=0---quadratic in standard form and it can be factored
(x+7)(x-3)=0
x=-7 and
x=+3---------------------------answer (problem calls for positive number)
CK
3^2+7^2=58
9+49=58
58=58
Hope this helps--ptaylor

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Let x=Jill's age now; x+10=Jill's age 10 years from now
Then x+14=Jack's age now; x+24=Jack's age 10 years from now
Now we are told that:
x+24=2(x+10) simplify
x+24=2x+20 subtract x and also 20 from each side
x=4---Jill's age now
x+14=18 Jack's age now
Then in 5 years Jack will be 18+5=23 years old
Hope this helps----ptaylor

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Let x=amt of pure silver needed
Now we know that the amount of PURE silver that exists before the mixture takes place has to equal the amount of PURE silver that exists after the mixture takes place:
Pure silver before the mixture takes place: x+180*0.75=x+135
Pure silver after the mixture takes place: (180+x)*0.80=144+0.80x
Sooooooooooooo:
x+135=0.80x+144 subtract 0.80x and also 135 from both sides
x-0.80x+135-135=0.80x-0.80x+144-135 collect like terms
0.20x=9
x=45 oz-----------------amt of pure silver needed
CK
45+135=144+0.80*45
180=180
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
t=35 min or 35/60=7/12 hr
d=1.5 mi
soooooooooooooo:
r=d/t=1.5 divided by 7/12 or 1.5/7/12. Lets multiply the numerator and denominator by 12/7. That will get rid of the complex fraction. When we do this, we get:
(1.5)(12/7) divided by (7/12)(12/7) and this reduces to: 18/7/1 or 18/7=
2.571 mi/hr-----------------ans
CK
d=rt; 1.5=(18/7)(7/12)=18/12=3/2=1.5
1.5=1.5 CK
Hope this helps---ptaylor

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Let x=smaller number
And let y=larger number
Now we are told that:
x+y=3x or
y=2x-------------------eq1
and
y-x=x or
y=2x-------------------eq2
This system of equations is consistent and dependent which means there is an infinite number of solutions. You can check this out by choosing a number for x then solve for y and see if the two numbers work.
Hope this helps---ptaylor

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Each drain empties at the rate of (1/9)V per hour where V is the volume of each pool
When the second drain is opened, the first pool has emptied 2*(1/9)V leaving V-(2/9)V=(7/9)V yet to be emptied.
Let t=number of hours it takes AFTER the second drain is opened for the second pool to have three time the remaining volume of the first pool
Amount of water remaining in the first pool after t hours=(7/9)V-(t/9)V
Amount of water remaining in the second pool after t hours V-(t/9)V and this is three time the remaining volume of the first pool, soooo:
3{(7/9)V-(t/9)V}=V-(t/9)V simplify
(21/9)V-(3t/9)V=V-(t/9)V
(21/9)V-V=(3t/9)V-(t/9)V
(12/9)V=(2t/9)V divide each side by V and multiply each side by 9
2t=12
t=6 hr--number of hours it takes AFTER the second drain is opened
Second drain is opened at 8 am; 8 am plus 6 hours=2 pm --the time at which the second pool has 3 times the volume of the first pool
CK
In 6 hours the first pool has drained 6+2=8/9 its volume, leaving (1/9)V
In 6 hours the second pool has drained 6/9 its volume leaving (3/9)V and this is three times more that (1/9)V
Hope this helps---ptaylor

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Let x=number of gal of 6% butterfat needed
Then 246-x=number of gal of 3% butterfat needed
Now we know that the amount of pure butterfat in the 6% mixture (0.06x) plus the amount of pure butterfat in the 3% mixture(0.03(246-x)) has to equal the amount of pure butterfat in the final mixture(0.05*246), soooooo:
0.06x+0.03(246-x)=0.05*246 simplify
0.06x+7.38-0.03x=12.3
0.03x=12.3-7.38
0.03x=4.92
x=164 gal------amount of 6% butterfat needed
246-164=82----amount of 3% butterfat needed
CK
164*0.06+82*0.03=0.05*246
9.84+2.46=12.3
12.3=12.3
Hope this helps---ptaylor

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Let x=B's current age; x-9=B's age 9 years ago
Then x+2=A's current age; x+2-9=x-7=A's age 9 years ago
And we are told that:
x-7=3(x-9)
x-7=3x-27
2x=20
x=10 --------------B's current age
x+2=10+2=12 A's current age
CK
12-9=3*(10-9)
3=3
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 s=speed of the current
Emily's speed upstream=14-s
Her speed downstream=14+s
time required to go 6 miles upstream=6/(14-s)
time required to go 8 miles downstream=8/(14+s)
and we are told that these times are the same, sooooo
6/(14-s)=8/(14+s) multiply each term by (14-s)(14+s) or just cross-multiply and we get:
6(14+s)=8(14-s) or
3(14+s)=4(14-s)
42+3s=56-4s
7s=14
s=2 mph----speed of current
CK
Emily's speed upstream=14-2=12mph
Her speed downstream=14+2=16 mph
time required to go 6 mi upstream=6/12=1/2 hr
time required to go 8 mi downstream=8/16=1/2 hr
Hope this helps ---ptaylor

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Let x=the smaller number; 2x=smaller number doubled
Then x+6=the larger number; 4(x+6)=4 times the larger number
Sooo our equation to solve is:
2x+4(x+6)=90
2x+4x+24=90 subtract 24 from each side
2x+4x+24-24=90-24 collect like terms
6x=66
x=11-------------------smaller number
x+6=11+6=17 larger number
CK
2*11+4*17=90
22+68=90
90=90
Hope this helps--ptaylor

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Let x=amount of time it takes Skyler to do the job
Then (x-3) = amount of time it takes George to do the job
Skyler works at the rate of 1/x job per hour
George works at the rate of 1/(x-3) job per hour
Together they work at the rate of 1/2 of the job per hour
sooooo:
1/x +1/(x-3)=1/2 multiply each side by 2x(x-3)
2(x-3)+2x=x(x-3) simplify
2x-6+2x=x^2-3x collect like terms
4x-6=x^2-3x
x^2-7x+6=0 quadratic in standard form and it can be factored
(x-6)(x-1)=0
x=6
and
x=1 Clearly x=1 does not work because George would have do do negative work to do the job in 3 hours fewer time, sooooo
x=6 hours time time it takes Skyler to do the job
x-3=6-3=3 hours time it takes George to do the job
CK
1/6+1/3=1/2
1/6 + 2/6=1/2
1/2=1/2
Hope this helps---ptaylor

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Let x=amount of time it takes Bill to paint the house by himself
Then Bill paints at the rate of 1/x house per hour
Sid paints at the rate of 1/10 house per hour
Sid and Bill together paints at the rate of 1/4 house per hour
soooooo:
1/10+1/x=1/4 multiply each term by 40x
4x+40=10x
6x=40
x=6 two thirds hours-------time it takes Bill to paint the house by himself
CK
1/10+1/(20/3)=1/4
1/10+3/20=1/4
2/20+3/20=1/4
5/20=1/4
1/4=1/4
Hope this helps---ptaylor

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Let x=amount of time it takes Steve to clean the room working alone
Then Steve works at the rate of 1/x room per hour
Beverly works at the rate of 1/4 room per hour
Beverly and Steve working together works at the rate of 1/(4/5) =5/4 room per hour (48 min=4/5 hr and we are working in hours)
Soooo, our equation to solve is:
(1/4)+(1/x)=5/4 multiply each term by 4x
x+4=5x
4x=4
x=1 hour----amount of time it takes Steve to clean the room working alone.
CK
in 4/5 of an hour Steve cleans 4/5 th of the room
in 4/5 of an hour Beverly cleans(1/4)(4/5) =1/5 of the room
1/5+4/5=1 (room that is)
Hope this helps---ptaylor

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-4(k-7)-(-5k-7)=8 get rid of parens
-4k+28+5k+7=8 collect like terms
-4k+5k+28+7=8
k+35=8 subtract 35 from each side
k=-35+8
k=-27
CK
-4(-27-7)-((-5*-27)-7)=8
-4*(-34)-(135-7)=8
136-128=8
8=8
Hope this helps---ptaylor

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OK. Now we know that for every 7+4=11 pieces of gold found, Jim finds 4 and Bob finds 7. In other words, Jim finds 4/11 of the gold and Bob finds 7/11 of the gold.
(4/11)*55 =20----amount found by Jim
(7/11)*55=35---amount found by Bob
CK
20/35=4/7
Hope this helps---ptaylor

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Let x=number of 1 dollar bills
And let y=number of 5 dollar bills
Now we are told that:
x+y=29------------------------------eq1
and
x+5y=73-------------------------------eq2
subtract eq1 from eq2:
4y=44
y=11 number of 5 dollar bills
from eq 1:
x+11=29
x=18---------------number of 1 dollar bills
CK
18+55=73
73=73
and
11+18=29
29=29

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 bike A and Bike B are approaching each other at the rate of 10+5=15 mph
Time it takes bikes to collide=t=d/r=15/15=1 hour
Sooooo, the fly has 1 hour of flying time before it gets squashed and in the one hour, it travels a distance of d=rt=20*1=20 mi

Hope this helps--ptaylor

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Well, Dan shovels at the rate of 1/45 driveway per min
Janice shovels at the rate of 1/75 driveway per min
soooo, ratio of Dan's shoveling time to Janice's shoveling time =(1/45)/(1/75)= (1/45)(75/1)=75/45=5/3. In other words, Dan can shovel 5 driveways to Janice's 3.
Hope this helps---ptaylor

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Let x=amount of liquid fertilizer that needs to be added
Now we know that the amount of pure fertilizer that exists before the mixture takes place has to equal the amount of pure fertilizer that exists after the mixture takes place, sooooo:
x=0.25(600+x)
x=150+0.25x subtract 0.25x from each side
x-0.25x=150
0.75x=150
x=200 ml----amount of liquid fertilizer needed
CK
200ml=0.25*800ml
200ml=200ml
Hope this helps---ptaylor

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You need to recheck the wording of the problem. Too many is's

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Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
1 kilometer=1000 meters
4.1 min=4.1*60=246 sec
d=1000 meters
time=246 sec
soooo, velocity=1000 meters/246 sec West=4.065 meters/sec WEST
P.S. Velocity is a vector. It is made up of both speed and direction
Hope this helps---ptaylor