Waste less time on Facebook — follow Brilliant.
×

find the process these simple but critical equations

x-y = -1 & (a/x + b/y) = (a/2 + b/3) show the process to find x & y by Addition / Elimination / Substitution

I know the answer is x=2 & y=3. But I want to see the process.

Note by Fahim Rahman
4 years, 4 months ago

No vote yet
6 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

\[ x - y = -1 \implies y = x + 1\]

\[ \frac{a}{x} + \frac{b}{y} = \frac{a}{2} + \frac{b}{3} \implies \frac{ay + bx}{xy} = \frac{3a + 2b}{6} \implies \frac{a(x+1) + bx}{x(x+1)} = \frac{3a + 2b}{6} \]

So,

\[ \begin{align*} 6 \left( a(x+1) + bx \right) &= x(x+1)(3a+2b) \\ 6ax + 6a + 6bx &= 3ax + 2b + 3ax^2 + 2bx^2 \\ 3ax + 6a + 4bx &= 3ax^2 + 2bx^2 \\ 3a(x+1)(x-2) + 2bx(x-2) &= 0 \\ (x-2) \left( 3a(x+1) + 2bx \right) &= 0 \end{align*} \]

Hence

\[ x = 2, \quad y = x+1 = 3 \] or \[ 3a(x+1) + 2bx \implies x(3a+2b) = -1 \implies x = \frac{-1}{3a+2b}, \quad y = x+1 = 1 - \frac{1}{3a+2b} \]

So, the only value of \((x,y)\) not dependent on \(a\) and \(b\) is

\[ (x,y) = (2,3). \]

Tim Vermeulen - 4 years, 4 months ago

Log in to reply

Good job! We should remember to exclude \( x = 0, -1 \) from the solution, due to conditions in the question.

It is interesting that the solution set is a line and a (typically isolated) point. Why is this the case?

Calvin Lin Staff - 4 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...