MHB How to Solve Simultaneous Equations with Multiplication Symbols?

AI Thread Summary
The discussion focuses on solving a system of simultaneous equations involving multiplication symbols. Participants clarify that the dot represents multiplication and attempt to manipulate the equations to find solutions. One user derives expressions for y in terms of x and substitutes them into another equation. However, it is concluded that the system has no real solutions, raising questions about the accuracy of the problem statement. The conversation emphasizes the importance of correctly interpreting and solving simultaneous equations.
AstroBoy1
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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 :)
 

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AstroBoy said:
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
 
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?:)
 
Hello, AstroBoy!

I agree with anemone.

\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}

\begin{array}{cccccccc}4\times[1]& x+y \,=\,4-2xy \\ 12\times[2] & 16xy - 9x + 3y \,=\,36 \end{array}We have: .\begin{array}{cccc}x + y + 2xy &=& 4 & [3] \\ 9x - 3y - 16xy &=& \text{-}36 & [4] \end{array}
\begin{array}{cccccc}8\times[3] & 8x + 8y + 16xy &=& 32 \\ \text{Add [4]} & 9x - 3y - 16xy &=& \text{-}36 \end{array}

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

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

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

And we have: .34x^2 + 20x + 24 \:=\:0But .17x^2 + 10x + 12 \:=\:0 .has no real roots.
 
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