Pair of simultaneous equations.

• Feb 1st 2011, 03:13 PM
mandarep
Pair of simultaneous equations.
Solve the following pair of simultaneous equations for x and y.
ax+y=c
x+by=d
I don't know why I can't get this one, because it seems pretty simple.
So far I've done this:
I multiplied 1 and 2 and got abx+by=cb
Then I added that and 2 together and got abx+x+2by=cd+d
x(ab+1)+2by=cd+d
(cb+d)/(ab+1)=x+2by
(cd+d)/(ab+1)-2by=x
However the answer in the back of the book is: x=(d-bc)/(1-ab).
I'm just not too sure where I went wrong.
Thanks.
• Feb 1st 2011, 03:19 PM
Quote:

Originally Posted by mandarep
Solve the following pair of simultaneous equations for x and y.
ax+y=c
x+by=d
I don't know why I can't get this one, because it seems pretty simple.
So far I've done this:
I multiplied 1 and 2 and got abx+by=cb
Then I added that and 2 together and got abx+x+2by=cd+d
x(ab+1)+2by=cd+d
(cb+d)/(ab+1)=x+2by
(cd+d)/(ab+1)-2by=x
However the answer in the back of the book is: x=(d-bc)/(1-ab).
I'm just not too sure where I went wrong.
Thanks.

You could proceed

$y=c-ax$

and substitute to eliminate y and solve for x

$x+b(c-ax)=d$

$x+bc-abx=d$

Now factor out "x" and divide
• Feb 1st 2011, 03:21 PM
topsquark
You have solved the equations for x in terms of y, not in terms of just a,b ,c ,d.

Start here:
ax+y=c
x+by=d

Multiply the second equation by a:
ax+y=c

Now subtract the second equation from the first:
(ax+y) - (ax + aby) = c - ad

Take it from here.

-Dan
• Feb 1st 2011, 03:44 PM
mandarep
Thanks so much guys.
Now when I sub that into one of the other equations, can I do it into any of them?
Cause I tried with subbing it in to the first equation.
I got stuck again with my algebra this time.
What should I do first? The a or something else.
Thanks again though guys. Your great helpers.
• Feb 1st 2011, 04:00 PM
Quote:

Originally Posted by mandarep
Thanks so much guys.
Now when I sub that into one of the other equations, can I do it into any of them?
Cause I tried with subbing it in to the first equation.
I got stuck again with my algebra this time.
$\displaystyle\ ax=c-\frac{ad-c}{ab-1}=c+\frac{ad-c}{1-ab}=\frac{c(1-ab)+(ad-c)}{1-ab}$