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The method requires you to multiply the first equation by some number, the second equation by some number, in order that either the x coefficients are the same or that the y coefficients are the same. Sometimes, you get lucky and a pair of coefficients, either the x's or the y's are already equal in which case the jobs already done, or it's sufficient to multiply just one of the equations by a number in order to make a pair of coefficients equal.

With these equations, you have a choice. Multiply the first equation by 3 to get 6x + 9y = -3, multiply the second equation by 2 to get 6x + 8y = -8, and now subtract one equation from the other. The x's cancel out. That allows you to find the value of y, and by substituting that into one of the earlier equations allows you to find the value of x.

Alternatively, multiply the first equation by 4 to get 8x + 12y = -4, the second equation by 3 to get

9x + 12y = -12,  and subtract one from the other. That gets you the value of x, and by substituting that into one of the earlier equations, the value of y.

You should do it both ways, as an exercise, to show that you get the same results, either way.

There will be sets of equations where the x's or y's are opposite in sign in which case the equations have to be added rather than subtracted.

Yes i am an aclaimed artist.   LOL

I just haven't been discovered yet. :))

I'm going to cheat slightly and refer to Alan's rather nice diagram. (I drew it on paper).

The angle that OB makes with the negative y-axis is 15 deg which means that the angle AOB is 120 deg, which in turn means that the angle OBQ is 60 deg.

OB and BQ being the same length implies that the triangle OBQ is equilateral in which case the length of OQ is 6 and it's at an angle of 45 deg to the negative y-axis.

That means that Q has polar co-ordinates (6, -45deg).

here is a slightly different way.

I would have multiplied  both sides by 10

10(1/2x + 4/5x) =10*2

5x+8x=20

13x=20

x=20/13

CPhill has the 100 in his equation too Rosala.

he used   r=0.034  and you used r=3.4/100  

see, they are the same.  :)

I think yours was just a careless error.

you had P=1200 instead of P =2200    

It's not an equation. For it to be an equation, something has to be equal to something else.

There is nothing to solve, this is just an expression that can be written in different ways.

I am sorry anon but your answer is not correct.  Can you work our why?

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{2}}\right)} = {\mathtt{63.434\: \!948\: \!822\: \!922^{\circ}}}$$

general solution.   Where n is an integer

$$x=(180n+63.43)^0$$

Look at the sticky notes on the right of your page.

Can you be more specific please - I don't understand what you are asking.

Make U the subject in:S= U*V / U + V (this being a fraction).

$$\small{\text{$
S= \dfrac{ U\cdot V } { U + V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U + V } { U\cdot V }
$}}\\\\
\small{\text{$
\dfrac{1}{S}= \dfrac{ U } { U\cdot V } + \dfrac{ V } { U\cdot V }
$}}\\\\
\boxed{\small{\text{$
\dfrac{1}{S}= \dfrac{ 1 } { V } + \dfrac{ 1 } { U }
$}}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ 1 } { S } - \dfrac{ 1 } { V }
$}}\\\\
\small{\text{$
\dfrac{1}{U}= \dfrac{ V-S } { V\cdot S }
$}}\\\\
\boxed{
\small{\text{$
U= \dfrac{ V\cdot S } { V-S }
$}}}$$

multtiply each term,

x³+3x²-4x-3x²-4x-3x²-9x+12

combine like terms.

Infinity doesn't behave like an ordinary number, and shouldn't be considered as an ordinary number.

Some infinities are bigger than other infinities, in fact one infinity can be infinitely larger than another infinity. Infinity/infinity or even infinity/(infinity - 1) is an indeterminate quantity and can be equal to anything between zero and infinity, without further information you just don't know, that's why it's called indeterminate.

this is the second time you have posted this .

Please follow the instructions for reposting.   

what other information are you given?

No, a regular calculator will do.

$${\frac{{\mathtt{3}}}{{\mathtt{8}}}}$$

$${\frac{{\mathtt{3}}}{{\mathtt{8}}}} = {\mathtt{0.375}}$$

In a polar coordinate system ,O is the pole.The polar coordinates of A and B are(6,15) and (6,255) respectively.P is a moving point in the system such that PA=PB.If Q is a point lying on P such that OABQ is a rhombus , find polar coordinates of Q

$$\small{\text{$
\vec{Q}=\vec{A}+\vec{B}
$, because $|\vec{A}| = |\vec{B}| = 6 = r \quad$
Q lies on the bisectors of an angle by O
}}\\\\
\small{\text{
$\vec{A}=
\left(
\begin{array}{c}r\cdot \cos{(15)}\\r\cdot \sin{(15)}\end{array}
\right)
=
\left(
\begin{array}{c}6\cdot \cos{(15)}\\6\cdot \sin{(15)}\end{array}
\right)$
}}\\
\small{\text{
$\vec{B}=
\left(
\begin{array}{c}r\cdot \cos{(255)}\\r\cdot \sin{(255)}\end{array}
\right)
=
\left(
\begin{array}{c}6\cdot \cos{(255)}\\6\cdot \sin{(255)}\end{array}
\right)
$
}}\\\\
\small{\text{$
\vec{Q}=\vec{A}+\vec{B}
=
\left(
\begin{array}{c}6\cdot \cos{(15)}\\6\cdot \sin{(15)}\end{array}
\right)
+
\left(
\begin{array}{c}6\cdot \cos{(255)}\\6\cdot \sin{(255)}\end{array}
\right)
=
\left(
\begin{array}{c}
6\cdot \cos{(15)} + 6\cdot \cos{(255)}
\\ 6\cdot \sin{(15)} + 6\cdot \sin{(255)}
\end{array}
\right)
$}}\\\\
\small{\text{$
=
\left(
\begin{array}{c} 4.24264068712 \\ -4.24264068712 \end{array}
\right)
=
\left(
\begin{array}{c} \sqrt{18} \\ - \sqrt{18} \end{array}
\right)
$}}$$

polar coordinates of Q

$$\small{\text{
$
\vec{Q}= \left(\begin{array}{c} Q_x =\sqrt{18}\\ Q_y=-\sqrt{18} \end{arry} \right)
\qquad r = \sqrt{ Q_x^2 + Q_y^2 } = \sqrt{18+18} = 6
$
}}\\\\
\small{\text{
$
\tan{(\varphi)} = \dfrac{Q_y}{ Q_x} = \dfrac{-\sqrt{18}}{\sqrt{18}} = -1 \qquad\varphi = -45\ensurement{^{\circ}} $ or $ \varphi = 315\ensurement{^{\circ}} $
}}\\\\
\vec{Q}=(6, 315\ensurement{^{\circ}} )$$

Since the result is 2, it must mean that the opposite side divided by the djacent side equals 2. This only occurs whens the oppostie side is twice the adjacent side. Therefore it must be at an angle of 30 degrees. If you draw the 30-60-90 triangle this can be verified.

I have no idea where to start.

Some of the permutations are obvious but how you would count them I have no idea

zero, but you can never get there.

I did it exactly the same way - that is comforting for me anyway 

Here is a video on it.

19

aalthough I do not know why you do not recogognize the sume of the terms.

$${\frac{{\mathtt{3}}}{{\mathtt{7}}}} = {\mathtt{0.428\: \!571\: \!428\: \!571\: \!428\: \!6}}$$