I'm probably making a silly mistake or Wolfram Alpha is lying to me.

Question: Find the value of c and d.

$3d=13-2c$
$\frac{3c+d}{2}=8$

Rearranged, simplified and multiply each equation by 2:

$6d+4c=26$
$d+3c=16$

Now find the common multiple which in my case I will use 12:

$18d+12c=78$
$-4d-12c=-64$

Then add them and find what d is worth:

$14d=14$

$d=1$

Now when I plug this back into the equation, I will use the first one:

$3(1)+2c=13$
$3+2(c)=13$
$c=5$

$d=1, c=5$

What am I doing wrong? Sorry if this is the long winded way to do it.

2. ## Re: Linear equations addition

You aren't doing anything wrong, that solution is correct. Perhaps you entered the fraction incorrectly into Wolfram Alpha

3. ## Re: Linear equations addition

You could have avoided one step and that is multiplying the first equation by 2.
Re arrange the equations and we have
2c + 3d = 13 ---- (1)
3c + d = 16 ------(2)
Multiply the second equation by -3 and re write the equations
2c + 3d = 13 ---- (1)
-9c -3 d = -48 ------(3)
Add the equations and you have:
-7c = -35 i.e., c = 5. Now get d from equation (2) we get d = 1