dr. math.. please help me to answer the algebra worded problems?

Question

1. The sum of two numbers is 36. If the largers is divided by the smaller the
quotient is 2 and the remainder is 3. Find the number.

2. The denominator of a fraction is 4 less than the numerator. If the denominator is doubled and the numerator is increased by 6, the sum of the original fraction and the new one is 3. Find the original fraction..

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please help me.. i need the solutions..

Answer 1

1. Let x = the larger number, and y = the smaller number

Given:x+y=36 a) based on first given condition.
x/y =2+3/y b) based on second given condition.

Note: If you have a fraction 5/3 the quotient is 1 remainder 2, right? To express this in the form of an equation, we write: 5/3=1+2/3. Note the number 3. In the same way if x/y gives a quotient of 2 remainder 3, to express this in the form of an equation, we write:
x/y=2+3/y as shown above.
Note where the y goes on the right side of the equation.

From equation a): x = 36-y

Substitute this value of x in equation b):

(36-y)/y=2+3/y

Simplify this equation and solve for y:

(36-y)/y=(2y+3)/y

The two y on each side of the equation will cancel out. Do you agree? Therefore,

36-y=2y+3
-y-2y=3-36
-3y=-33

The negative sign on each side of the equation will cancel out. Solve for y by dividing each side of the equation by 3:
y=33/3
y =11

To get the value of x substitute y=11 in equation a):

x+11=36
x=36-11
x=25

Check this answer by dividing 25 by 11:

25/11= 2 remainder 3.


So the values of x and y of 25 and 11, respectively, must
be correct because they satisfy the given condition.


2. Let x= the numerator

Given: x-4= the denominator

Therefore the original fraction is x/(x-4).

Also given: (x+6)/2(x-4)+x/(x-4)=3

Simplify and solve for x by first multiplying each term by 2(x-4):

(x+6)+x(2)=3(2)(x-4)
x+6+2x=6x-24
x+2x-6x=-24-6
-3x=-30
Divide each side of the equation by 3. Note the negative
sign will cancel out.
x=30/3
x=10 (the numerator)
x-4=10-4
=6 (the denominator)

Check the answer: the original fraction is 10/6
the new fraction is (10+6)/2(6)
The sum of the two fractions equals 3, as shown below:
10/6+(10+6)/2(6)=
10/6+16/12=
20/12+16/12=36/12
=3

The original fraction must be 10/6 because it satisfies the given condition.

Answer 2

Q1> LEt the larger no be 'x' and the smaller no be 'y'
Given (x+y)=36 and
x/y = 2(3/y) {in mixed fraction format}.....
In other words (x = 2y +3)
Substituting this value of x in the first eqn we get
2y + 3 + y = 36
3y +3 = 36
y + 1 = 12 (Dividing throughout by 3)
y = 11
=> x = 25.....

Note: The second eqn can be understood thus..... the larger is divided by the smaller and gives the quotiient as 2 and remainder 3.....That is...twice the smaller no plus three gives the larger no....


2> let the numerator (nr.) of the original fraction be x
Implies original fraction = x / (x-4)
Therfore New fraction = (x+6) / [2*(x-4)]

Given Original Fraction + New Fraction = 3
[x / (x-4)] + (x+6) / [2*(x-4)] = 3
Taking LCM of Dr. which is [2*(x-4)]
(2x + x + 6) / [2*(x-4)] = 3
(3x +6) = 3*2*(x-4)
3x + 6 = 6x - 24
30 = 3x
x = 10

Therefore the Original Fraction is 10 / 6

Answer 3

1. let the larger no. be x & smaller be y
from statement(sum of no. is 36):x+y=36
when x is divided by y them quotient is 2&reminder is 3
x=2*y+3
so x-2y=3
multiply x+y=36 by 2 =>2x+2y=72
now add last expression &senod one
3x=75
so x=75/3=25
so y=36-x=36-25=11
answer :1st no is25&2nd number is 11

2.let denominator b x&numerator be y
so x=y-4 =>
so fraction is y/(y-4)
when denominator is doubled it will b 2(y-4)
and numerator in increased by 6 it will become y+6
now fraction will b (y+6)/2(y-4)
finally,
y/(y-4)+(y+6)/2(y-4)=3
we get the equation y*y-14y+40=0
if we solve for y we get the answer as y=10 or y=4
y cant b 4 as the fractrion will become 0
therefore y should be 10
at the end the fraction will be 10/6

Answer 4

1.) 25+11= 36
25 divided by 11= 2 r 3

2.) x= the numerator
x+6
------------- = 3 x = 6
2(x-4)

Answer 5

1. p+q=36
let p>q
therefore p-2q=3
solving these equations we get ...
p=25 and q=11.

2. let the numerator be p.
Therefore by the given condition.

(p/(p-4))+((p+6)/2(p-4))=3
solving for p
we get p=10
as the original fraction is (p/(p-4))
the original fraction is 10/6=5/3.

Answer 6

number 1:
x + y = 36 (i)
x = 2y + 3 (ii)
substitute (ii) into (i) and you get:
2y + 3 + y = 36
solving: y = 11 substitute into (i) and you get x = 25

number 2:
(x + 6)/(2x-8) + x/(x-4) = 3

solve for x = 10
therefore original fraction was: 10/6

Answer 7

a million. this is what you're able to do.... get how lots activity the $4 hundred makes at 12 % 4 hundred x 0.12 = $40 8 your objective yeild is extra desirable than $sixty seven.. sixty seven - 40 8 = $19 ---that's what you're able to earn from the relax $2 hundred.... so the question is at what fee? discover out how many of $2 hundred is $19... (19/2 hundred) x a hundred% = .0.5 x a hundred% = 9.5% 9.5% is the fee you're able to invest the $2 hundred to get a yeild of $sixty seven..... which you will get a yeild of extra desirable than $sixty seven, Lane has to invest the $2 hundred to a fee extra desirable than 9.5% 2. x^3 = 10x divide the two facets by making use of x, you get x^2 = 10 x = sqrt(10), x is sq. root of 10..

Answer 8

1.
x+y=36
x>y
x=2y+3

that gives x = 25 and y = 11

2.

d=x-4

(x/x-4) + {(x+6)/2(x-4)} = 3

This gives x = 10

so original frcation is 10/6

Thanks

Answer 9

11 & 25

Answer 10

1. p+q=36
p>q
p=2q+3

2. d=n-4
n/d+(n+6)/2/d =3
so (1.5*n+3)/(n-4)=3

Best of luck