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Simultaneous Equations

AstroBoy

New member
Feb 25, 2014
1
Can someone solve this, i know its not very hard but for me it is :/
Untitled.png
The dot . is meaned to be * (multiplication)

can someone help me :)
 
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Amer

Active member
Mar 1, 2012
275
Can someone solve this, i know its not very hard but for me it is :/
View attachment 2014
The dot . is meaned to be * (multiplication)

can someone help me :)
You want to find $ \dfrac{x}{4} + \dfrac{x}{4} = \dfrac{x}{2} $ we want half of x

$ \dfrac{x+y}{4} = 1 - \dfrac{xy}{2} $

$ \dfrac{2(2xy)}{3} - \dfrac{3x-y}{4} = 3 $

From the first equation write y with respect to x, first i would like to multiply it with 4 to eliminate the denominator

$ x+y = 4 - 2xy \Rightarrow y( 1 + 2x) = 4-x \Rightarrow y = \dfrac{4 -x }{1+2x} $
Sub it in the second and solve it for x
Tell us what you get...
I supposed that (1+2x) =/= 0
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,677
Hey Astroboy,

I noticed the system of equations that you cited has no solution over the real numbers. Are you sure you have copied the problem correctly?:)
 

soroban

Well-known member
Feb 2, 2012
409
Hello, AstroBoy!

I agree with anemone.


[tex]\begin{array}{cccc}\tfrac{1}{4}(x+y) \;=\;1 - \tfrac{1}{2}xy & [1] \\ \tfrac{2}{3}(2xy) - \tfrac{1}{4}(3x-y) \;=\;3 & [2] \end{array}[/tex]

[tex]\begin{array}{cccccccc}4\times[1]& x+y \,=\,4-2xy \\ 12\times[2] & 16xy - 9x + 3y \,=\,36 \end{array}[/tex]


We have: .[tex]\begin{array}{cccc}x + y + 2xy &=& 4 & [3] \\ 9x - 3y - 16xy &=& \text{-}36 & [4] \end{array}[/tex]



[tex]\begin{array}{cccccc}8\times[3] & 8x + 8y + 16xy &=& 32 \\ \text{Add [4]} & 9x - 3y - 16xy &=& \text{-}36 \end{array}[/tex]

We have: .[tex]17x + 5y \:=\:\text{-}4 \quad\Rightarrow\quad y \:=\:\text{-}\frac{17x+4}{5}[/tex]

Substitute into [3]: .[tex]x - \frac{17x+4}{5} + 2x\left(\text{-}\frac{17x+4}{5}\right) \:=\:4[/tex]

Multiply by 5: .[tex]5x - 17x - 4 - 34x^2 - 8x \:=\:20[/tex]

And we have: .[tex]34x^2 + 20x + 24 \:=\:0[/tex]


But .[tex]17x^2 + 10x + 12 \:=\:0[/tex] .has no real roots.