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Let number = x
(1) 5 decreased by the product of a number and 3 = 3x-5

(2) The sum of -9 times a number and 6 is 72 = 6-9x=72

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Distance = Speed * time
Distance = 50 * 3
Distance = 150 miles

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or
or

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Let number = x
Twice a number = 2x
Increased by 7 = 2x+7
2x+7=19
2x=19-7
2x=12
2x/2=12/2
x=6
Number is = x = 6

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Let
Smaller Number = x
Larger Number = y

Half the sum of two numbers is 20
(x+y)/2=20
x+y=2*20
x+y=40...............(1)

Three times their difference is 18
3(y-x)=18
3(y-x)/3=18/3
y-x=6..............(2)
Add (1) and (2)
x+y=40...............(1)
y-x=6..............(2)
---------------------
2y=46
2y/2=46/2
y=23
Put the value of y in (1)
x+y=40...............(1)
x+23=40
x=40-23
x=17

Smaller Number = x = 17
Larger Number = y = 23

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Let
Edz present age = e
Noe present age = n

Edz is twice as old as Noe
e=2n..............(1)
The sum of their ages is 66
e+n=66............(2)
Put the value of e in (2)
2n+n=66
3n=66
3n/3=66/3
n=22
Put the value of n in (1)
e=2n..............(1)
e=2(22)
e=44


Edz present age = e = 44
Noe present age = n = 22

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Let
Speed of slower train = x
Speed of faster train = x+10.3

Speed of slower train + Speed of faster train = Sum of speeds
x+x+10.3=723.7
2x+10.3=723.7
2x=723.7-10.3
2x=713.4
2x/2=713.4/2
x=356.7


Speed of slower train = x = 356.7
Speed of faster train = x+10.3 = 356.7+10.3=367

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In 80 mins...............7 miles
In 80/80 mins............7/80 miles
In 1 mins............7/80 miles
In (32)*1 mins.........(7/80)*(32)miles
In 32 mins.........(7/80)(32)miles
In 32 mins.........(7/5)(2)miles
In 32 mins.........14/5miles
In 32 mins.........2.8miles

Or
In 1 minute keiko runs = 7/80 = 0.0875 miles
In 32 minutes keiko run = 0.875*32 = 2.8 miles

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Let x and y be the two numbers
x+y=2.7
x+y=27/10
10(x+y)=27
10x+10y=27.............(1)

xy=1.8
xy=18/10
x=18/10y...............(2)

Put the value of x from (2) to (1)
10(18/10y)+10y=27
18/y+10y=27
(18+10Y^2)/y=27
18+10Y^2=27y
10y^2-27y+18=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=9 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.5, 1.2. Here's your graph:

So
y=1.5 or 1.2

Put the values of y in (1)
10x+10y=27.............(1)
Put y=1.5
10x+10(1.5)=27
10x+15=27
10x=27-15
10x=12
x=12/10
x=1.2

Put the values of y in (1)
10x+10y=27.............(1)
Put y=1.2
10x+10(1.2)=27
10x+12=27
10x=27-12
10x=15
x=15/10
x=1.5
So numbers are 1.2 and 1.5


Check
=====
xy=1.8
(1.2)(1.5)=1.8
1.8=1.8

x+y=2.7
1.2+1.5=2.7
2.7=2.7

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Let
Present age of mother = m years old
Present age of father = f years old
Present age of John = j years old

Two years hence
mother will be = m+2 years old
John will be = j+2 years old
Then
mother will be twice his age
m+2=2(j+2)
m+2=2j+4
m=2j+4-2
m=2j+2.................(1)


Five years hence
father will be = f+5 years old
John will be = j+5 years old
Then
father will be thrice his present age
f+5=3(j+5)
f+5=3j+15
f=3j+15-5
f=3j+10..................(2)

the sum of all the three ages is 105
m+j+f=105
Put the value of m and f from (1) and (2) to above equation
2j+2+j+3j+10=105
6j+12=105
6j=105-12
6j=93
6j/6=93/6
j=31/2
Put the value of j in (1)
m=2j+2.................(1)
m=2(31/2)+2
m=31+2
m=33
Put the value of j in (2)
f=3j+10..................(2)
f=3(31/2)+10
f=93/2+10
f=(93+20)/2
f=113/2
f=56.5
Difference b/w parents ages = f-m=56.5-33= 23.5 years.

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Lauren Purchased adult tickets = x
Lauren Purchased student tickets = 6x
Cost of a Adult ticket = 8.50
Cost of a student ticket = 4.75
Total Cost of all tickets = 111
(Lauren Purchased adult tickets)(Cost of a Adult ticket)+(Lauren Purchased student tickets)(Cost of a student ticket) = Total Cost of all tickets
(x)(8.50)+(6x)(4.75)=111
8.50x+28.50=111
37x=111
37x/37=111/37
x=3
Lauren Purchased adults tickets = x = 3
Lauren Purchased students tickets = 6x = 6*3 = 18


Check
======
Total Cost of Adults tickets = 8.50*3=25.50
Total Cost of students tickets= 4.75*18=85.50
Total Cost = 25.5+85.5=111

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Equation of the line slope-intercept form
y=mx+b
Given
Equation of the Given line
y=6
Compare with the equation of line slope-intercept form
Slope of the given = m = 0
Since lines are parallel their slope will be same
Slope of the required line = m = 0
Point (5, 7)
We have a point and slope we can get easily the require line equation.
Use Equation of the line point-slope form
m = y2-y1/x2-x1
Put the values in above equation, we have
m=(y-2)/(x-5)
0(x-5)=(y-7)
0=y-7
y=7
Above equation is the equation of the required line

Graph
======
y=6
y=7

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Let
Present age of Paul = p years old
Present age of Wendy = w years old

Paul is 11 years younger than Wendy.
w=p+11............(1)
In 10 years,
Paul will be = p+10 years old
Wendy will be = w+10 years old
Then
Wendy will be twice as old as Paul
w+10=2(p+10)
w+10=2p+20
w=2p+20-10
w=2p+10................(2)
Put the value of w from (1) to (2)
p+11=2p+10
p-2p=10-11
-p=-1
p=1
Put the value of p in (1)
w=p+11............(1)
w=1+11
w=12

Present age of Paul = p = 1 year old
Present age of Wendy = w = 12 years old

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Cream A
=======
Amount = x
Cream = 5%=0.05

Cream B
=======
Amount = 75-x
Cream = 0.5%=0.005

Mixture Cream
=============
Amount = 75g
Cream = 2%=0.02

(Amount)(cream A)+(Amount)(Cream B)= (Amount)(Mixture Cream)
0.05x+(75-x)0.005=(75)(0.02)
0.05x+(75-x)0.005=(0.02)(75)
0.05x+(75-x)0.005=1.5
0.05x+0.375-0.005x=1.5
0.045x+0.375=1.5
0.045x=1.5-0.375
0.045x=1.125
0.045x/0.045=1.125/0.045
x=25



Cream A
=======
Amount = x = 25g
Cream = 5%=0.05
Cream B
=======
Amount = 75-x=75-25=50g
Cream = 0.5%=0.005
Mixture Cream
=============
Amount = 75g
Cream = 2%=0.02


25g of a 5% cream and 50g of a 0.5% cream would be required to compound 75 g of a 2% cream.

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Solution A
===========
Amount = x gallons
Concentration = 20%=0.2



Solution B
===========
Amount = 30-x gallons
Concentration = 80%=0.8

Mixture
=======
Amount = 30 gallons
Concentration = 60% = 0.6


(Amount of Solution A)(Concentration of Solution A)+(Amount of Solution B)(Concentration of Solution B)=(Amount of Mixture)(Concentration of Mixture)
(x)(0.2)+(30-x)(0.8)=(30)(0.6)
0.2x+24-0.8x=18
0.2x-0.8x=18-24
-0.6x=-6
-0.6x/-0.6=-6/-0.6
x=10


10 gallon of 20% antifreeze solution must be mixed with 20 gallons of 80% antifreeze solution to produce 30 gallons of a 60% disinfectant solution.

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Solution A
===========
Amount = x gallons
Concentration = 61%=0.61



Solution B
===========
Amount = 68-x
Concentration = 27%=0.27

Mixture
=======
Amount = 68 gallons
Concentration = 44% = 0.44


(Amount of Solution A)(Concentration of Solution A)+(Amount of Solution B)(Concentration of Solution B)=(Amount of Mixture)(Concentration of Mixture)
(x)(0.61)+(68-x)(0.27)=(68)(0.44)
0.61x+18.36-0.27x=29.92
0.34x+18.36=29.92
0.34x=29.92-18.36
0.34x=11.56
0.34x/0.34=11.56/0.34
x=34



Solution A
===========
Amount = x = 34 gallons
Concentration = 61%=0.61
Solution B
===========
Amount = 68-x = = 68-34 = 34 gallons
Concentration = 27%=0.27
Mixture
=======
Amount = 68 gallons
Concentration = 44% = 0.44


34 gallons of each should be used.

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Let
Increase on all sides = x

Area of rectangle = length * width
A=L*W
A=15*30
A=450
After Increase
A=450+144
A=594
Length = 30+x
Width = 15+x
594=(15+x)(30+x)
594=450+15x+30x+4x^2
594=450+45x+x^2
594-450=45x+x^2
144=45x+x^2
x^2+45x-144=0
x^2+48x-3x-144=0
x(x+48)-3(x+48)=0
(x-3)(x+48)=0
x-3=0 or x+48=0
x=3 or x=-48 (unacceptable)
x=3


Each dimension increased by 3cm
Length = 30+x = 30+3 = 33cm
Width = 15+x = 15+3 = 18cm

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=
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=
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Let
Present age of Julies = j years old
Present age of Father = f years old

the sum of julies age and her fathers age is 58
j+f=58
j=58-f.............(1)
julies father is 10 years younger than triple Julie's age
3j=f-10............(2)
Put the value of j from (1) to (2)
3j=f-10
3(58-f)=f-10
174-3f=f-10
-3f-f=-10-174
-4f=-164
-4f/-4=-164/-4
f=41
Put the value of f in (1)
j=58-f.............(1)
j=58-41
j=17



Present age of Julies = j = 17 years old
Present age of Father = f = 41 years old

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if you question is as follows:

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Let
Present age of Katie = k years old

Six years ago
Katie was = k-6 years old
In Six years
Katie will be = k+6 years old
The square of katie's age six year ago is equal to her age in six years time
(k-6)^2=k+6
k^2-12k+36=k+6
k^2-13k+30=0
Solve for K
k^2-10k-3k+30=0
k(k-10)-3(k-10)=0
(k-3)(k-10)=0
k-3=0 or k-10=0
k=3 (unacceptable) or k=10
k=10

Present age of Katie = k = 10 years old

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You were = 4 years
Your brother was = 2 years (half of your age)
You are = (4+46) =50 years old now
Your brother is = (2+46) = 48 years old now

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You were = 4 years
Your brother was = 2 years (half of your age)
You are = (4+46) =50 years old now
Your brother is = (2+46) = 48 years old now

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New Per credit compensation = Old compensation + increase%
New Per credit compensation = 950+(950)(4/100)
New Per credit compensation = 950+(950)(1/25)
New Per credit compensation = 950+38
New Per credit compensation = 988

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=
=
=
=

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8y-1=x..............(1)
3x=2y...............(2)
Substitute the value of x from (1) to (2)
3(8y-1)=2y
24y-3=2y
24y-2y=3
22y=3
22y/22=3/22
y=3/22
Put the value of y in (2)
3x=2(3/22)
3x=3/11
3x/3=(3/11)/3
x=1/11
x,y=1/11,3/22


Check
=====
Put the values of x and y in (1)
8y-1=x..............(1)
8(3/22)-1=1/11
24/22-1=1/11
(24-22)/22=1/11
2/22=1/11

Put the values of x and y in (1)
3x=2y...............(2)
3(1/11)=2(3/22)
3/11=6/22

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Adults tickets sold = x
Children tickets sold= 3x
Total Amount = $700
(Cost of a children ticket)(Children tickets sold)+(Cost of an adult ticket)( Adult tickets sold)= Total Amount
3(3x)+5(x)=700
9x+5x=700
14x=700
14x/14=700/14
x=50


Adults tickets sold = x = 50
Children tickets sold= 3x = 3*50 = 150



Check
=====
3*150+5*50=700
450+250=700
700=700

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Let
Present age of of Father = f years old
Present age of Son = s years old
Father is 3 times as old as his son
f=3s...............(1)
12 years ago
father was = f-12 years old
Son was = s-12 years old
Then
f-12=6(s-12)
f-12=6s-72
f=6s-72+12
f=6s-60..............(2)
Compare (1) and (2)
3s=6s-60
3s-6s=-60
-3s=-60
-3s/-3=-60/-3
s=20
Put the value of s in (1)
f=3s
f=3(20)
f=60
Present age of of Father = f = 60 years old
Present age of Son = s = 20 years old

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0.05x+0.1x+0.25x=6
0.4x=6
x=6/0.4
x=15
15 of coins of each nickels, dimes,quarters are there.