Two numbers differ by 3. The sum of their reciprocals is seven tenths. How do you find the numbers?

1 Answer
Jun 17, 2015

Answer:

Tthere are two solutions to a problem:
#(x_1, y_1)=(5,2)#
#(x_2, y_2)=(6/7,-15/7)#

Explanation:

This is a typical problem that can be solved using a system of two equations with two unknown variables.

Let the first unknown variable be #x# and the second #y#.

The difference between them is #3#, which results in the equation:
(1) #x-y=3#

Their reciprocals are #1/x# and #1/y#, the sum of which is #7/10#, which results in the equation:
(2) #1/x+1/y=7/10#
Incidentally, existence of reciprocals necessitates the restrictions:
#x!=0# and #y!=0#.

To solve this system, let's use the method of substitution.
From the first equation we can express #x# in terms of #y# and substitute into the second equation.

From equation (1) we can derive:
(3) #x = y+3#

Substitute it into equation (2):
(4) #1/(y+3) + 1/y = 7/10#
Incidentally this necessitates another restriction:
#y+3!=0#, that is #y!=-3#.

Using common denominator #10y(y+3)# and considering only numerators, we transform equation (4) into:
#10y+10(y+3)=7y(y+3)#

This is a quadratic equation that can be rewritten as:
#20y+30=7y^2+21y# or
#7y^2+y-30=0#

Two solutions to this equation are:
#y_(1,2)=(-1+-sqrt(1+840))/14#
or
#y_(1,2)=(-1+-29)/14#

So, we have two solutions for #y#:
#y_1=2# and #y_2=-30/14=-15/7#

Correspondingly, using #x=y+3#, we conclude that there are two solutions to a system:
#(x_1, y_1)=(5,2)#
#(x_2, y_2)=(6/7,-15/7)#

In both cases #x# is greater than #y# by #3#, so the first condition of a problem is satisfied.
Let's check the second condition:
(a) for a solution #(x_1, y_1)=(5,2)#:
#1/5+1/2=(2+5)/(5*2)=7/10# - checked
(b) for a solution #(x_2, y_2)=(6/7,-15/7)#:
#7/6-7/15=70/60-28/60=42/60=7/10# - checked

Both solutions are correct.