Solve the following system of equations.

Mathematics

1 answer:

Elan Coil [88]1 year ago

5 0

Answer:

(2,-3)

Step-by-step explanation:

I am not sure if you meant the first equation to be y or -y. I solved it as y.

y = x-5 -x -3y =7

I am going to take the second equation and write it as x =

-x - 3y = 7 Give equation

-x = 3y +7 Add 3y to both sides

x = -3y-7 Multiplied each term in the equation by -1 so that x could be positive

I am going to substitute -3y-7 for x in the first equation up above

y = x - 5

y = -3y -7 - 5 Substitute -3y-7 for x

y = -3y -12 Combined -7-5

4y = -12 Added 3y to both sides

y = -3 Divided both sides by 4.

I now know that y is -3, I will plug that into x = -3y-7 to solve for x

x = -3(-3) -7

x = 9-7 A negative times a negative is a positive

x = 2

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What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$