Operations & Word Problems

Depth · one concept from first principles · step 2 of 4

Rename repeated reciprocals to get a linear system

Solve the equation 5x+y−2x−y=1x−y−3x+y=1\frac{5}{x+y}-\frac{2}{x-y}=\frac{1}{x-y}-\frac{3}{x+y}=1.

Expressions
Your answer
How to solve
Strategy Convert to Algebra — Reorganize the chain A=B=C into two equations that each equal 1. Only two building blocks appear, 1/(x+y) and 1/(x-y), so giving each a new name turns the problem into an ordinary linear system. Then break the rest into two small subproblems: undo the renaming, and solve for x and y.
1STEP 1

Split the chain into two equations

Each of the first two expressions equals the last one, 1.

5/(x+y)-2/(x-y)=1, 1/(x-y)-3/(x+y)=1
2STEP 2

Rename the repeated reciprocals

Let A=1/(x+y) and B=1/(x-y). Both equations become linear in A and B.

5A - 2B = 1, -3A + B = 1
3STEP 3

Solve for A and B

From the second equation B=1+3A. Substitute into the first equation.

5A - 2(1+3A) = 1 → -A - 2 = 1 → A = -3, B = 1 + 3 × (-3) = -8
4STEP 4

Undo the renaming

If 1/(x+y)=-3 then x+y=-1/3, and if 1/(x-y)=-8 then x-y=-1/8.

x + y = -1/3, x - y = -1/8
5STEP 5

Add and subtract to get x and y

Adding the two equations gives 2x; subtracting them gives 2y.

2x = -8/24 - 3/24 = -11/24 → x = -11/48, 2y = -8/24 + 3/24 = -5/24 → y = -5/48
Answer
x=-11/48, y=-5/48
Now x+y=-16/48=-1/3 and x-y=-6/48=-1/8. Then 5/(x+y)-2/(x-y)=-15+16=1 and 1/(x-y)-3/(x+y)=-8+9=1, so every part of the chain equals 1.
Takeaway

When the same messy piece keeps showing up, give it a new name and the problem often becomes one you already know how to solve.

  • Split the chain into two equations
  • Rename the repeated reciprocals
  • Solve for A and B
  • Undo the renaming
  • Add and subtract to get x and y
Where next?
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