Does the system have no solution?

Mathematics

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

$\mathrm{A)\:}-6$

Step-by-step explanation:

Notice that in the second equation, the coefficient of $y$ is $2$ times greater than in the first. Therefore, by making the coefficient of $x$ in the second equation $2$ times greater than in the first, it's impossible to only isolate one variable.

Therefore, when $a=-3\cdot 2=\fbox{-6}$ , there are no solutions to this system.

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First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].

$x=2t+3\iff t=\dfrac x2-\dfrac32\implies2\mathrm dt=\mathrm dx$
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$\displaystyle\int_{x=1}^{x=5}\sin x^2\,\mathrm dx=\displaystyle2\int_{t=-1}^{t=1}\sin(2t+3)^2\,\mathrm dt$

Let $f(t)=2\sin(2t+3)^2$ . With $n=4$ , we're looking for coefficients $c_i$ and nodes $x_i$ , with $1\le i\le4$ , such that

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx c_1f(x_1)+\cdots+c_4f(x_4)$

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).

Using the quadrature, we then have

$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.3749f(-0.8611)+0.6521f(-0.3400)+0.6521f(0.3400)+0.3749f-0.8611)$
$\displaystyle\int_{-1}^1f(t)\,\mathrm dt\approx0.5790$