Asked by ash
A store sells televisions for $360 and video cassette recorders for $270. at the beginning of the week its entire stock is worth $45,990. During the week it sells three quarters of the televisions and one third of the video cassette recorders for a total of $27,630. How many televisions and video cassette recorders did it have in its stock at the beginning of the week?
Answers
Answered by
drwls
You need to write two equations in two unknowns, and then solve for them both.
Let x be the original number of TVs and y be the original number of VCRs.
Here is what you know:
360 x + 270 y = 45,990
(3/4)*360 x + (1/3)*270 y = 27,630
which can be simplified to:
270 x + 90 y = 27,630
Tripe the last equation and subtract the first.
810 x + 270 y = 82,890
360 x + 270 y = 45,990
______________________
450 x = 36,900
x = 820
Use any of the original equations to solve for y.
Let x be the original number of TVs and y be the original number of VCRs.
Here is what you know:
360 x + 270 y = 45,990
(3/4)*360 x + (1/3)*270 y = 27,630
which can be simplified to:
270 x + 90 y = 27,630
Tripe the last equation and subtract the first.
810 x + 270 y = 82,890
360 x + 270 y = 45,990
______________________
450 x = 36,900
x = 820
Use any of the original equations to solve for y.
Answered by
Lena
I've been working on this for a while but I cant seem to get anywhere with this. I created this equation.
360(x) + 270 (y) = $ 45,990
360 (3x/4) + 270 (x/3) $27, 630
There's a starting point. If I figure out how to solve this I'll post.
360(x) + 270 (y) = $ 45,990
360 (3x/4) + 270 (x/3) $27, 630
There's a starting point. If I figure out how to solve this I'll post.