Algebraic Manipulation with Simultaneous Equations

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Find $a+b+c$, given that $x+yneq -1$ and beginalign*
ax+by+c&=x+7,\
a+bx+cy&=2x+6y,\
ay+b+cx&=4x+y.
endalign*




linear-algebra systems-of-equations




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  • What have you done?
    – saulspatz
    Jul 26 at 3:10










  • Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
    – dxiv
    Jul 26 at 3:17















up vote
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down vote

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Find $a+b+c$, given that $x+yneq -1$ and beginalign*
ax+by+c&=x+7,\
a+bx+cy&=2x+6y,\
ay+b+cx&=4x+y.
endalign*




linear-algebra systems-of-equations




share|cite|improve this question





















  • What have you done?
    – saulspatz
    Jul 26 at 3:10










  • Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
    – dxiv
    Jul 26 at 3:17













up vote
0
down vote

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1









up vote
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Find $a+b+c$, given that $x+yneq -1$ and beginalign*
ax+by+c&=x+7,\
a+bx+cy&=2x+6y,\
ay+b+cx&=4x+y.
endalign*




linear-algebra systems-of-equations




share|cite|improve this question













Find $a+b+c$, given that $x+yneq -1$ and beginalign*
ax+by+c&=x+7,\
a+bx+cy&=2x+6y,\
ay+b+cx&=4x+y.
endalign*





linear-algebra systems-of-equations





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share|cite|improve this question




share|cite|improve this question








edited Jul 30 at 12:49









Harry Peter

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5,45311438









asked Jul 26 at 3:03









Vivek

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  • What have you done?
    – saulspatz
    Jul 26 at 3:10










  • Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
    – dxiv
    Jul 26 at 3:17

















  • What have you done?
    – saulspatz
    Jul 26 at 3:10










  • Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
    – dxiv
    Jul 26 at 3:17
















What have you done?
– saulspatz
Jul 26 at 3:10




What have you done?
– saulspatz
Jul 26 at 3:10












Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
– dxiv
Jul 26 at 3:17





Random hint: $;1+2+4=6+1=7,$. But really, you should add some context, and show what you have tried. See How to ask a good question.
– dxiv
Jul 26 at 3:17











1 Answer
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Add the three equations altogether to get:



$(ax+by+c)+(a+bx+cy)+(ay+b+cx)=(ax+bx+cx)+(ay+by+cy)+(a+b+c)= (a+b+c)(x+y+1)=7x+7y+7$



Since $x+y$ is not equal to $-1$, we can divide both sides of the equation by $x+y+1$ to get $a+b+c=(7x+7y+7)/(x+y+1)=7$ which gives us what we wanted.






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    Add the three equations altogether to get:



    $(ax+by+c)+(a+bx+cy)+(ay+b+cx)=(ax+bx+cx)+(ay+by+cy)+(a+b+c)= (a+b+c)(x+y+1)=7x+7y+7$



    Since $x+y$ is not equal to $-1$, we can divide both sides of the equation by $x+y+1$ to get $a+b+c=(7x+7y+7)/(x+y+1)=7$ which gives us what we wanted.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      Add the three equations altogether to get:



      $(ax+by+c)+(a+bx+cy)+(ay+b+cx)=(ax+bx+cx)+(ay+by+cy)+(a+b+c)= (a+b+c)(x+y+1)=7x+7y+7$



      Since $x+y$ is not equal to $-1$, we can divide both sides of the equation by $x+y+1$ to get $a+b+c=(7x+7y+7)/(x+y+1)=7$ which gives us what we wanted.






      share|cite|improve this answer























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Add the three equations altogether to get:



        $(ax+by+c)+(a+bx+cy)+(ay+b+cx)=(ax+bx+cx)+(ay+by+cy)+(a+b+c)= (a+b+c)(x+y+1)=7x+7y+7$



        Since $x+y$ is not equal to $-1$, we can divide both sides of the equation by $x+y+1$ to get $a+b+c=(7x+7y+7)/(x+y+1)=7$ which gives us what we wanted.






        share|cite|improve this answer













        Add the three equations altogether to get:



        $(ax+by+c)+(a+bx+cy)+(ay+b+cx)=(ax+bx+cx)+(ay+by+cy)+(a+b+c)= (a+b+c)(x+y+1)=7x+7y+7$



        Since $x+y$ is not equal to $-1$, we can divide both sides of the equation by $x+y+1$ to get $a+b+c=(7x+7y+7)/(x+y+1)=7$ which gives us what we wanted.







        share|cite|improve this answer













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        share|cite|improve this answer











        answered Jul 26 at 3:16









        user573497

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