Author Topic: Simultaneous equations. (Read 831 times) Tweet Share

cindyy · « **on:** January 04, 2010, 12:54:44 am »

This is probably really simple, but i cant seem to get the answer! so heres the question.

For the simultaneous equations:
x/a + y/b = 1
x/b + y/a = 1, show that x = y = ab/(a+b)

thanks in advance.

TrueTears · « **Reply #1 on:** January 04, 2010, 12:58:50 am »

Stroodle · « **Reply #2 on:** January 04, 2010, 01:13:47 am »

Another method is to multiply the first equation by $a$ and the second by $b$ then subtract one equation from the other and solve for $y$

cindyy · « **Reply #3 on:** January 04, 2010, 01:30:15 am »

Quote from: TrueTears on January 04, 2010, 12:58:50 am

$x = a\left(1 - \frac{y}{b}\right)$

$\implies \frac{a\left(1 - \frac{y}{b}\right)}{b} + \frac{y}{a} = 1$

$\therefore y = \frac{ab}{a+b}$

that is what i have done, but its the parts in the middle that just dont work out for me. thanks anyways

GerrySly · « **Reply #4 on:** January 04, 2010, 01:41:21 am »

Quote from: cindyy on January 04, 2010, 01:30:15 am

that is what i have done, but its the parts in the middle that just dont work out for me. thanks anyways

Dunno if you haven't got it yet, but I thought i'd elaborate it for others

$\begin{align*} 1&=\frac{y}{a}+\frac{y}{b}\\ &=y\left (\frac{1}{a}+\frac{1}{b}\right )\\ &=y\left (\frac{b+a}{ab}\right )\\ y&=\frac{1}{\frac{b+a}{ab}}\\ &=\frac{ab}{a+b} \end{align*}$

davidle_10 · « **Reply #5 on:** January 04, 2010, 05:26:25 pm »

$\implies \frac{a\left(1 - \frac{y}{b}\right)}{b} + \frac{y}{a} = 1$

Can't figure out the answer when I expand this.

brightsky · « **Reply #6 on:** January 04, 2010, 05:49:12 pm »

Following on from what TT has done using substitution method:

$\frac {x}{a} + \frac {y}{b} = 1$

$\implies x = \left(1 - \frac{y}{b}\right)a$

Subsituting x back into second equation:

$\frac {\left(1 - \frac{y}{b}\right)a}{b} + \frac {y}{a} = 1$

Multiply ab to both sides to eliminate tedious fractions:

$a^2 \left(1 - \frac{y}{b}\right) + by = ab$

$a^2 - \frac {a^2y}{b} + by = ab$

$a^2 + \frac{b^2y - a^2y}{b} = ab$

$a^2 + \frac {y(b^2 - a^2)}{b} = ab$

$\frac {y(b-a)(b+a)}{b} = ab - a^2$

$y(b+a)(b-a) = ab(b - a)$

$y = \frac {ab(b-a)}{(b+a)(b-a)} = \frac {ab}{b+a} = \frac {ab}{a+b}$

GerrySly · « **Reply #7 on:** January 04, 2010, 05:55:44 pm »

Quote from: davidle_10 on January 04, 2010, 05:26:25 pm

$\implies \frac{a\left(1 - \frac{y}{b}\right)}{b} + \frac{y}{a} = 1$

Can't figure out the answer when I expand this.

I found a better approach was to find that $x=y$ then just sub that into one of the first equations and solve.

$\begin{align*} \frac{x}{a}+\frac{y}{b}&=\frac{x}{b}+\frac{y}{a}\\ \frac{x}{a}-\frac{x}{b}&=\frac{y}{a}-\frac{y}{b}\\ x\left (\frac{1}{a}-\frac{1}{b}\right )&=y\left (\frac{1}{a}-\frac{1}{b}\right )\\ \therefore x&=y \end{align*}$

Then just sub

$\begin{align*} 1&=\frac{x}{a}+\frac{x}{b}\\ &=x\left (\frac{1}{a}+\frac{1}{b}\right )\\ &=x\left (\frac{a+b}{ab}\right )\\ x&=\frac{1}{\frac{a+b}{ab}}\\ \therefore x&=\frac{ab}{x+b} \end{align*}$

cindyy · « **Reply #8 on:** January 04, 2010, 06:46:41 pm »

thank you, they are both very good ways

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Author Topic: Simultaneous equations. (Read 831 times) Tweet Share

cindyy

Simultaneous equations.

TrueTears

Re: Simultaneous equations.

Stroodle

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cindyy

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GerrySly

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davidle_10

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brightsky

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GerrySly

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cindyy

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