How do you solve these 2 equations?












0














$$xy = 1/6$$
$$y+x = 5xy$$
I tried solving them using all methods - substitution, elimination and graphing - but can't get the solutions






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question
























  • What were "all those methods" that you tried?
    – Parcly Taxel
    Nov 18 '18 at 6:07










  • Substitution, elimination and graphing.
    – Jullian Santos
    Nov 18 '18 at 6:11










  • Take the first equation and insert it into the second
    – Fakemistake
    Nov 18 '18 at 6:11










  • Actually I can't use the graphing so what I meant was substitution and elimination.
    – Jullian Santos
    Nov 18 '18 at 6:12










  • Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
    – PradyumanDixit
    Nov 18 '18 at 6:12
















0














$$xy = 1/6$$
$$y+x = 5xy$$
I tried solving them using all methods - substitution, elimination and graphing - but can't get the solutions






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question
























  • What were "all those methods" that you tried?
    – Parcly Taxel
    Nov 18 '18 at 6:07










  • Substitution, elimination and graphing.
    – Jullian Santos
    Nov 18 '18 at 6:11










  • Take the first equation and insert it into the second
    – Fakemistake
    Nov 18 '18 at 6:11










  • Actually I can't use the graphing so what I meant was substitution and elimination.
    – Jullian Santos
    Nov 18 '18 at 6:12










  • Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
    – PradyumanDixit
    Nov 18 '18 at 6:12














0












0








0







$$xy = 1/6$$
$$y+x = 5xy$$
I tried solving them using all methods - substitution, elimination and graphing - but can't get the solutions






systems-of-equations quadratics symmetric-polynomials







share|cite|improve this question















$$xy = 1/6$$
$$y+x = 5xy$$
I tried solving them using all methods - substitution, elimination and graphing - but can't get the solutions






systems-of-equations quadratics symmetric-polynomials




systems-of-equations quadratics symmetric-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 11:59









Harry Peter

5,46111439




5,46111439










asked Nov 18 '18 at 6:06









Jullian SantosJullian Santos

6




6












  • What were "all those methods" that you tried?
    – Parcly Taxel
    Nov 18 '18 at 6:07










  • Substitution, elimination and graphing.
    – Jullian Santos
    Nov 18 '18 at 6:11










  • Take the first equation and insert it into the second
    – Fakemistake
    Nov 18 '18 at 6:11










  • Actually I can't use the graphing so what I meant was substitution and elimination.
    – Jullian Santos
    Nov 18 '18 at 6:12










  • Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
    – PradyumanDixit
    Nov 18 '18 at 6:12


















  • What were "all those methods" that you tried?
    – Parcly Taxel
    Nov 18 '18 at 6:07










  • Substitution, elimination and graphing.
    – Jullian Santos
    Nov 18 '18 at 6:11










  • Take the first equation and insert it into the second
    – Fakemistake
    Nov 18 '18 at 6:11










  • Actually I can't use the graphing so what I meant was substitution and elimination.
    – Jullian Santos
    Nov 18 '18 at 6:12










  • Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
    – PradyumanDixit
    Nov 18 '18 at 6:12
















What were "all those methods" that you tried?
– Parcly Taxel
Nov 18 '18 at 6:07




What were "all those methods" that you tried?
– Parcly Taxel
Nov 18 '18 at 6:07












Substitution, elimination and graphing.
– Jullian Santos
Nov 18 '18 at 6:11




Substitution, elimination and graphing.
– Jullian Santos
Nov 18 '18 at 6:11












Take the first equation and insert it into the second
– Fakemistake
Nov 18 '18 at 6:11




Take the first equation and insert it into the second
– Fakemistake
Nov 18 '18 at 6:11












Actually I can't use the graphing so what I meant was substitution and elimination.
– Jullian Santos
Nov 18 '18 at 6:12




Actually I can't use the graphing so what I meant was substitution and elimination.
– Jullian Santos
Nov 18 '18 at 6:12












Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
– PradyumanDixit
Nov 18 '18 at 6:12




Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another.
– PradyumanDixit
Nov 18 '18 at 6:12










3 Answers
3






active

oldest

votes


















2














You can solve this equation by using substitution of variables.



Using the first equation $xy=1/6$, we can rewrite it as $y=frac1{6x}$.



Plugging this in to the second equation, we get $$x+frac1{6x}=5frac{x}{6x}$$



Simplify the right-hand side to $frac56$ and multiply both sides by $x$ to obtain



$$x^2+frac16=frac56x to x^2-frac56x+frac16=0$$



Using the quadratic formula, we now have $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$



Our solutions for $x$ are $frac13$ and $frac12$. Using $y=frac1{6x}$, we get the coordinate pairs to be $(frac13,frac12)$ and $(frac12,frac13)$.






share|cite|improve this answer

















  • 2




    In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
    – Henno Brandsma
    Nov 18 '18 at 6:27










  • Thanks for the answer!
    – Jullian Santos
    Nov 18 '18 at 6:27



















0














1) Substitute $;xy=frac{1}{6}$ into the 2nd equation to get $;x+y=frac{5}{6}$



2) Solving $;xy=frac{1}{6}$ and $;x+y=frac{5}{6};$ is equivalent to finding the zeros of the function $f(z)=z^2+frac{5}{6}z+frac{1}{6}=0,;$ so using the quadratic formula, $;x=-frac{1}{3};text{and}; y=-frac{1}{2};$ or $;y=-frac{1}{3};text{and};x=-frac{1}{2}.$






share|cite|improve this answer























  • Would you check your quadratic equation? The signs.
    – user376343
    Dec 29 '18 at 10:00



















0














begin{array}{c}
xy = frac 16 \
x + y = 5xy = frac 56
end{array}



Let $x = frac{5}{12} + t$ and $y = frac{5}{12}-t$. Then $x+y=frac 56$.



begin{align}
xy &= frac 16 \
bigg(frac{5}{12} + t bigg) bigg(frac{5}{12} - tbigg) &= frac 16 \
frac{25}{144} - t^2 &= frac{24}{144} \
t^2 &= frac{1}{144} \
t &= pm frac{1}{12} \
end{align}



$frac{5}{12} + frac{1}{12} = frac 12$ and $frac{5}{12} - frac{1}{12} = frac 13$



So $(x,y) = left{big(frac 12,frac 13 big),big(frac 13,frac 12 big)right}$






share|cite|improve this answer























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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You can solve this equation by using substitution of variables.



    Using the first equation $xy=1/6$, we can rewrite it as $y=frac1{6x}$.



    Plugging this in to the second equation, we get $$x+frac1{6x}=5frac{x}{6x}$$



    Simplify the right-hand side to $frac56$ and multiply both sides by $x$ to obtain



    $$x^2+frac16=frac56x to x^2-frac56x+frac16=0$$



    Using the quadratic formula, we now have $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$



    Our solutions for $x$ are $frac13$ and $frac12$. Using $y=frac1{6x}$, we get the coordinate pairs to be $(frac13,frac12)$ and $(frac12,frac13)$.






    share|cite|improve this answer

















    • 2




      In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
      – Henno Brandsma
      Nov 18 '18 at 6:27










    • Thanks for the answer!
      – Jullian Santos
      Nov 18 '18 at 6:27
















    2














    You can solve this equation by using substitution of variables.



    Using the first equation $xy=1/6$, we can rewrite it as $y=frac1{6x}$.



    Plugging this in to the second equation, we get $$x+frac1{6x}=5frac{x}{6x}$$



    Simplify the right-hand side to $frac56$ and multiply both sides by $x$ to obtain



    $$x^2+frac16=frac56x to x^2-frac56x+frac16=0$$



    Using the quadratic formula, we now have $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$



    Our solutions for $x$ are $frac13$ and $frac12$. Using $y=frac1{6x}$, we get the coordinate pairs to be $(frac13,frac12)$ and $(frac12,frac13)$.






    share|cite|improve this answer

















    • 2




      In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
      – Henno Brandsma
      Nov 18 '18 at 6:27










    • Thanks for the answer!
      – Jullian Santos
      Nov 18 '18 at 6:27














    2












    2








    2






    You can solve this equation by using substitution of variables.



    Using the first equation $xy=1/6$, we can rewrite it as $y=frac1{6x}$.



    Plugging this in to the second equation, we get $$x+frac1{6x}=5frac{x}{6x}$$



    Simplify the right-hand side to $frac56$ and multiply both sides by $x$ to obtain



    $$x^2+frac16=frac56x to x^2-frac56x+frac16=0$$



    Using the quadratic formula, we now have $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$



    Our solutions for $x$ are $frac13$ and $frac12$. Using $y=frac1{6x}$, we get the coordinate pairs to be $(frac13,frac12)$ and $(frac12,frac13)$.






    share|cite|improve this answer












    You can solve this equation by using substitution of variables.



    Using the first equation $xy=1/6$, we can rewrite it as $y=frac1{6x}$.



    Plugging this in to the second equation, we get $$x+frac1{6x}=5frac{x}{6x}$$



    Simplify the right-hand side to $frac56$ and multiply both sides by $x$ to obtain



    $$x^2+frac16=frac56x to x^2-frac56x+frac16=0$$



    Using the quadratic formula, we now have $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$



    Our solutions for $x$ are $frac13$ and $frac12$. Using $y=frac1{6x}$, we get the coordinate pairs to be $(frac13,frac12)$ and $(frac12,frac13)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 18 '18 at 6:20









    Christopher MarleyChristopher Marley

    1,020115




    1,020115








    • 2




      In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
      – Henno Brandsma
      Nov 18 '18 at 6:27










    • Thanks for the answer!
      – Jullian Santos
      Nov 18 '18 at 6:27














    • 2




      In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
      – Henno Brandsma
      Nov 18 '18 at 6:27










    • Thanks for the answer!
      – Jullian Santos
      Nov 18 '18 at 6:27








    2




    2




    In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
    – Henno Brandsma
    Nov 18 '18 at 6:27




    In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4cdot 6 = 1$ so we get a nice result.
    – Henno Brandsma
    Nov 18 '18 at 6:27












    Thanks for the answer!
    – Jullian Santos
    Nov 18 '18 at 6:27




    Thanks for the answer!
    – Jullian Santos
    Nov 18 '18 at 6:27











    0














    1) Substitute $;xy=frac{1}{6}$ into the 2nd equation to get $;x+y=frac{5}{6}$



    2) Solving $;xy=frac{1}{6}$ and $;x+y=frac{5}{6};$ is equivalent to finding the zeros of the function $f(z)=z^2+frac{5}{6}z+frac{1}{6}=0,;$ so using the quadratic formula, $;x=-frac{1}{3};text{and}; y=-frac{1}{2};$ or $;y=-frac{1}{3};text{and};x=-frac{1}{2}.$






    share|cite|improve this answer























    • Would you check your quadratic equation? The signs.
      – user376343
      Dec 29 '18 at 10:00
















    0














    1) Substitute $;xy=frac{1}{6}$ into the 2nd equation to get $;x+y=frac{5}{6}$



    2) Solving $;xy=frac{1}{6}$ and $;x+y=frac{5}{6};$ is equivalent to finding the zeros of the function $f(z)=z^2+frac{5}{6}z+frac{1}{6}=0,;$ so using the quadratic formula, $;x=-frac{1}{3};text{and}; y=-frac{1}{2};$ or $;y=-frac{1}{3};text{and};x=-frac{1}{2}.$






    share|cite|improve this answer























    • Would you check your quadratic equation? The signs.
      – user376343
      Dec 29 '18 at 10:00














    0












    0








    0






    1) Substitute $;xy=frac{1}{6}$ into the 2nd equation to get $;x+y=frac{5}{6}$



    2) Solving $;xy=frac{1}{6}$ and $;x+y=frac{5}{6};$ is equivalent to finding the zeros of the function $f(z)=z^2+frac{5}{6}z+frac{1}{6}=0,;$ so using the quadratic formula, $;x=-frac{1}{3};text{and}; y=-frac{1}{2};$ or $;y=-frac{1}{3};text{and};x=-frac{1}{2}.$






    share|cite|improve this answer














    1) Substitute $;xy=frac{1}{6}$ into the 2nd equation to get $;x+y=frac{5}{6}$



    2) Solving $;xy=frac{1}{6}$ and $;x+y=frac{5}{6};$ is equivalent to finding the zeros of the function $f(z)=z^2+frac{5}{6}z+frac{1}{6}=0,;$ so using the quadratic formula, $;x=-frac{1}{3};text{and}; y=-frac{1}{2};$ or $;y=-frac{1}{3};text{and};x=-frac{1}{2}.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 29 '18 at 9:58









    user376343

    3,0032823




    3,0032823










    answered Nov 18 '18 at 6:23









    Muchang BahngMuchang Bahng

    663




    663












    • Would you check your quadratic equation? The signs.
      – user376343
      Dec 29 '18 at 10:00


















    • Would you check your quadratic equation? The signs.
      – user376343
      Dec 29 '18 at 10:00
















    Would you check your quadratic equation? The signs.
    – user376343
    Dec 29 '18 at 10:00




    Would you check your quadratic equation? The signs.
    – user376343
    Dec 29 '18 at 10:00











    0














    begin{array}{c}
    xy = frac 16 \
    x + y = 5xy = frac 56
    end{array}



    Let $x = frac{5}{12} + t$ and $y = frac{5}{12}-t$. Then $x+y=frac 56$.



    begin{align}
    xy &= frac 16 \
    bigg(frac{5}{12} + t bigg) bigg(frac{5}{12} - tbigg) &= frac 16 \
    frac{25}{144} - t^2 &= frac{24}{144} \
    t^2 &= frac{1}{144} \
    t &= pm frac{1}{12} \
    end{align}



    $frac{5}{12} + frac{1}{12} = frac 12$ and $frac{5}{12} - frac{1}{12} = frac 13$



    So $(x,y) = left{big(frac 12,frac 13 big),big(frac 13,frac 12 big)right}$






    share|cite|improve this answer




























      0














      begin{array}{c}
      xy = frac 16 \
      x + y = 5xy = frac 56
      end{array}



      Let $x = frac{5}{12} + t$ and $y = frac{5}{12}-t$. Then $x+y=frac 56$.



      begin{align}
      xy &= frac 16 \
      bigg(frac{5}{12} + t bigg) bigg(frac{5}{12} - tbigg) &= frac 16 \
      frac{25}{144} - t^2 &= frac{24}{144} \
      t^2 &= frac{1}{144} \
      t &= pm frac{1}{12} \
      end{align}



      $frac{5}{12} + frac{1}{12} = frac 12$ and $frac{5}{12} - frac{1}{12} = frac 13$



      So $(x,y) = left{big(frac 12,frac 13 big),big(frac 13,frac 12 big)right}$






      share|cite|improve this answer


























        0












        0








        0






        begin{array}{c}
        xy = frac 16 \
        x + y = 5xy = frac 56
        end{array}



        Let $x = frac{5}{12} + t$ and $y = frac{5}{12}-t$. Then $x+y=frac 56$.



        begin{align}
        xy &= frac 16 \
        bigg(frac{5}{12} + t bigg) bigg(frac{5}{12} - tbigg) &= frac 16 \
        frac{25}{144} - t^2 &= frac{24}{144} \
        t^2 &= frac{1}{144} \
        t &= pm frac{1}{12} \
        end{align}



        $frac{5}{12} + frac{1}{12} = frac 12$ and $frac{5}{12} - frac{1}{12} = frac 13$



        So $(x,y) = left{big(frac 12,frac 13 big),big(frac 13,frac 12 big)right}$






        share|cite|improve this answer














        begin{array}{c}
        xy = frac 16 \
        x + y = 5xy = frac 56
        end{array}



        Let $x = frac{5}{12} + t$ and $y = frac{5}{12}-t$. Then $x+y=frac 56$.



        begin{align}
        xy &= frac 16 \
        bigg(frac{5}{12} + t bigg) bigg(frac{5}{12} - tbigg) &= frac 16 \
        frac{25}{144} - t^2 &= frac{24}{144} \
        t^2 &= frac{1}{144} \
        t &= pm frac{1}{12} \
        end{align}



        $frac{5}{12} + frac{1}{12} = frac 12$ and $frac{5}{12} - frac{1}{12} = frac 13$



        So $(x,y) = left{big(frac 12,frac 13 big),big(frac 13,frac 12 big)right}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 21:32

























        answered Dec 29 '18 at 10:36









        steven gregorysteven gregory

        17.7k32257




        17.7k32257






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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