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

Question

Isaiah Anthony · Accepted Answer

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

Jameson Allen · Answer

undefined Let the two numbers be $x$ and $x + 3$.

The sum of their reciprocals is:

$\frac{1}{x} + \frac{1}{x + 3} = \frac{7}{10}$

To solve this equation, first find a common denominator:

$\frac{x + 3}{x(x + 3)} + \frac{x}{x(x + 3)} = \frac{7}{10}$

Combine the fractions:

$\frac{x + 3 + x}{x(x + 3)} = \frac{7}{10}$

Simplify:

$\frac{2x + 3}{x(x + 3)} = \frac{7}{10}$

Cross multiply:

$10(2x + 3) = 7x(x + 3)$

Expand:

$20x + 30 = 7x^2 + 21x$

Rearrange into a quadratic equation:

$7x^2 + 21x - 20x - 30 = 0$

Simplify:

$7x^2 + x - 30 = 0$

Factor or use the quadratic formula to find the solutions for $x$. Once you find $x$, you can find the other number by adding 3 to it.