Question 68·Medium·Nonlinear Equations in One Variable; Systems in Two Variables
The positive real numbers , , and satisfy
Which equation correctly expresses in terms of and ?
For equations involving sums of reciprocals, first isolate the reciprocal of the variable you care about (here, ) by moving other terms to the opposite side. Then combine the remaining fractions using a common denominator, simplify to a single fraction, and finally take the reciprocal to solve for the original variable. Pay close attention to the order of subtraction and signs, since small sign errors often create tempting but incorrect answer choices.
Hints
Get alone first
Try to move to the other side of the equation so that is by itself.
Subtract the fractions carefully
Once you have , use a common denominator of to combine the two fractions into one.
Turn into
After you find a single fraction equal to , think about what operation gives you itself from .
Desmos Guide
Set values for A and k
In Desmos, define two sliders: A = 5 and k = 2 (or any positive values with ). These will stand in for the parameters in the problem.
Compute B from the original equation numerically
Type B = 1 / (1/k - 1/A) into Desmos. This uses the rearranged form to give a numerical value for based on your chosen and .
Test each answer choice against that B value
For each option, enter the right-hand side as a separate expression, for example B1 = k*A/(k - A), B2 = (A - k)/(k*A), B3 = k*A/(A - k), B4 = (k - A)/(k*A). Compare the numeric values of B1, B2, B3, and B4 to the earlier B. The expression whose value matches B for your chosen and corresponds to the correct formula.
Step-by-step Explanation
Isolate
Start from the given equation:
Subtract from both sides to get an expression for :
Combine the fractions on the right-hand side
To subtract and , use the common denominator :
So we have
Solve for by taking the reciprocal
If , then is the reciprocal of that fraction:
So the correct expression for in terms of and is