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klasskru [66]
2 years ago
6

Two student clubs were selling t-shirts and school notebooks to raise money for an upcoming school event. In the first few minut

es, club A sold 3 t-shirts and 2 notebooks, and made $19. Club B sold 1 t-shirt and 1 notebook, for a total of $8. Use the given matrix equation to solve for the cost of t-shirts and notebooks that were sold. Explain the steps that you took to solve this problem.
A matrix with 2 rows and 2 columns, where row 1 is 3 and 2 and row 2 is 1 and 1, is multiplied by matrix with 2 rows and 1 column, where row 1 is x and row 2 is y, equals a matrix with 2 rows and 1 column, where row 1 is 19 and row 2 is 8.
Mathematics
1 answer:
mixas84 [53]2 years ago
7 0

Using a system of equations, it is found that the cost of a t-shirt is of $3 and the cost of a notebook is of $5.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

The variables are given as follows:

  • Variable x: Cost of a t-shirt.
  • Variable y: Cost of a notebook.

Considering the costs of the purchases of Clubs A and B, the matrices give the equations as follows:

  • 3x + 2y = 19.
  • x + y = 8 -> y = (8 - x).

Hence, replacing the second equation into the first:

3x + 2(8 - x) = 19

x = 3.

y = 8 - x = 8 - 3 = 5.

The cost of a t-shirt is of $3 and the cost of a notebook is of $5.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

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Elanso [62]

Answer:

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3 years ago
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What is the solution of the linear-quadratic system of equations? { y = x 2 + 5 x − 3 y − x = 2
photoshop1234 [79]

Try substituting those values, and you will see they do not fit either equation.

looking at the 2nd equation, y = x+2, so

x+2 = x^2+5x-3

x^2 + 4x - 5 = 0

(x+5)(x-1) = 0

x = -5 or 1

so, y = -3 or 3

the solutions are thus (-5,-3)(1,3)

Hope this helps

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8 0
3 years ago
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A hardware store clerk mixed nuts worth $1.32 per kilogram with bolts worth $1.20 per kilogram. If he made 108 kilograms of a mi
AlladinOne [14]

Answer:

the way to work these types of problems is to create a table. The table will have three rows and three columns.

The table:

Let x represent the kilograms of nuts, then 138 - x will represent the kilograms of bolts.

Product     Amount        Cost            Total

--------------------------------------------------------

  Nuts          x               .60         .60 * x

  Bolts      138 - x          .66         .66 * (138 - x)

  Mix          138             .65         .65 * (138)

Step-by-step explanation:

The equation:

The nuts plus the bolts must equal the mix or:

.60 * x + .66 * (138 - x) = .65 * 138

I do not like to work with decimals, so multiply each term by 100:

60 * x + 66 * (138 - x) = 65 * 138

Remove the parenthesis and combine like terms:

-6x = -66 * 138 + 65 * 138

-6x = -138

Divide both sides by -6:

x = 23

Solution:

The mixture contains 23 kilograms of nuts.

The mixture contains 138 - 23, or 115 kilograms of bolts.

8 0
2 years ago
There were 63 equal piles of plantain fruit put together and 7 single fruits. They were divided evenly among 23 travelers. What
antiseptic1488 [7]

Answer:

There are 28 fruits in each pile.

Step-by-step explanation:

We are given the following in the question:

Let x denote the number of fruits in each pile and y denote the number of fruits that every traveler receive.

Consider the diophantine equation:

63x + 7 = 23y\\63x-23y = -7

We obtain the greatest common divisor of (63,-23).

The greatest common divisor f 63 and -23 is 1.

Therefore, there exist \alpha,\beta such that

-23\alpha+63\beta=1

By applying extended Euclidean Algorithm, we have,

1 = (1\times 6)+(-1\times 5)\\1 = (1\times 17)+(3\times 6)\\1 = (3\times 23)+(-4\times 17)\\1 = (-4\times 40)+(7\times 23)\\1 = (7\times 63)+(-11\times 40)\\1 = (-11\times -23)+(-4\times 63)

1 = (-11\times -23) + (-4\times 63)

Multiplying -7 on both sides, we get,

-7 = (77\times -23) + 28\times 63)

Thus, we get,

x = 28\\y=77

Thus, there are 28 fruits in each pile.

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