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45 - 5 + 80 + n = 200
40 + 80 + n = 200
120 + n = 200
n = 200 - 120 = 80

n = 80

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Let the two consecutive integers be n, n+1.

three times the smaller integer = 3n
four times the larger integer = 4(n+1)

3n + 4(n+1) = 123
3n + 4n + 4 = 123
7n + 4 = 123
7n = 123 - 4 = 119
n = 119/7 =17

smaller integer is 17

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The perimeter of rectangle is 2(l+b) where l is the length and b is the width
2(l+b) = 38
2(7.5 + b) = 38
15 + 2b = 38
2b = 38 - 15
2b = 23
b = 11.5

Width is 11.5cm

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1kg = 1,000,000 milligrams
Therefore 23 kg = 23, 000,000 mg

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Let the number be n

if 4 is added to a number and the sum is tripled the result is 28 more than the number.

4 is added to the number and the sum is tripled
3 * ( n + 4)

3 * ( n + 4) = n + 28
Solving the equation
3n + 12 = n + 28
3n - n = 28 -12
2n = 16
n =8

Number is 8

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30-[9+4(8-5)]

Start solving from innermost parenthesis

30-[9+4(3)]
30-[9+12]
30-21
9

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M^2+8m+7
8m is split as m +7m
m^2 + m + 7m +7
Group first two terms and last two terms together
(m^2+M) + (7m + 7)
m(m+1) + 7(m+1)

take the common term m+1
(m+1)(m+7)

M= -1, -7

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The sum of the digits of a two digit number is 11.
The 2 digit numbers whose sum of digits add to 11 are 29, 38, 47, 56, 65, 74, 83, 92

The original two-digit number is 5 more than 3 times the number with its digits reversed

From the above statement, it is understood that the original two-digit number is greater than the reversed number

so the number should be either 65, 74, 83 or 92

original number = 5 + 3(reversed number)
Let's apply the formula to 65, 74, 83, 92 to see which is correct


65 = 5 + 3(56) = 5 + 177 = 182 -------not correct
74 = 5 + 3(47) = 5 + 141 = 146 -------not correct
83 = 5 + 3(38) = 5 + 114 = 119 -------not correct

92 = 5 + 3(29) = 5 + 87 = 92 -------correct


92 is the number

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5x - 3y = 16......eq 1
4x + 5y = -2......eq 2

Multiply eq1 by 5 and eq2 by 3 to cancel out 'y'
25x - 15y = 80
12x - 15y = -6

Add the above equations
37x = 74
x = 2

Substitute x = 2 in eq 1
5x - 3y = 16
5(2) - 3y = 16
10 - 3y = 16
-3y = 16 - 10
-3y = 6

y = -2

x = 2, y = -2

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5x + 3y = 8......eq 1
8x - 2y = 6......eq 2

Multiply eq1 by 2 and eq2 by 3 to cancel out 'y'
10x + 6y = 16
24x - 6y = 18

Add the above equations
34x = 34
x = 1

Substitute x = 1 in eq 1
5x + 3y = 8
5(1) + 3y = 8
5 + 3y = 8
3y = 8 - 5

3y = 3
y = 1

x = 1, y = 1

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x + 9 = 12
Keep the variable on one side and numbers on the other side
x = 12 - 9 = 3
x = 3

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m^2 - 16 = 0
m^2 = 16

Take squareroot on both sides
m = +4, -4

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Let boy's age be b
Let father's age be f

The sum of a boy and his father's age is 63 years
f + b = 63 -----------------------eq. 1

The difference between their ages is 33 years
f - b = 33 -----------------------eq. 2


Solving the 2 equations(Add the 2 equations)
f + b = 63
f - b = 33

2f = 96
f = 96/2 = 48

Father's age is 48
Substitute f = 48 in eq 1

f + b = 63
48 + b = 63
b = 63 - 48
b = 15

Boy's age is 15
Father's age is 48

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Let Percy's age be p
Let Ginny's age be g

Percy is 8 years older than Ginny
g + 8 = p -----------------------eq. 1

In ten years time he will be three times Ginny's present age
In ten years Percy will be p + 10 yrs old
Three times Ginny's age = 3g
p + 10 = 3g --------------eq 2

Substitue eq1 in eq2
p + 10 = 3g
g + 8 + 10 = 3g
g + 18 = 3g
2g = 18

g = 9

Ginny's age is 9 years
Substitute g = 9 in eq 1

g + 8 = p
9 + 8 = p

P = 17

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Let the 3 colors be w,g,b for white, green and blue

The number of blues was 7 less than the sum or the whites and the greens
w + g -7 = b
w + g - b = 7 ...... eq 1

The number of greens was 1 greater than the sum of the blues and the whites
g = 1 + b + w
b + w -g = -1 .........eq 2

three times as many greens as blues
3b = g ------eq 3

Substitute eq 3 in eq 1
w + g - b = 7
w + 3b - b = 7
w + 2b = 7 -------eq 4
Substitute eq 3 in eq 2
b + w - g = -1
b + w - 3b = -1
w - 2b = -1 .........eq5
Add eq4 and eq 5
w + 2b = 7
w - 2b = -1
2w = 6
w = 3

Substitute w = 3 in eq 4 to get the value of b
w + 2b = 7
3 + 2b = 7
2b = 7-3 = 4
b = 2

Substitute w = 3, b = 2 in eq 1 to get the value of g
w + g - b = 7
3 + g - 2 = 7
1 + g = 7
g = 6

There are 3 whites, 2 blues and 6 greens

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Let the 2 numbers be x,y
the sum of twice one number and three times a second number is 84
2x + 3y =84 ......eq 1

sum of thrice the first number and five times the second is 136
3x +5y =136 ......eq 2

Multiply eq1 by 3 and eq 2 by 2 to cancel out 'x'
6x + 9y =252
6x +10y =272

Subtract the above equations
-y = -20
y = 20

Substitute y = 20 in eq 1
2x + 3y =84
2x + 3(20) =84
2x + 60 =84
2x =84 -60
2x = 24
x =12
x = 12, y = 20

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3x -y =2 ......eq 1
x +2y =10 ......eq 2

Multiply eq1 by 2
6x -2y =4
x +2y =10

Add the above equations
7x =14
x=2

Substitute x = 2 in eq 1
3x -y =2
3(2) -y =2
6 - y =2
-y = -4
y = 4

x = 2, y = 4

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5x-2y=-5
y-5x=3

Rearranging the equation
5x -2y = -5 .......eq 1
-5x +y = 3 ....... eq 2

Adding Eq1 and eq2 you get
-y = -2
y = 2

Substitute y = 2 in eq 1 oe eq 2
5x -2y = -5
5x -2(2) = -5
5x -4 = -5
5x = -1
x = -1/5

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Let Marivic's age be x
Let Michelle's age be y


Marivic is five years older than Michelle
x = y + 5 -----------------------eq. 1

In three years, the product of her age and Michelle's age five years ago will be 90 years
In three years marvic will be x + 3 yrs old
Michelle's age five years ago was y - 5 yrs old

(x + 3)(y - 5) = 90 --------------eq 2

Substitue eq1 in eq2
(y + 5 + 3) (y - 5) = 90
(y + 8) (y - 5) = 90
y^2 + 3y -40 = 90
y^2 + 3y -130 = 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=529 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 10, -13. Here's your graph:

Answers for y= 10,-13
Age is a positive integer, so Michelle's age is 10

x = y + 5
x= 15
so Marivic's age is 15

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Let the number be x
The difference between twice a number and 9 is 16
2x - 9 = 16
2x = 16 + 9 = 25
x = 12.5

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Let the number be x
2(4 + x) = 34
4 + x = 34/2 = 17
x = 17 - 4 = 13

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3x^2 = 11x + 4
3x^2 - 11x - 4 = 0
3x^2 - 12x + x - 4 = 0
3x(x - 4) + 1(x - 4) = 0
(3x + 1)(x - 4)=0
x = -1/3, 4

Using the formula

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=169 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 4, -0.333333333333333. Here's your graph:

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square route (1 + x^3 )= 3
Square on both sides
1 + x^3 = 3^2
1 + x^3 = 9
x^3 = 8
x = 2

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The numbers were truncated in the above solution, so writing the equations
x + x + 237 = 17459
2x + 237 = 17459
2x = 17222
x = 8611

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18 = 9 -(3 - x)
18 = 9 -3 + x
18 = 6 + x
x= 12

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a)
Cost for 3 days is 25 cents i.e., 3*.25 = .75
Cost for 4 days is 0.75 + 0.05 = 0.80
Cost for 5 days is 0.75 + 0.05 + 0.05 = 0.85
...
Formula would be c = 0.05(d-3) + 0.75 if the days are >=3
Formula would be c = 0.25d if the days is < 3 (i.e., 1 or 2 days)

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b) Substitute d = 7 in the above formula
c = 0.05(d-3) + 0.75
c = 0.05(7-3) + 0.75
c = 0.20 + 0.75
c = 0.95 ( 95 cents)

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Let Ann's age be A and Bill's age be B
ann's age is seven-tenths of Bill's age.
A = 7/10(B) ....eq1

In 4 years Ann's age will be eight-elevenths of Bill's age.
A + 4 = 8/11(B+4).....eq2
Substitute eq1 in eq2

(7/10(B)) + 4 = 8/11(B+4)
Cross multiply
11(7B + 40) = 80(B + 4)
77B + 440 = 80B + 320
-3B = -120
B=40

Bill's age is 40
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Let the ten digit be T and units digit U
The number will be 10T + U

The Ten's digit of a two-digit number is 3 more than the unit's digit.
T = 3 + U .....eq1

The number is 8 more than six times the sum of the digits.
10T + U = 8 + 6(T + U)
10T + U = 8 + 6T + 6U
10T + U - 6T - 6U = 8
4T - 5U = 8
SUBSTITUTE T = 3 + U
4(3 + U) - 5U = 8
12 + 4U -5U =8
-U = -4
U = 4( UNIT'S DIGIT IS 4)

Since Ten's digit is 3 more than the unit's digit, it 7
Number is 74

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Find a value for k so that 9m^2 - kn^2 will have the factors 3m + 7n, and 3m - 7n.


(a + b)(a-b) = a^2 - b^2
The factors are 3m + 7n, and 3m - 7n. Let's apply the above formula
a = 3m and b = 7n

(3m + 7n)(3m - 7n) = 9m^2 - 49n^2

Comparing 9m^2 - 49n^2 with 9m^2 - kn^2, we get k = 49

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3[2+v] - 4v = v + 16
6 + 3v -4v -v -16 = 0
-2v - 10 = 0
-2v = 10
v = -5

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5x + 8 + 3x - x + 5 = 6x - 3
5x + 8 + 3x - x + 5 - 6x + 3 = 0
x + 16 = 0
x = -16

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b + (b+5) -2(b-5) = 18 + b
b + b + 5 -2b + 10 - 18 - b = 0
-b -3 = 0
b = -3