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4 years ago

13

What is the answer to y+2x+3 y-2x=-1

2 answers:

Dennis_Churaev [7]4 years ago

6 0

$\left \{ {{y+2x=3} \atop {y-2x=-1}} \right.$
$y+y+2x+(-2x)=3+(-1)\\\\ 2y+2x-2x=3-1\\\\ 2y=2 \ |:(-2)\\\\ y=1$
----------
$y+2x=3$

$1+2x=3 \ |-1\\\\2x=2 \ |:2\\\\ x=1$
$\left \{ {{x=1} \atop {y=1}} \right.$

SpyIntel [72]4 years ago

6 0

Hello :
y+2x=3 ...(1)
y-2x=-1...(2)
by (1) : y =-2x+3
by(2) : y=2x-1
2x-1=-2x+3
2x+2x=3+1
4x=4
x=1
y= 2(1)-1=1

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3 years ago

Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r >

Ira Lisetskai [31]

Answer:

The other pairs are:

$(a)\ (2, \frac{5\pi}{6}) \to$ $(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3}) \to$ $(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4}) \to$ $(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

See attachment for plots

Step-by-step explanation:

Given

$(a)\ (2, \frac{5\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

Solving (a): Plot a, b and c

See attachment for plots

Solving (b): Find other pairs for $r > 0$ and $r < 0$

The general rule is that:

The other points can be derived using

$(r, \theta) = (r, \theta + 2n\pi)$

and

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

Let $n =1$ ---- You can assume any value of n

So, we have:

$(r, \theta) = (r, \theta + 2n\pi)$

$(r, \theta) = (r, \theta + 2*1*\pi)$

$(r, \theta) = (r, \theta + 2\pi)$

$(r, \theta) = (-r, \theta + (2n + 1)\pi)$

$(r, \theta) = (-r, \theta + (2*1 + 1)\pi)$

$(r, \theta) = (-r, \theta + (2 + 1)\pi)$

$(r, \theta) = (-r, \theta + 3\pi)$

$(a)\ (2, \frac{5\pi}{6})$

$r = 2\ \ \ \ \theta = \frac{5\pi}{6}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi}{6} + 2\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (2, \frac{5\pi+12\pi}{6})$

$(2, \frac{5\pi}{6}) = (2, \frac{17\pi}{6})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi}{6} + 3\pi)$

Take LCM

$(2, \frac{5\pi}{6}) = (-2, \frac{5\pi+18\pi}{6})$

$(2, \frac{5\pi}{6}) = (-2, \frac{23\pi}{6})$

The other pairs are:

$(2, \frac{17\pi}{6})$ and $(-2, \frac{23\pi}{6})$

$(b)\ (1, -\frac{2\pi}{3})$

$r = 1\ \ \ \theta = -\frac{2\pi}{3}$

So, the pairs are:

$(r, \theta) = (r, \theta + 2\pi)$

$(1, -\frac{2\pi}{3}) = (1, -\frac{2\pi}{3} + 2\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (1, \frac{-2\pi+6\pi}{3})$

$(1, -\frac{2\pi}{3}) = (1, \frac{4\pi}{3})$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(1, -\frac{2\pi}{3}) = (-1, -\frac{2\pi}{3} + 3\pi)$

Take LCM

$(1, -\frac{2\pi}{3}) = (-1, \frac{-2\pi+9\pi}{3})$

$(1, -\frac{2\pi}{3}) = (-1, \frac{7\pi}{3})$

The other pairs are:

$(1, \frac{4\pi}{3})$ and $(-1, \frac{7\pi}{3})$

$(c)\ (-1, \frac{5\pi}{4})$

$r = -1 \ \ \ \ \theta = \frac{-5\pi}{4}$

So, the pairs are

$(r, \theta) = (r, \theta + 2\pi)$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi}{4} + 2\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (-1, \frac{-5\pi+8\pi}{4} )$

$(-1, \frac{-5\pi}{4}) = (-1, \frac{3\pi}{4} )$

And

$(r, \theta) = (-r, \theta + 3\pi)$

$(-1, \frac{-5\pi}{4}) = (-(-1), \frac{-5\pi}{4}+ 3\pi)$

Take LCM

$(-1, \frac{-5\pi}{4}) = (1, \frac{-5\pi+12\pi}{4})$

$(-1, \frac{-5\pi}{4}) = (1, \frac{7\pi}{4})$

So, the other pairs are:

$(-1, \frac{3\pi}{4} )$ and $(1, \frac{7\pi}{4})$

5 0

3 years ago