Thread: 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.

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



and substitute to eliminate y and solve for x





Now factor out "x" and divide

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
ax+aby=ad

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

Take it from here.

-Dan

Thanks so much guys.
I ended up getting y=(ad-c)/(ab-1).
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.
So its ax+(ad-c)/(ab-1)=c.
What should I do first? The a or something else.
Thanks again though guys. Your great helpers.

Originally Posted by mandarep

Thanks so much guys.
I ended up getting y=(ad-c)/(ab-1).
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.
So its ax+(ad-c)/(ab-1)=c.
What should I do first? The a or something else.
Thanks again though guys. Your great helpers.

You could find the equation for x seperately.

However, substituting your y, to get x as given...



which comes out as given.

Ok, cool. Thanks so much. =D