# Thread: making correct linear equation

1. ## making correct linear equation

this problem is hurting me. It's from the National University of Colombia 2007 entrance exam (translated from Spanish), and I need to understand it's principles for their test in two days.

When one adds a hard drive to a new personal computer, the new system costs $2900 x 10^3. One knows that 1/3 of the computer, plus 1/5 of the hard drive adds up to$870 x 10^3. If x represents the value of the computer and y the hard drive, a system of linear equations that permits calculating the value of the hard drive is:

A.
x - y = 290 x 10^4
5x + 3y = 1305 x 10^4

B.
x+y=290 x 10^4
3x + 5y = 87 x 10^4

C.
x - y = 290 x 10^4
5x + 3y = 87 x 10^4

D.
x + y = 290 x 10^4
5x + 3y = 1305 x 10^4

The answer is D, but I'm not sure why. Please explain it if you can. Thank you

2. Originally Posted by mpoderoso
this problem is hurting me. It's from the National University of Colombia 2007 entrance exam (translated from Spanish), and I need to understand it's principles for their test in two days.

When one adds a hard drive to a new personal computer, the new system costs $2900 x 10^3. One knows that 1/3 of the computer, plus 1/5 of the hard drive adds up to$870 x 10^3. If x represents the value of the computer and y the hard drive, a system of linear equations that permits calculating the value of the hard drive is:

A.
x - y = 290 x 10^4
5x + 3y = 1305 x 10^4

B.
x+y=290 x 10^4
3x + 5y = 87 x 10^4

C.
x - y = 290 x 10^4
5x + 3y = 87 x 10^4

D.
x + y = 290 x 10^4
5x + 3y = 1305 x 10^4

The answer is D, but I'm not sure why. Please explain it if you can. Thank you
D is correct as obviously a computer (x) plus a hard drive (y) = 2900 X 10 ^3 therefore $\displaystyle x + y = 290\times 10^4$

Now one thrid of the computer plus one fifth of the hard drive costs 870X10^3.

The equation is

$\displaystyle \frac{1}{3}x + \frac{1}{5}y = 87\times 10^4$

If you multiply both sides of the equation through by 5 then by 3 you will get

$\displaystyle 5x + 3y = 1305\times 10^4$

and the system becomes...

$\displaystyle x + y = 290\times 10^4$

$\displaystyle 5x + 3y = 1305\times 10^4$

3. Thanks so much! It's perfectly clear now.