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Dovator [93]
2 years ago
7

Tell whether the ordered pair is a solution of the given system. (-2,-4) {y=1/2x -3, y=-2x-8}

Mathematics
1 answer:
kompoz [17]2 years ago
8 0

Answer:

The ordered pair (-2,\, -4) is indeed a solution to the system:

\left\lbrace\begin{aligned} & y = \frac{1}{2}\, x - 3 \\ & y = -2\, x - 8\end{aligned}\right..

Step-by-step explanation:

Consider a system of equations about variables x and y. An ordered pair (x_{0},\, y_{0}) (where x_{0} and y_{0} are constant) is a solution to that system if and only if all equations in that system hold after substituting in x = x_{0} and y = y_{0}.

For the system in this question, (-2,\, -4) would be a solution only if both equations in the system hold after replacing all x in equations of the system with (-2) and all y with (-4).

The \texttt{LHS} of the equation y = (1/2)\, x - 3 would become (-4). The \texttt{RHS} of that equation would become (1/2) \, (-2) - 3. The two sides are indeed equal.

Similarly, the \texttt{LHS} of the equation y = -2\, x - 8 would become (-4). The \texttt{RHS} of that equation would become (-2)\, (-2) - 8. The two sides are indeed equal.

Thus, x = (-2) and y = (-4) simultaneously satisfy both equations of the given system. Therefore, the ordered pair (-2,\, -4) would indeed be a solution to that system.

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Let L, W, F, H, P represent, respectively, the length, width, number of floors, height, and perimeter of the building. Here are some equations that model the given information.

  • L = W + 40
  • H = 12×F
  • foundation cost = 10×L×W
  • cost of floors = 35000×F
  • total cost = foundation cost + cost of floors
  • total cost = 1,180,000
  • P = 2×(L+W)
  • P = 3×H

We can rewrite these into a system of equations involving L, W, F.

Let's look first at the relation between the perimeter and height. The two versions of perimeter are equal to each other, so we have (after substituting for H) ...

... 3H = 2(L+W)

... 3(12F) = 2((W +40) +W) . . . . substitute for H and L

... 36F = 4W +80 . . . . . . . . . . . simplify

... 9F = W +20 . . . . . . . . . . . . . divide by 4

Now, we can look at the cost equations.

... 1180000 = 35000F + 10W(W+40) . . . . . substitute for the various cost terms and for L

... 118000 = 3500F +W² +40W . . . . . . . . . divide by 10

The two simplified equations in W and F are ...

  • 9F = W +20
  • 3500F = 118000 -40W -W²

You can solve the first for F and substitute for F in the second. Then you have a quadratic equation in W that can be solved by the usual methods. (Likely, the quadratic formula would be the best choice.) Here, we have elected to let a graphing calculator show the two solutions. One solution has negative dimensions, so is clearly infeasible.

There is one (nearly feasible) solution:

... W ≈ 180.788

... F ≈ 22.31

There is no combination of floor dimensions and integral numbers of floors that will match <em>all</em> of the problem requirements.

_____

A solution that comes about as close as possible is ...

  • length = 221 ft
  • width = 181 ft
  • perimeter = 804 ft
  • # of floors = 22
  • height = 264 ft . . . . . (3×height = 792 ft vs. 804 ft)
  • cost = $1,170,010 . . . (vs. 1,180,000)

4 0
3 years ago
Read 2 more answers
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