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nekit [7.7K]
3 years ago
10

Find the solution of the system of equations. – 2 – 8y = -30 2x + 8y = 36

Mathematics
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

x= 1.5

y= 4.125

Step-by-step explanation:

Step one:

given the two equations

-2x - 8y = -30-------1\\\\2x + 8y = 36--------2

Step two:

let us subtract the equations

-2x - 8y = -30-------1\\\\ -(2x + 8y = 36)--------2\\\\0x-16y=-66

-16y=-66\\\\16y=66

divide both sides by 16

y=66/16\\\\y=4.125

put y= 4.125 in equation 2

2x + 8(4.125)= 36\\\\2x+33=36\\\\2x=36-33\\\\2x=3\\\\x= 3/2\\\\x= 1.5

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Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

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Second equation:

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\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

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Calculate the sum of the first five terms:

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Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

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