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.

#Algebra #MathProblem #Math

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Math	Appears 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

$x - y = -1 \implies y = x + 1$

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

So,

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

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 - 7 years, 9 months ago

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 - 7 years, 9 months ago

Math	Appears 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}`	$\boxed{123}$