solve this plzzz..it is a mathematical question.from linear equations in 2 variables?

Question

solve this plzzz..it is a mathematical question.from linear equations in 2 variables?

Answer 1

*Let m = man's present age

*Let s = son's present age





*2 yrs. ago a man's age was 3 times the square of his son's age.

m - 2 = 3(s - 2)^2



(s - 2)^2= (s-2)(s-2)

m - 2 = 3(s^2 - 4s + 4)

m - 2 = 3s^2 - 12s + 12

m = 3s^2 - 12s + 12 + 2

m = 3s^2 - 12s + 14...........(1)



*3 yrs. hence his age will be 4 times his son's age.

m + 3 = 4(s + 3)

m + 3 = 4s + 12

m = 4s + 12 - 3

m = 4s + 9



Replace m with (4s+9) in the 1st equation

4s + 9 = 3s^2 - 12s + 14

0 = 3s^2 - 12s - 4s + 14 - 9

0 = 3s^2 - 16s + 5

Factors to

(3s - 1)(s - 5) = 0

s = 5 yrs is the reasonable answer for son's present age

:

Find m, using m = 4s + 9

m = 4(5) + 9

m = 29 yrs, man's present age

Answer 2

1.1 kg wt 3/16 cm 15 kg-- 1 5x( 3/16) =2.812 5 c Answer:: 80 +2.81 = 82.81 cm 2. q, First 3 Hrs $100 balance 21x$ 35=$735 whole day charge =$ ( 100+ 735 ) =$835 3,water per liter 75cents: for 100 liters first 3 hrs blane in a day 21 hrs to bepaid.extra 20 liter =20x0.75 $= $15; bill =$240 fixed charges= $(240-15) =$ 225 4. 2010 chips ordered. excesd than 100 $17.50 first 100 chips ---$2,500 balance 2010-100=1910 chips cost=1910x17.50= $ 33,425 first 100 charge = $ 2500 total $ 35.925

Answer 3

let 2years ago son's age = x years.

2 years ago man's age = 3x^2 years

son's current age = (x +2 )years

man's current age = (3x^2 +2) years.

three years hence son's age =( x+2 +3) =(x+5) years.

three years hence man's agee =(3x^2 +2+3) =(3x^2+5) years

according to quetion

( 3x^2+5) = 4(x+5)

or 3x^2+5 =4x +20

or 3x^2 -4x -15 = 0

or 3x^2 -9x +5x -15 = 0

or 3x(x-3)+5(x-3) = 0

or (x-3) (x+5) =0

or (x-3) =0 or (x+5) = 0

or x = 3 or x = -5 but age can't be negative

therefore x =3 accepted

present age of son = x+2 = 3+2 = 5 years ans

present age of man = 3x^2+2 = 3(3)^2 +2= 29 years ans.

Answer 4

hi grl..its bin more than 2 years since i messed with such math junk..yet lemme try ma best..

present ages...

man=x son=y

2years ago

man=x-2 son=y-2

x-2=3(y-2) the whole square.........(1)

3years frm now

man=x+3 son=y+3

x+3=4(y+3).............(2)

now solve eqn (1) and (2)

u vl get the answer...hopefully...

Answer 5

m - 2 = 3(s - 2)^2

m + 3 = 4(s + 3)

m = 3s^2 - 12s + 14

m = 4s+9

3s^2 - 12s + 14 = 4s + 9

3s^2 - 16s + 5 = 0

(s - 5)(3s - 1) = 0

s =1/3 → reject

s = 5

m = 29

the man is 29 and his son is 5

Answer 6

lets father=x son=y

According to 1st logic

x=3(y2-4y+4)+2------------condition (1)



x=4(y+4)-3....condition(2)



so,



3y2-12y+12+2=4y+16-3

3y2-16y+5=0

y=5,y=1/3

x=29





27=3*9=27

32=4*8=32

3y2-16y+3=0

Answer 7

quadratic, not linear

---------------------------

2 yrs ago::::: son : x , father 3x^2

3 yrs hence: son x+5, father 4(x+5)

4(x+5) - 3x^2 = 5

open out & collect terms

3x^2 - 4x - 15 = 0

factorise

(3x+5)(x-3) = 0

take +ve solution

x = 3

2 yrs ago, son was 3, father was 27

present ages : son = 5 yrs, father = 29 yrs

-------------------

IF THE ABOVE ANS IS WRONG,

IT'S BECAUSE OF AMBIGUOUS USE OF "HENCE"

ANOTHER INTERPRETATION BELOW



2 yrs ago::::: son : x , father 3x^2

3 yrs hence: son x+1, father 4(x+1)



4(x+1) - 3x^2 = 3



open out & collect terms

3x^2 - 4x - 7 = 0



factorise

(3x-7)(x+1) = 0



take +ve solution

x = 7/3



2 yrs ago, son was 2 1/3, father was 49



present ages : son = 4 1/3 yrs, father = 51 yrs

-------------------

Answer 8

this onezz eazzy dudette..

take X as fathers age

and y as sons age..



so accordin to the ques..

2 years ago : (X-2) = 3(y-2)

on simplifyin.., we get X-3y= -4 this is equation one.. - 1





3 years hence : (X+3) = 4(y+3)

on simplifyin again.. we get X-4y= 9 this is eqution 2



so one solvin we get .. X=11 and y=5



note: plzz chk the answers are ryt im not too sure as iam a 10th grader too..



if the anszz is ryt and u have some doubts ..add additional details here or mail me



wen u say u have doubt here.. i mean add additional data.. ill giveya my id.. ^_^

Answer 9

man's present age=x

son's=y

(x-2)=3(y-2)^2

(x+3)=4(y+3)

x-2=3(y^2-4y+4)

x=3y^2-12y+12+2

so..

3y^2-12y+14+3=4y+12

3y^2-16y+5=0

(3y-1)(y-5)=0

y=1/3 or 5

1/3 cant be..

so y=5

x=29

hope it helps XDD

Answer 10

Best question and too interesting.



Jet the present age of son be S



Let the present age of Father be F



F+3 = 4(S+3)

F-3 = 3(S-3)^2



Solve further

Answer 11

We put father current age is x and son is y

According to first condition

(x-2)=3(y-2)2

(x-2)= 3(y2-4y+4)

x-2 =3y2-12y+12

x=3y2-12y+12+2

x=3y2-12y+14 ……..(i)





According to second condition

(x+3)=4(y+3)

x+3=4y+12

x=4y+12-3

x=4y+9 ……………(ii)



From Equation (i) and (ii)

3y2-12y+14 =4y+9

3y2-16y+5=0

3y2-15y-y+5=0

3y(y-5)-1(y-5)=0

(3y-1)(y-5)=0

y-5=0

y=5

And from equation (ii)

x= 4(5)+9

x=20+9

x=29

Final Answer is x=29 and y=5

Answer 12

Let the current age of the man be x years.

Let the current age of the son be y years.



According to first condition:

(x-2) =3(y-2)^2

x-2=3(y^2-4y+4)

x-2=3y^2-12y+12

x=3y^2-12y+14-------equation 1



According to the 2nd condition:

x+3= 4(y+3)

x+3=4y+12

x=4y+9----------equation 2



Subtracting equation 2 from 1,we'll get



3y^2-16y+5=0----------equation 3.

(3y-1)(y-5)=0

y=1/3 or 5



Since, age cannot be 1/3. So y=5.

Substituting the value of y in equation 2, we'll get the value of x as 29.

Therefore, the age of father and son is 29 and 5 years respectively. I hope this is clear enough.