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NISA [10]
3 years ago
13

Solve the simultaneous equations 3 2x - y = 7 xy = 15

Mathematics
1 answer:
ehidna [41]3 years ago
8 0

Answer:

(5, 3 ) and (- \frac{3}{2}, - 10 )

Step-by-step explanation:

Given the 2 equations

2x - y = 7 → (1)

xy = 15 → (2)

Rearrange (1) expressing y in terms of x , that is

y = 2x - 7 → (3)

Substitute y = 2x - 7 into (2)

x(2x - 7) = 15

2x² - 7x = 15 ( subtract 15 from both sides )

2x² - 7x - 15 = 0 ← in standard form

(x - 5)(2x + 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = - \frac{3}{2}

Substitute these values into (3) for corresponding values of y

x = 5 : y = 2(5) - 7 = 10 - 7 = 3 ⇒ (5, 3 )

x = - \frac{3}{2} : y = 2(- \frac{3}{2} ) - 7 = - 3 - 7 = - 10 ⇒ (- \frac{3}{2}, - 10 )

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Answer:

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Step-by-step explanation:

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For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
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The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

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f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
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Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

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(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}

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