I'm probably making a silly mistake or Wolfram Alpha is lying to me. Question: Find the value of c and d. [itex]3d=13-2c[/itex] [itex]\frac{3c+d}{2}=8[/itex] Rearranged, simplified and multiply each equation by 2: [tex]6d+4c=26[/tex] [tex]d+3c=16[/tex] Now find the common multiple which in my case I will use 12: [tex]18d+12c=78[/tex] [tex]-4d-12c=-64[/tex] Then add them and find what d is worth: [tex]14d=14[/tex] [tex]d=1[/tex] Now when I plug this back into the equation, I will use the first one: [tex]3(1)+2c=13[/tex] [tex]3+2(c)=13[/tex] [tex]c=5[/tex] [tex]d=1, c=5[/tex] What am I doing wrong? Sorry if this is the long winded way to do it.
Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.
So Wolfram is lying to me it seems? Wolfram says the answer is [tex]c=\frac{35}{16}, d=\frac{23}{8}[/tex]
If you meant this: $$ \frac{3c + d}{2}$$ you should have written it as (3c + d)/2. Also, there was some wasted effort when you multiplied the first equation by 2. You didn't need to do that.