[SOLVED] solving for x and making x the subject (yr 11 maths)

**smmmc** · February 14th 2009, 08:43 PM

solving for x (i solved for y already, but subbing y into one of the equations seems to be too long of a process, is there another way?)
a)x(b-c)+by-c=0 and y(c-a)-ax+c=0
b)axy +b=(a+c)y and bxy+a=(b+c)y

make x the subject
x+a/x-c + x+c/x-a = 2 SOLVED

thanks

**earboth** · February 14th 2009, 11:07 PM

Originally Posted by smmmc

sol...

make x the subject
x+a/x-c + x+c/x-a = 2

thanks

I assume that you mean (x+a)/(x-c) + (x+c)/(x-a)=2

If so:
1. Multiply both sides of the equation by (x-c)(x-a). You'll get:

$(x+a)(x-a)+(x+c)(x-c)=2(x-c)(x-a)$

2. Expand the brackets:

$x^2-a^2+x^2-c^2=2(x^2-ax-cx+ab)$

3. Collect like terms. Move all terms containing x to the LHS, all terms with constants to the RHS:

$2ax+2cx=a^2+2ac+c^2~\implies~2(a+c)x=(a+c)^2$

4. Divide by the leading factor of x:

$x = \frac12\cdot (a+c)$

**smmmc** · February 14th 2009, 11:36 PM

Originally Posted by earboth

3. Collect like terms. Move all terms containing x to the LHS, all terms with constants to the RHS:

$2ax+2cx=a^2+2ac+c^2~\implies~2(a+c)x=(a+c)^2$

thanks earboth. for the quoted above how did you get the (a+c)^2 ? from 2ac+c^2+a^2 ??

**earboth** · February 14th 2009, 11:38 PM

Originally Posted by smmmc

thanks earboth. for the quoted above how did you get the (a+c)^2 ? from 2ac+c^2+a^2 ??

Correct. If you expand

$(a+c)^2=a^2+2ac+c^2$

I only used the reverse gear

**smmmc** · February 14th 2009, 11:48 PM

ohhokay thanks. Anyone know of how to do the 2 remaining questions?

**earboth** · February 15th 2009, 12:29 AM

Originally Posted by smmmc

solving for x (i solved for y already, but subbing y into one of the equations seems to be too long of a process, is there another way?)
a)x(b-c)+by-c=0 and y(c-a)-ax+c=0
...

To solve such a system of simultaneous equations I would use the Cramer rule. If you have the system of linear equations:

$\begin{array}{l}a_1x+b_1y = c_1\\a_2x+b_2y = c_2\end{array}$

then the the solution is:

$x = \dfrac{c_1\cdot b_2 - c_2\cdot b_1}{a_1\cdot b_2 - a_2\cdot b_1}$ ..... and ........ $y = \dfrac{a_1\cdot c_2 - a_2\cdot c_1}{a_1\cdot b_2 - a_2\cdot b_1}$

With you example you get:

$x = \dfrac{c^2-ac+bc}{ac+bc-c^2}$ ..... and ........ $y = \dfrac{-bc+c^2+ac}{ac+bc-c^2}$ ........ which simplifies to:

$x = \dfrac{c-a+b}{a+b-c}$ ..... and ........ $y = \dfrac{a-b+c}{a+b-c}$

**smmmc** · February 15th 2009, 02:01 AM

cramers rule uses matrixes right? i haven't learnt matrixes so this rule would be difficult for me..

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