1. ## Simultaneous equations

Solve the following pairs of simultaneous equations for x and y:
(a + b)x + cy = bc
(b + c)y + ax = -ab

unsure how to start any help would be greatful!

Also write s in terms of a in the following pairs of equations
1) s = ah
h = 2a + 1

2) s = h^2 + ah
h = 3a^2

2. ## then

solve simultaniously for x and y

$(a + b)x + cy = bc$
$(b + c)y + ax = -ab$

3. I have to solve for x and y not a and b though

4. do you mean like this:
${x = c (b-y)/(a+b),y = -a(x-b)/(b+c)}$

5. Sorry but I don't know hat you have done there!
x = c
y = -a
I just don't know how to get there!

6. Originally Posted by scubasteve94
1) s = ah
h = 2a + 1
Just replace h in the first equation with what h is in the second

$s = ah$

$s = a(2a + 1)$

7. the answers are seen in the x= and y= (post #4)
but don't know how the rest cancels
just hope another reply comes to rescue

was wondering if your text had an example of a problem like this

8. Pickslides: thank you

No the text does not have an example

9. Originally Posted by scubasteve94
Solve the following pairs of simultaneous equations for x and y:
(a + b)x + cy = bc
(b + c)y + ax = -ab

$(a + b)x + cy = bc$...(1)

$(b + c)y + ax = -ab$...(2)

Working on (1) to make x the subject

$(a + b)x + cy = bc$

$(a + b)x = bc-cy$

$x = \frac{bc-cy}{a+b}$

Now sub this into (2) for x

$(b + c)y + ax = -ab$

$(b + c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab$

exapnding in the a and separately dividing the denominators

$(b + c)y + \left(\frac{abc}{a+b}\right) - \left(\frac{acy}{a+b}\right)= -ab$

Grouping the y terms together

$(b + c)y - \left(\frac{acy}{a+b}\right)+\left(\frac{abc}{a+b} \right) = -ab$

Taking out y as a common factor

$y\left((b + c) - \left(\frac{ac}{a+b}\right)\right)+\left(\frac{abc }{a+b}\right) = -ab$

$y\left((b + c) - \left(\frac{ac}{a+b}\right)\right) = -ab-\left(\frac{abc}{a+b}\right)$

and finally dividing

$y = \frac{-ab-\left(\frac{abc}{a+b}\right)}{\left((b + c) - \left(\frac{ac}{a+b}\right)\right)}$

You can simplify further and sub back in to find x.

10. Thank you so much

11. Originally Posted by pickslides
$(a + b)x + cy = bc$...(1)

$(b + c)y + ax = -ab$...(2)

Working on (1) to make x the subject

$(a + b)x + cy = bc$

$(a + b)x = bc-cy$

$x = \frac{bc-cy}{a+b}$

Now sub this into (2) for x

$(b + c)y + ax = -ab$

$(b + c)y + a\left(\frac{bc-cy}{a+b}\right) = -ab$

exapnding in the a and separately dividing the denominators

$(b + c)y + \left(\frac{abc}{a+b}\right) - \left(\frac{acy}{a+b}\right)= -ab$

Grouping the y terms together

$(b + c)y - \left(\frac{acy}{a+b}\right)+\left(\frac{abc}{a+b} \right) = -ab$

Taking out y as a common factor

$y\left((b + c) - \left(\frac{ac}{a+b}\right)\right)+\left(\frac{abc }{a+b}\right) = -ab$

$y\left((b + c) - \left(\frac{ac}{a+b}\right)\right) = -ab-\left(\frac{abc}{a+b}\right)$

and finally dividing

$y = \frac{-ab-\left(\frac{abc}{a+b}\right)}{\left((b + c) - \left(\frac{ac}{a+b}\right)\right)}$

You can simplify further and sub back in to find x.
I'd be inclined to use the elimination method:

Multiply equation (1) by a and equation (2) by (a + b), then substract one equation from the other and make y the subject.

Then, multiply equation (1) by (b + c) and equation (2) by c, then substract one equation from the other and make x the subject.