Finding x and y from two linear equations

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-1
down vote

favorite












I have two linear equations:



$110x + y = 10$



and



$95x + y = 0$



The solution given in the course I have is $x = 2/3$ and $y = -63.33$
No workings are given.



I am not sure how to solve for $x$ and $y$. I tried rearranging the first equation to find $x$:



$x = 10-y/110$



Then do I substitute this for $x$ into the other equation?



$95 (10-y/110) + d = 10$



Not sure if I'm on the right track.




linear-algebra




share|cite|improve this question





















  • What is $d$? Is that $y$?
    – Dave
    2 days ago










  • Sorry, yes $d$ is actually meant to be $y$.
    – axiom111
    2 days ago










  • I have now corrected this.
    – axiom111
    2 days ago














up vote
-1
down vote

favorite












I have two linear equations:



$110x + y = 10$



and



$95x + y = 0$



The solution given in the course I have is $x = 2/3$ and $y = -63.33$
No workings are given.



I am not sure how to solve for $x$ and $y$. I tried rearranging the first equation to find $x$:



$x = 10-y/110$



Then do I substitute this for $x$ into the other equation?



$95 (10-y/110) + d = 10$



Not sure if I'm on the right track.




linear-algebra




share|cite|improve this question





















  • What is $d$? Is that $y$?
    – Dave
    2 days ago










  • Sorry, yes $d$ is actually meant to be $y$.
    – axiom111
    2 days ago










  • I have now corrected this.
    – axiom111
    2 days ago












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











I have two linear equations:



$110x + y = 10$



and



$95x + y = 0$



The solution given in the course I have is $x = 2/3$ and $y = -63.33$
No workings are given.



I am not sure how to solve for $x$ and $y$. I tried rearranging the first equation to find $x$:



$x = 10-y/110$



Then do I substitute this for $x$ into the other equation?



$95 (10-y/110) + d = 10$



Not sure if I'm on the right track.




linear-algebra




share|cite|improve this question













I have two linear equations:



$110x + y = 10$



and



$95x + y = 0$



The solution given in the course I have is $x = 2/3$ and $y = -63.33$
No workings are given.



I am not sure how to solve for $x$ and $y$. I tried rearranging the first equation to find $x$:



$x = 10-y/110$



Then do I substitute this for $x$ into the other equation?



$95 (10-y/110) + d = 10$



Not sure if I'm on the right track.





linear-algebra





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked 2 days ago









axiom111

93




93











  • What is $d$? Is that $y$?
    – Dave
    2 days ago










  • Sorry, yes $d$ is actually meant to be $y$.
    – axiom111
    2 days ago










  • I have now corrected this.
    – axiom111
    2 days ago
















  • What is $d$? Is that $y$?
    – Dave
    2 days ago










  • Sorry, yes $d$ is actually meant to be $y$.
    – axiom111
    2 days ago










  • I have now corrected this.
    – axiom111
    2 days ago















What is $d$? Is that $y$?
– Dave
2 days ago




What is $d$? Is that $y$?
– Dave
2 days ago












Sorry, yes $d$ is actually meant to be $y$.
– axiom111
2 days ago




Sorry, yes $d$ is actually meant to be $y$.
– axiom111
2 days ago












I have now corrected this.
– axiom111
2 days ago




I have now corrected this.
– axiom111
2 days ago










3 Answers
3






active

oldest

votes

















up vote
1
down vote













Subtract second from first you get:
$$15x = 10$$
which means $x = frac23$
So use any equation of the two to find $y$ as
$$y = -95x = -95frac23 = -frac1903$$






share|cite|improve this answer




























    up vote
    0
    down vote













    We have $$15x=10$$ or
    $$x=frac23.$$
    Thus, $$y=-frac95cdot23.$$






    share|cite|improve this answer




























      up vote
      0
      down vote













      Don't rush your steps.



      When you solve for $x$, you have:
      $$110x = 10 - y$$
      $$Rightarrow x = frac10colorred110 - fracy110$$
      $$Rightarrow x = frac111 - fracy110$$



      Substituting into $95x + y = 0$ we have:



      $$95 left(frac111 - fracy110 right)+y = colorred0$$
      $$Rightarrow frac9511 - frac95110y + y = 0$$
      $$Rightarrow frac9511 - frac95110y + frac110110y = 0$$
      $$Rightarrow - frac95110y + frac110110y = 0 - frac9511$$
      $$Rightarrow frac15110y = -frac9511$$
      $$Rightarrow y = -frac9511 cdot frac11015$$
      $$= -frac191 cdot frac103$$
      $$= -frac1903$$



      Now you can substitute $y$ into any of the two equations to find $x$.



      Note: Of course, eliminating $y$ from both equations is much simpler.






      share|cite|improve this answer



















      • 1




        @Moo I've edited my answer.
        – Toby Mak
        2 days ago










      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );








       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871731%2ffinding-x-and-y-from-two-linear-equations%23new-answer', 'question_page');

      );

      Post as a guest

















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote













      Subtract second from first you get:
      $$15x = 10$$
      which means $x = frac23$
      So use any equation of the two to find $y$ as
      $$y = -95x = -95frac23 = -frac1903$$






      share|cite|improve this answer

























        up vote
        1
        down vote













        Subtract second from first you get:
        $$15x = 10$$
        which means $x = frac23$
        So use any equation of the two to find $y$ as
        $$y = -95x = -95frac23 = -frac1903$$






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Subtract second from first you get:
          $$15x = 10$$
          which means $x = frac23$
          So use any equation of the two to find $y$ as
          $$y = -95x = -95frac23 = -frac1903$$






          share|cite|improve this answer













          Subtract second from first you get:
          $$15x = 10$$
          which means $x = frac23$
          So use any equation of the two to find $y$ as
          $$y = -95x = -95frac23 = -frac1903$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 2 days ago









          Ahmad Bazzi

          2,162417




          2,162417




















              up vote
              0
              down vote













              We have $$15x=10$$ or
              $$x=frac23.$$
              Thus, $$y=-frac95cdot23.$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                We have $$15x=10$$ or
                $$x=frac23.$$
                Thus, $$y=-frac95cdot23.$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  We have $$15x=10$$ or
                  $$x=frac23.$$
                  Thus, $$y=-frac95cdot23.$$






                  share|cite|improve this answer













                  We have $$15x=10$$ or
                  $$x=frac23.$$
                  Thus, $$y=-frac95cdot23.$$







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 2 days ago









                  Michael Rozenberg

                  86.9k1576178




                  86.9k1576178




















                      up vote
                      0
                      down vote













                      Don't rush your steps.



                      When you solve for $x$, you have:
                      $$110x = 10 - y$$
                      $$Rightarrow x = frac10colorred110 - fracy110$$
                      $$Rightarrow x = frac111 - fracy110$$



                      Substituting into $95x + y = 0$ we have:



                      $$95 left(frac111 - fracy110 right)+y = colorred0$$
                      $$Rightarrow frac9511 - frac95110y + y = 0$$
                      $$Rightarrow frac9511 - frac95110y + frac110110y = 0$$
                      $$Rightarrow - frac95110y + frac110110y = 0 - frac9511$$
                      $$Rightarrow frac15110y = -frac9511$$
                      $$Rightarrow y = -frac9511 cdot frac11015$$
                      $$= -frac191 cdot frac103$$
                      $$= -frac1903$$



                      Now you can substitute $y$ into any of the two equations to find $x$.



                      Note: Of course, eliminating $y$ from both equations is much simpler.






                      share|cite|improve this answer



















                      • 1




                        @Moo I've edited my answer.
                        – Toby Mak
                        2 days ago














                      up vote
                      0
                      down vote













                      Don't rush your steps.



                      When you solve for $x$, you have:
                      $$110x = 10 - y$$
                      $$Rightarrow x = frac10colorred110 - fracy110$$
                      $$Rightarrow x = frac111 - fracy110$$



                      Substituting into $95x + y = 0$ we have:



                      $$95 left(frac111 - fracy110 right)+y = colorred0$$
                      $$Rightarrow frac9511 - frac95110y + y = 0$$
                      $$Rightarrow frac9511 - frac95110y + frac110110y = 0$$
                      $$Rightarrow - frac95110y + frac110110y = 0 - frac9511$$
                      $$Rightarrow frac15110y = -frac9511$$
                      $$Rightarrow y = -frac9511 cdot frac11015$$
                      $$= -frac191 cdot frac103$$
                      $$= -frac1903$$



                      Now you can substitute $y$ into any of the two equations to find $x$.



                      Note: Of course, eliminating $y$ from both equations is much simpler.






                      share|cite|improve this answer



















                      • 1




                        @Moo I've edited my answer.
                        – Toby Mak
                        2 days ago












                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      Don't rush your steps.



                      When you solve for $x$, you have:
                      $$110x = 10 - y$$
                      $$Rightarrow x = frac10colorred110 - fracy110$$
                      $$Rightarrow x = frac111 - fracy110$$



                      Substituting into $95x + y = 0$ we have:



                      $$95 left(frac111 - fracy110 right)+y = colorred0$$
                      $$Rightarrow frac9511 - frac95110y + y = 0$$
                      $$Rightarrow frac9511 - frac95110y + frac110110y = 0$$
                      $$Rightarrow - frac95110y + frac110110y = 0 - frac9511$$
                      $$Rightarrow frac15110y = -frac9511$$
                      $$Rightarrow y = -frac9511 cdot frac11015$$
                      $$= -frac191 cdot frac103$$
                      $$= -frac1903$$



                      Now you can substitute $y$ into any of the two equations to find $x$.



                      Note: Of course, eliminating $y$ from both equations is much simpler.






                      share|cite|improve this answer















                      Don't rush your steps.



                      When you solve for $x$, you have:
                      $$110x = 10 - y$$
                      $$Rightarrow x = frac10colorred110 - fracy110$$
                      $$Rightarrow x = frac111 - fracy110$$



                      Substituting into $95x + y = 0$ we have:



                      $$95 left(frac111 - fracy110 right)+y = colorred0$$
                      $$Rightarrow frac9511 - frac95110y + y = 0$$
                      $$Rightarrow frac9511 - frac95110y + frac110110y = 0$$
                      $$Rightarrow - frac95110y + frac110110y = 0 - frac9511$$
                      $$Rightarrow frac15110y = -frac9511$$
                      $$Rightarrow y = -frac9511 cdot frac11015$$
                      $$= -frac191 cdot frac103$$
                      $$= -frac1903$$



                      Now you can substitute $y$ into any of the two equations to find $x$.



                      Note: Of course, eliminating $y$ from both equations is much simpler.







                      share|cite|improve this answer















                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 days ago


























                      answered 2 days ago









                      Toby Mak

                      2,4301922




                      2,4301922







                      • 1




                        @Moo I've edited my answer.
                        – Toby Mak
                        2 days ago












                      • 1




                        @Moo I've edited my answer.
                        – Toby Mak
                        2 days ago







                      1




                      1




                      @Moo I've edited my answer.
                      – Toby Mak
                      2 days ago




                      @Moo I've edited my answer.
                      – Toby Mak
                      2 days ago












                       

                      draft saved


                      draft discarded


























                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871731%2ffinding-x-and-y-from-two-linear-equations%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more