# Thread: extra credit problem i got :(

1. ## im stuck - its for some extra credits

x+y+z+a = 16480

x=y+ 1\2z
y=a-z
z=3x
a=y+z
find a,x,y,z

See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.

2. ## extra credit problem i got :(

x+y+z+a = 16480

x=y+ 1\2z
y=a-z
z=3x
a=y+z
find a,x,y,z

See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.

3. Originally Posted by spiderz
x+y+z+a = 16480

x=y+ 1\2z
y=a-z
z=3x
a=y+z
find a,x,y,z

See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.

Can you please verify is x:
$\displaystyle x = y + \frac{1}{{2z}}$
or
$\displaystyle x = \frac{{y + 1}}{{2z}}$

4. Originally Posted by OReilly
Can you please verify is x:
$\displaystyle x = y + \frac{1}{{2z}}$
or
$\displaystyle x = \frac{{y + 1}}{{2z}}$
i meant one half times Z or .5z

so it should say Y plus half of Z

5. Originally Posted by spiderz
i meant one half times Z or .5z

so it should say Y plus half of Z
You mean
$\displaystyle x = y + \frac{z}{2}$
?

6. If
$\displaystyle x = y + \frac{z}{2}$
is correct than this is solution:

$\displaystyle \begin{array}{l} x = y + \frac{z}{2} \\ x = y + \frac{{3x}}{2} \\ x = y + \frac{{3x}}{2} \\ x = \frac{{2y + 3x}}{2} \\ 2x = 2y + 3x \\ 2x - 3x = 2y \\ - x = 2y \\ x = - 2y \\ y = - \frac{x}{2} \\ \end{array}$

$\displaystyle \begin{array}{l} a = - \frac{x}{2} + z \\ a = - \frac{x}{2} + 3x \\ a = \frac{{5x}}{2} \\ \end{array}$

$\displaystyle \begin{array}{l} x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\ 6x = 16480 \\ x = \frac{{16480}}{6} = \frac{{8240}}{3} \\ \end{array}$

$\displaystyle \begin{array}{l} z = 3\frac{{8240}}{3} \\ z = 8240 \\ \end{array}$

$\displaystyle \begin{array}{l} y = - \frac{{8240}}{6} \\ \\ a = - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\ \end{array}$

So finally we can fill all variables in equation x+y+z+a=16480:
$\displaystyle \frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480$

$\displaystyle \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$

7. Please, don't post twice same message.

8. Originally Posted by OReilly
If
$\displaystyle x = y + \frac{z}{2}$
is correct than this is solution:

$\displaystyle \begin{array}{l} x = y + \frac{z}{2} \\ x = y + \frac{{3x}}{2} \\ x = y + \frac{{3x}}{2} \\ x = \frac{{2y + 3x}}{2} \\ 2x = 2y + 3x \\ 2x - 3x = 2y \\ - x = 2y \\ x = - 2y \\ y = - \frac{x}{2} \\ \end{array}$

$\displaystyle \begin{array}{l} a = - \frac{x}{2} + z \\ a = - \frac{x}{2} + 3x \\ a = \frac{{5x}}{2} \\ \end{array}$

$\displaystyle \begin{array}{l} x + \frac{{5x}}{2} + \frac{{5x}}{2} = 16480 \\ 6x = 16480 \\ x = \frac{{16480}}{6} = \frac{{8240}}{3} \\ \end{array}$

$\displaystyle \begin{array}{l} z = 3\frac{{8240}}{3} \\ z = 8240 \\ \end{array}$

$\displaystyle \begin{array}{l} y = - \frac{{8240}}{6} \\ \\ a = - \frac{{8240}}{6} + 8240 = \frac{{41200}}{6} = \frac{{20600}}{3} \\ \end{array}$

So finally we can fill all variables in equation x+y+z+a=16480:
$\displaystyle \frac{{8240}}{3} - \frac{{8240}}{6} + 8240 + \frac{{20600}}{3} = 16480$

$\displaystyle \frac{{16480 - 8240 + 49440 + 41200}}{6} = 16480$
thank you *bows*
does this show that y = z?

9. Originally Posted by spiderz
thank you *bows*
does this show that y = z?
No,
$\displaystyle y = - \frac{{8240}}{6}$
and
$\displaystyle z = 8240$

10. Originally Posted by spiderz
x+y+z+a = 16480

x=y+ 1\2z
y=a-z
z=3x
a=y+z
find a,x,y,z

See i try subbing into the top but i always have too many variable and anything else at the point undos something. I think i am missing a place where i can simplify. Please help. Any tips or anything will be appreciated.
Equations y=a-z and a=y+z are just the csame equations.
The trick is to express everything in terms of a variable you want. I choose z.
z=3x hence x=z/3
find the value of using the equation x= y + 1/2z
find a =y+z
then substitute value of x,y and a (in terms of z) in the first equation and the value of z.

11. (Moderator Note: I merged the threads, I was going to delete it but malaygoel made a post and I considered it to mean to delete it and have his post gone forever).