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3 years ago

−5x+4y=38 −x+2y=10 solve by elimination

Mathematics

1 answer:

tatyana61 [14]3 years ago

8 0

Answer:

(-6, 2)

Step-by-step explanation:

-5x + 4y - 38 = 0 (1) | -x +2y - 10 = 0 (2)

First i multiply the 2nd equation by -5, because i want the magnitude of the x coefficient to be the same but with opposite signs, because when i'll add the equations the x's are going to cancel.

(2) * (-5):

5x - 10y + 50 = 0 (2')

(1) + (2'):

-5x + 4y - 38 + 5x - 10y + 50 = 0

-6y + 12 = 0

y = 2

Use the second equation, before multiplying to find the x - solution.

-x + 2(2) - 10 = 0

-x + 4 - 10 = 0 | add 'x' to both sides

x = -6

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The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

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