Linear equations using addition

uperkurk

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.

jbunniii

Your answer is correct, as you can verify by plugging ##d = 1## and ##c = 5## into the two given equations.

uperkurk

So Wolfram is lying to me it seems?

Wolfram says the answer is [tex]c=\frac{35}{16}, d=\frac{23}{8}[/tex]

jbunniii

It seems more likely that you didn't enter the problem correctly into Wolfram Alpha.

AlephZero

I think you told Wolfram the second equation was
$$3c + \frac d 2 = 8$$

uperkurk

Yes, looking back that is what is shows under "Input Result" How would I input the correct format?

Mark44

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.

micromass

Use correct brackets.