How to solve this simple equation $frac{46}{y} + y = 25$?












1












$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22
















1












$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22














1












1








1





$begingroup$


How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus







share|cite|improve this question











$endgroup$




How do I solve this simple equation? $frac{46}{y} + y = 25$



I know that the answer is $2$, but how do I arrive at that?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 13:28









6005

37k751127




37k751127










asked Jan 5 at 13:18









brilliantbrilliant

30739




30739








  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22














  • 1




    $begingroup$
    Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
    $endgroup$
    – Wuestenfux
    Jan 5 at 13:19










  • $begingroup$
    Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
    $endgroup$
    – Thomas Shelby
    Jan 5 at 13:20










  • $begingroup$
    $2$ is not the only solution - the quadratic formula will give you another.
    $endgroup$
    – Mark Bennet
    Jan 5 at 13:21






  • 1




    $begingroup$
    What aobut $y=23$?
    $endgroup$
    – Yanko
    Jan 5 at 13:22








1




1




$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19




$begingroup$
Multiply by $yne 0$ and you get a quadratic equation which can be easily solved.
$endgroup$
– Wuestenfux
Jan 5 at 13:19












$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20




$begingroup$
Multiplying both sides by $y $ will give you a quadratic equation. Then use quadratic formula.
$endgroup$
– Thomas Shelby
Jan 5 at 13:20












$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21




$begingroup$
$2$ is not the only solution - the quadratic formula will give you another.
$endgroup$
– Mark Bennet
Jan 5 at 13:21




1




1




$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22




$begingroup$
What aobut $y=23$?
$endgroup$
– Yanko
Jan 5 at 13:22










4 Answers
4






active

oldest

votes


















3












$begingroup$

$$frac{46}{y}+y = 25$$



Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



$$46+y^2 = 25y$$



$$y^2-25y+46 = 0$$



Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



$$(y+y_1)(y+y_2) = 0$$



The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



$$(y-23)(y-2) = 0$$



Setting either factor equal to $0$ yields




  • $$y-23 = 0 iff y = 23$$

  • $$y-2 = 0 iff y = 2$$


You could also use the Quadratic Formula if desired:




$$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



$$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



$$y = 23; quad y = 2$$







share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $$y^2-25y+46=0implies y=2,23$$
    you can use middle term method or the quadratic formula.
    the formula is...
    if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



      begin{align*}
      left(frac{46}{y} + yright) cdot y &= 25 cdot y \
      46 + y^2 &= 25y
      end{align*}



      Now we bring all the terms over to one side:
      $$
      y^2 - 25y + 46 = 0
      $$
      so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



      If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Multiply both sides by $y$ and you get
        $$46 + y^2 = 25y$$
        $$Leftrightarrow 0 = y^2 -25y + 46$$
        Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
        $$y = frac{25 pm sqrt{21^2}}{2}$$
        So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
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          active

          oldest

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          4 Answers
          4






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          active

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          active

          oldest

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          3












          $begingroup$

          $$frac{46}{y}+y = 25$$



          Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



          $$46+y^2 = 25y$$



          $$y^2-25y+46 = 0$$



          Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



          $$(y+y_1)(y+y_2) = 0$$



          The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



          $$(y-23)(y-2) = 0$$



          Setting either factor equal to $0$ yields




          • $$y-23 = 0 iff y = 23$$

          • $$y-2 = 0 iff y = 2$$


          You could also use the Quadratic Formula if desired:




          $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



          $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



          $$y = 23; quad y = 2$$







          share|cite|improve this answer











          $endgroup$


















            3












            $begingroup$

            $$frac{46}{y}+y = 25$$



            Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



            $$46+y^2 = 25y$$



            $$y^2-25y+46 = 0$$



            Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



            $$(y+y_1)(y+y_2) = 0$$



            The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



            $$(y-23)(y-2) = 0$$



            Setting either factor equal to $0$ yields




            • $$y-23 = 0 iff y = 23$$

            • $$y-2 = 0 iff y = 2$$


            You could also use the Quadratic Formula if desired:




            $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



            $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



            $$y = 23; quad y = 2$$







            share|cite|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              $$frac{46}{y}+y = 25$$



              Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



              $$46+y^2 = 25y$$



              $$y^2-25y+46 = 0$$



              Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



              $$(y+y_1)(y+y_2) = 0$$



              The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



              $$(y-23)(y-2) = 0$$



              Setting either factor equal to $0$ yields




              • $$y-23 = 0 iff y = 23$$

              • $$y-2 = 0 iff y = 2$$


              You could also use the Quadratic Formula if desired:




              $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



              $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



              $$y = 23; quad y = 2$$







              share|cite|improve this answer











              $endgroup$



              $$frac{46}{y}+y = 25$$



              Here, assuming $y neq 0$, you can multiply both sides of the equation by $y$, yielding



              $$46+y^2 = 25y$$



              $$y^2-25y+46 = 0$$



              Here, you can solve the quadratic equation by factoring, completing the square, or the Quadratic Formula, but the first way is the simplest, considering $a = 1$:



              $$(y+y_1)(y+y_2) = 0$$



              The key here is to ask yourself: which two numbers multiply to give $+46$ and add to give $-25$? Clearly the two numbers must be negative, and you can figure out they’re $-23$ and $-2$. So, the factored equation becomes



              $$(y-23)(y-2) = 0$$



              Setting either factor equal to $0$ yields




              • $$y-23 = 0 iff y = 23$$

              • $$y-2 = 0 iff y = 2$$


              You could also use the Quadratic Formula if desired:




              $$ax^2+bx+c = 0 iff x = frac{-bpmsqrt{b^2-4ac}}{2a}$$



              $$y = frac{-(-25)pmsqrt{(-25)^2-4(1)(46)}}{2(1)} = frac{25pmsqrt{441}}{2} = frac{25pm 21}{2}$$



              $$y = 23; quad y = 2$$








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 5 at 13:34

























              answered Jan 5 at 13:27









              KM101KM101

              6,0901525




              6,0901525























                  2












                  $begingroup$

                  $$y^2-25y+46=0implies y=2,23$$
                  you can use middle term method or the quadratic formula.
                  the formula is...
                  if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    $$y^2-25y+46=0implies y=2,23$$
                    you can use middle term method or the quadratic formula.
                    the formula is...
                    if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      $$y^2-25y+46=0implies y=2,23$$
                      you can use middle term method or the quadratic formula.
                      the formula is...
                      if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$






                      share|cite|improve this answer











                      $endgroup$



                      $$y^2-25y+46=0implies y=2,23$$
                      you can use middle term method or the quadratic formula.
                      the formula is...
                      if $$ax^2+bx+c=0,then~~~ x=frac{-bpm sqrt{b^2-4ac}}{2a}$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 5 at 13:30

























                      answered Jan 5 at 13:22









                      Rakibul Islam PrinceRakibul Islam Prince

                      988211




                      988211























                          0












                          $begingroup$

                          Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                          begin{align*}
                          left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                          46 + y^2 &= 25y
                          end{align*}



                          Now we bring all the terms over to one side:
                          $$
                          y^2 - 25y + 46 = 0
                          $$
                          so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                          If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                            begin{align*}
                            left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                            46 + y^2 &= 25y
                            end{align*}



                            Now we bring all the terms over to one side:
                            $$
                            y^2 - 25y + 46 = 0
                            $$
                            so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                            If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                              begin{align*}
                              left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                              46 + y^2 &= 25y
                              end{align*}



                              Now we bring all the terms over to one side:
                              $$
                              y^2 - 25y + 46 = 0
                              $$
                              so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                              If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.






                              share|cite|improve this answer









                              $endgroup$



                              Since you see $frac{46}{y}$ in there, it's a good idea to multiply both sides by $y$:



                              begin{align*}
                              left(frac{46}{y} + yright) cdot y &= 25 cdot y \
                              46 + y^2 &= 25y
                              end{align*}



                              Now we bring all the terms over to one side:
                              $$
                              y^2 - 25y + 46 = 0
                              $$
                              so you have a quadratic equation, and you can solve it. (Quadratic formula or factoring!)



                              If you happen to get $y = 0$ as a solution to the equation, you have to throw it out, since at the beginning there was $frac{46}{y}$, which means that $y$ was not $0$. But here it turns out that $y = 0$ is not a solution.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 5 at 13:27









                              60056005

                              37k751127




                              37k751127























                                  0












                                  $begingroup$

                                  Multiply both sides by $y$ and you get
                                  $$46 + y^2 = 25y$$
                                  $$Leftrightarrow 0 = y^2 -25y + 46$$
                                  Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                  $$y = frac{25 pm sqrt{21^2}}{2}$$
                                  So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Multiply both sides by $y$ and you get
                                    $$46 + y^2 = 25y$$
                                    $$Leftrightarrow 0 = y^2 -25y + 46$$
                                    Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                    $$y = frac{25 pm sqrt{21^2}}{2}$$
                                    So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Multiply both sides by $y$ and you get
                                      $$46 + y^2 = 25y$$
                                      $$Leftrightarrow 0 = y^2 -25y + 46$$
                                      Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                      $$y = frac{25 pm sqrt{21^2}}{2}$$
                                      So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      Multiply both sides by $y$ and you get
                                      $$46 + y^2 = 25y$$
                                      $$Leftrightarrow 0 = y^2 -25y + 46$$
                                      Now calculating the discriminant $D = (-25)^2 -4cdot 46 = 625-184 = 441 = 21^2$ gives
                                      $$y = frac{25 pm sqrt{21^2}}{2}$$
                                      So $y = frac{25+21}{2} = 23$ or $y = frac{25-21}{2} = 2$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 5 at 13:29









                                      Jonas De SchouwerJonas De Schouwer

                                      3748




                                      3748






























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                                          Cabo Verde

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                                            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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                                          Gyllenstierna

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