MATHS...simultaneous equation

yan

2010-07-11 8:21 pm

1.Find two consecutive integers if the sum of their reciprocals is 7/12.

ANS is 3 and 4

2.If a fration is reduce to its simplest form,the numerator is less than the
denominator by 3. When both the numerator and denominator are increased
by 2,the fration is increased by 1/20. Find the original fration.

ANS is 7/10

3.x and y are two positive integers. The sum of two x and y is 5.The sum of x
and the reciprocal of y is 4/3 .Find the values of x and y.

ANS is x=1,y=3

THX!!!

回答 (1)

STEVIE-G™

2010-07-11 8:42 pm

✔ 最佳答案

1)
Let the two consecutive integers be x and (x+1)

if the sum of their reciprocals is 7/12
=>1/x+1/(x+1)=7/12
=>[(x+1)+x]/[x(x+1)]=7/12
=>12(2x+1)=7(x^2+x)
=>7x^2-17x-12=0
=>(x-3)(7x+4)=0
=>x=3 or x=-4/7 (rejected)

So, the two integers are 3 and 4.

2)
Let the original fraction be x/(x+3)

When both the numerator and denominator are increased
by 2,the fration is increased by 1/20
=> (x+2)/[(x+3)+2]-x/(x+3)=1/20
=> (x+2)/(x+5)-x/(x+3)=1/20
=> [(x+2)(x+3)-x(x+5)]/[(x+5)(x+3)]=1/20
=> (x^2+5x+6-x^2-5x)/[(x+5)(x+3)]=1/20
=> 6*20=(x+5)(x+3)
=> x^2+8x+15=120
=> x^2+8x-105=0
=> (x-7)(x+15)=0
=> x=7 or x=-15 (rejected)

So, the orginial one is 7/10.

3)
2x+y=5...(1)
x+(1/y)=4/3...(2)

(1)-(2)*2:
y-2/y=7/3
y^2-(7/3)y-2=0
3y^2-7y-6=0
(y-3)(3y+2)=0
y=3 or y=-2/3 (rejected)

So, x=1 and y=3.

👍 0 👎 0