Find x;y
- Thread starter 123
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123 said:
What methods have you been taught to solve these?
mmm4444bot
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Do they ask for all Complex solutions, or just the Real solutions ?
One could use graphing tools, to find the Real solutions.
Algebraically, one could find all solutions by first solving each given equation for y, followed by equating results.
There are other approaches, too.
Can you share your thoughts?
Hello, 123!
\(\displaystyle \begin{array}{ccccccc}\text{Multiply [1] by 2:} & 4x^2 - 6xy + 2y^2 &=& 6 \\ \text{Add [2]:} & x^2 + 2xy - 2y^2 &=& 6 \end{array}\)
\(\displaystyle \text{and we have: }\;5x^2 - 4xy \:=\:12 \quad\Rightarrow\quad y \:=\:\frac{5x^2-12}{4x}\;\;[3]\)
\(\displaystyle \text{Substitute into [1]: }\;2x^2 - 3x\left(\frac{5x^2-12}{4x}\right) + \left(\frac{5x^2-12}{4x}\right)^2 \;=\;3\)
. . . . . . . . . . . \(\displaystyle 2x^2 + \frac{-15x^2+36}{4} + \frac{25x^4 - 120x^2 + 144}{16x^2} \;=\;3\)
\(\displaystyle \text{Multiply by }16x^2\!:\;\;32x^4 - 60x^4 + 144x^2 + 25x^4 - 120x^2 + 144 \;=\;48x^2 \quad\Rightarrow\quad -3x^4 - 24x^2 + 144 \;=\;0\)
\(\displaystyle \text{Divide by -3:}\quadx^4 + 8x^2 - 48 \;=\;0 \quad\Rightarrow\quad (x^2 +12)(x^2-4) \:=\:0\)
\(\displaystyle \text{We have: }\;\begin{Bmatrix}x^2\:=\;-12 & & \text{no real roots} \\ x^2\:=\:4 & \Rightarrow & x \:=\:\pm2 \end{Bmatrix}\)
\(\displaystyle \text{Substitute into [3]: }\;y \;=\;\frac{5(\pm2)^2 - 12}{4(\pm2)} \;=\;\frac{8}{\pm8} \:=\:\pm1\)
\(\displaystyle \text{Answers: }\;(x,y) \;=\;(2,1),\;(-2,-1)\)
\(\displaystyle \begin{array}{ccccccc}\text{Multiply [1] by 2:} & 4x^2 - 6xy + 2y^2 &=& 6 \\ \text{Add [2]:} & x^2 + 2xy - 2y^2 &=& 6 \end{array}\)
\(\displaystyle \text{and we have: }\;5x^2 - 4xy \:=\:12 \quad\Rightarrow\quad y \:=\:\frac{5x^2-12}{4x}\;\;[3]\)
\(\displaystyle \text{Substitute into [1]: }\;2x^2 - 3x\left(\frac{5x^2-12}{4x}\right) + \left(\frac{5x^2-12}{4x}\right)^2 \;=\;3\)
. . . . . . . . . . . \(\displaystyle 2x^2 + \frac{-15x^2+36}{4} + \frac{25x^4 - 120x^2 + 144}{16x^2} \;=\;3\)
\(\displaystyle \text{Multiply by }16x^2\!:\;\;32x^4 - 60x^4 + 144x^2 + 25x^4 - 120x^2 + 144 \;=\;48x^2 \quad\Rightarrow\quad -3x^4 - 24x^2 + 144 \;=\;0\)
\(\displaystyle \text{Divide by -3:}\quadx^4 + 8x^2 - 48 \;=\;0 \quad\Rightarrow\quad (x^2 +12)(x^2-4) \:=\:0\)
\(\displaystyle \text{We have: }\;\begin{Bmatrix}x^2\:=\;-12 & & \text{no real roots} \\ x^2\:=\:4 & \Rightarrow & x \:=\:\pm2 \end{Bmatrix}\)
\(\displaystyle \text{Substitute into [3]: }\;y \;=\;\frac{5(\pm2)^2 - 12}{4(\pm2)} \;=\;\frac{8}{\pm8} \:=\:\pm1\)
\(\displaystyle \text{Answers: }\;(x,y) \;=\;(2,1),\;(-2,-1)\)
Thanks soroban!
____________________
And I need advice for this:
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\xy+x+y=5 \end{matrix}\right.\)
How I do...
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\xy+x+y=5 /\cdot (-xy) \end{matrix}\right.\)
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\-x^2y^2-x^2y-xy^2=-5xy \end{matrix}\right.\)
\(\displaystyle x^2y+xy^2-x^2y^2-x^2y-xy^2+5xy=6 \rightarrow -x^2y^2+5xy=6\)
\(\displaystyle x^2y^2-5xy+6=0\)
\(\displaystyle D=y^2=y\)
\(\displaystyle x_1=\frac{3}{y}\) and \(\displaystyle x_2=\frac{2}{y}\)
And I stuck from here. I can't solve it. Or it's bad solving.
____________________
And I need advice for this:
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\xy+x+y=5 \end{matrix}\right.\)
How I do...
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\xy+x+y=5 /\cdot (-xy) \end{matrix}\right.\)
\(\displaystyle \left\{\begin{matrix}x^2y+xy^2=6 \\-x^2y^2-x^2y-xy^2=-5xy \end{matrix}\right.\)
\(\displaystyle x^2y+xy^2-x^2y^2-x^2y-xy^2+5xy=6 \rightarrow -x^2y^2+5xy=6\)
\(\displaystyle x^2y^2-5xy+6=0\)
\(\displaystyle D=y^2=y\)
\(\displaystyle x_1=\frac{3}{y}\) and \(\displaystyle x_2=\frac{2}{y}\)
And I stuck from here. I can't solve it.
Or it's bad solving.x^2y + xy^2 = 6 [1]123 said:
xy + x + y = 5 [2]
From [2]:
xy + y = 5 - x
y(x + 1) = 5 - x
y = (5 - x) / (x + 1)
Substitute in [1]:
x^2[(5 - x) / (x + 1)] + x[(5 - x) / (x + 1)]^2 = 6
Solve for x.
QUESTION:
why did you change xy + x + y = 5 to xy + x + y = 5 / . (-xy) ?
makes no sense...
are you unfamiliar with the basics?
NOTE:
next time you have a NEW problem, start a new thread:
do not "tack on" as you did here.