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lawyer [7]
4 years ago
6

What is the solution to the systen of equations y=1/2x+2 and y=-2x-3

Mathematics
1 answer:
Naya [18.7K]4 years ago
5 0

For this case we must solve a system of two equations with two unknowns, given by "x" and "y".

We have:

y = \frac {1} {2} x + 2 (1)\\y = -2x-3 (2)

We multiply (2) by -1:

-y = 2x + 3 (3)

We add (1) and (3):

y-y = \frac {1} {2} x + 2x + 2 + 3\\0 = \frac {5} {2} x + 5\\-5 = \frac {5} {2} x

Clearing x:

-10 = 5x\\x = \frac {-10} {5}\\x = -2

We substitute x = -2 in (2)

y = -2 (-2) -3\\y = 4-3\\y = 1

Thus, the solution of the system is(x, y) = (- 2,1)

Answer:

the solution of the system is(x, y) = (- 2,1)


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the answer is incomplete, below is the complete question

"Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter your answer in terms of s.) r(t) = 3ti + (1 - 4t)j + (1 + 2t)k r(t(s)) ="

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Step-by-step explanation:

The step by step procedure is to first determine the differentiate the given vector function

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since s(t) is the arc length for r(t), which is define as

s(t)=\int\limits^t_0 {||r'(t)||} \, dt

if we substitute the value of r'(t) we arrive at

s(t)=\int\limits^t_0 {||r'(t)||} \, dt\\s(t)=\int\limits^t_0 {\sqrt{3^{2} +4^{2}+2^{2}} \, dt\\s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\

s(t)=\int\limits^t_ 0{\sqrt{29} } \, dx\\\\s(t)=\sqrt{29} t\\hence \\t(s)=\frac{s}{\sqrt{29} }

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r(t(s)) = \frac{3s}{\sqrt{29} } i + (1 -\frac{4s}{\sqrt{29} }t)j + (1 + \frac{2s}{\sqrt{29} })k

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thus    \bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
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-----------------------------\\\\

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\end{cases}

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