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?

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.

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.

To solve this problem, we will use a system of equations with two variables.

Let's assume that the store had x televisions and y video cassette recorders in its stock at the beginning of the week.

From the given information, we can set up the following equations:

1. The total value of televisions: 360x
2. The total value of video cassette recorders: 270y
3. The total stock value at the beginning of the week: 36990 (given)

According to the question, during the week, the store sold three quarters (3/4) of the televisions and one third (1/3) of the video cassette recorders for a total of $27,630.

Now, let's calculate the value of the sold televisions and video cassette recorders:

1. The value of the sold televisions = 3/4 * 360x = (3/4) * 360x = 270x
2. The value of the sold video cassette recorders = 1/3 * 270y = (1/3) * 270y = 90y

According to the question, the total value of the sold televisions and video cassette recorders is $27,630.

Therefore, we have the equation:

270x + 90y = 27,630 --- (Equation 1)

Since the store sold three-quarters (3/4) of the televisions, it had one-quarter (1/4) of the televisions remaining in its stock. Similarly, the store had two-thirds (2/3) of the video cassette recorders remaining.

So, we also have the following equations:

1. The value of the remaining televisions = 1/4 * 360x = (1/4) * 360x = 90x
2. The value of the remaining video cassette recorders = 2/3 * 270y = (2/3) * 270y = 180y

The total value of the remaining stock = the total stock value at the beginning of the week - the total value of the sold products.

Therefore, we have the equation:

90x + 180y = 45,990 - 27,630
90x + 180y = 18,360 --- (Equation 2)

Now, we have the following system of equations:

270x + 90y = 27,630 --- (Equation 1)
90x + 180y = 18,360 --- (Equation 2)

We can solve this system of equations using any method such as substitution, elimination, or matrices to find the values of x and y, which represent the number of televisions and video cassette recorders, respectively.

Let's use the elimination method to solve this system:

1. Multiply Equation 1 by 2:

540x + 180y = 55,260 --- (Equation 3)

2. Multiply Equation 2 by -3:

-270x - 540y = -55,080 --- (Equation 4)

Now, adding Equation 3 and Equation 4, we get:

(540x + 180y) + (-270x - 540y) = (55,260 + (-55,080))

Simplifying:

270x - 360y = 180 --- (Equation 5)

Now, we can solve Equations 5 and 2 using the elimination method again:

1. Multiply Equation 5 by 2:

540x - 720y = 360 --- (Equation 6)

2. Add Equation 6 and Equation 2:

(540x - 720y) + (90x + 180y) = (360 + 18,360)

Simplifying:

630x - 540y = 18,720 --- (Equation 7)

Now, we have the following system of equations:

630x - 540y = 18,720 --- (Equation 7)
90x + 180y = 18,360 --- (Equation 2)

We can solve this system of equations again using the elimination method:

1. Multiply Equation 7 by 1/630:

x - (6/7)y = 4/7 --- (Equation 8)

2. Multiply Equation 2 by -1/90:

-(1/7)x - 2y = -204 --- (Equation 9)

Now, add Equation 8 and Equation 9:

(x - (6/7)y) + (-(1/7)x - 2y) = (4/7 - 204)

Simplifying:

(-1/7)x - (20/7)y = -204 + (28/7)

-1/7x - 20/7y = -204 + 4

-1/7x - 20/7y = -200 --- (Equation 10)

Now, we have the following system of equations:

-1/7x - 20/7y = -200 --- (Equation 10)
90x + 180y = 18,360 --- (Equation 2)

We can solve this system of equations using the elimination method one more time:

1. Multiply Equation 10 by -7:

x + 20y = 1400 --- (Equation 11)

2. Multiply Equation 2 by 1/90:

x + 2y = 204 --- (Equation 12)

Now, add Equation 11 and Equation 12:

(x + 20y) + (x + 2y) = (1400 + 204)

Simplifying:

2x + 22y = 1604 --- (Equation 13)

Now, we have the following system of equations:

2x + 22y = 1604 --- (Equation 13)
x + 2y = 204 --- (Equation 12)

To solve this system, we'll use the elimination method by multiplying Equation 12 by -2:

-2(x + 2y) = -2(204)

Simplifying:

-2x - 4y = -408 --- (Equation 14)

Adding Equation 13 and Equation 14:

(2x + 22y) + (-2x - 4y) = (1604 - 408)

Simplifying:

18y = 1196

Now, we can solve for y:

y = 1196 / 18
y = 66.44

Since we cannot have a fraction of a television or video cassette recorder, we will round up to the nearest whole number:

y ≈ 66

Next, substitute the value of y into Equation 12 to solve for x:

x + 2(66) = 204
x + 132 = 204
x = 204 - 132
x = 72

Therefore, the store had approximately 72 televisions and 66 video cassette recorders in its stock at the beginning of the week.