Find $ad - bc$ if $ac = 18$ and $bd = 50$












0












$begingroup$


I have a math problem to solve that goes as follows:



--



$(ax + by)(cx - dy)$



In the expression above, $a$, $b$, $c$, and $d$ are non-zero constants and $ad = bc$. If $ac = 18$ and $bd = 50$, what is the value of the coefficient of the $xy$ term when the expression is expanded and like terms are collected?



--



My work is below:



$(ax + by)(cx - dy)$ = $acx^2 - adxy + bcxy - bdy^2$



$xy$ term: $bcxy - adxy$



Therefore, all I need to find is $bc - ad$, right? I did this:



$bc - ad = 0$



$bc - (frac{18}{c})(frac{50}{b}) = 0$



$bc - frac{900}{bc} = 0$



$b^2c^2 - 900 = 0$



$b^2c^2 = 900$



$bc = 30$



I did the same with $ad$, and got the result $ad = 30$ as well.
That would mean $bc - ad = 0$. It doesn't feel right, but my answer key does confirm the answer of $0$. If I use numbers other than $18$ and $50$, the answer still seems to come out to $0$. How is this possible?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
    $endgroup$
    – Jacky Chong
    Oct 18 '16 at 22:38










  • $begingroup$
    One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
    $endgroup$
    – Callus
    Oct 18 '16 at 22:39






  • 1




    $begingroup$
    I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
    $endgroup$
    – la flaca
    Oct 18 '16 at 22:40










  • $begingroup$
    I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
    $endgroup$
    – tarunbod
    Nov 24 '16 at 23:36
















0












$begingroup$


I have a math problem to solve that goes as follows:



--



$(ax + by)(cx - dy)$



In the expression above, $a$, $b$, $c$, and $d$ are non-zero constants and $ad = bc$. If $ac = 18$ and $bd = 50$, what is the value of the coefficient of the $xy$ term when the expression is expanded and like terms are collected?



--



My work is below:



$(ax + by)(cx - dy)$ = $acx^2 - adxy + bcxy - bdy^2$



$xy$ term: $bcxy - adxy$



Therefore, all I need to find is $bc - ad$, right? I did this:



$bc - ad = 0$



$bc - (frac{18}{c})(frac{50}{b}) = 0$



$bc - frac{900}{bc} = 0$



$b^2c^2 - 900 = 0$



$b^2c^2 = 900$



$bc = 30$



I did the same with $ad$, and got the result $ad = 30$ as well.
That would mean $bc - ad = 0$. It doesn't feel right, but my answer key does confirm the answer of $0$. If I use numbers other than $18$ and $50$, the answer still seems to come out to $0$. How is this possible?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
    $endgroup$
    – Jacky Chong
    Oct 18 '16 at 22:38










  • $begingroup$
    One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
    $endgroup$
    – Callus
    Oct 18 '16 at 22:39






  • 1




    $begingroup$
    I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
    $endgroup$
    – la flaca
    Oct 18 '16 at 22:40










  • $begingroup$
    I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
    $endgroup$
    – tarunbod
    Nov 24 '16 at 23:36














0












0








0





$begingroup$


I have a math problem to solve that goes as follows:



--



$(ax + by)(cx - dy)$



In the expression above, $a$, $b$, $c$, and $d$ are non-zero constants and $ad = bc$. If $ac = 18$ and $bd = 50$, what is the value of the coefficient of the $xy$ term when the expression is expanded and like terms are collected?



--



My work is below:



$(ax + by)(cx - dy)$ = $acx^2 - adxy + bcxy - bdy^2$



$xy$ term: $bcxy - adxy$



Therefore, all I need to find is $bc - ad$, right? I did this:



$bc - ad = 0$



$bc - (frac{18}{c})(frac{50}{b}) = 0$



$bc - frac{900}{bc} = 0$



$b^2c^2 - 900 = 0$



$b^2c^2 = 900$



$bc = 30$



I did the same with $ad$, and got the result $ad = 30$ as well.
That would mean $bc - ad = 0$. It doesn't feel right, but my answer key does confirm the answer of $0$. If I use numbers other than $18$ and $50$, the answer still seems to come out to $0$. How is this possible?






algebra-precalculus







share|cite|improve this question











$endgroup$




I have a math problem to solve that goes as follows:



--



$(ax + by)(cx - dy)$



In the expression above, $a$, $b$, $c$, and $d$ are non-zero constants and $ad = bc$. If $ac = 18$ and $bd = 50$, what is the value of the coefficient of the $xy$ term when the expression is expanded and like terms are collected?



--



My work is below:



$(ax + by)(cx - dy)$ = $acx^2 - adxy + bcxy - bdy^2$



$xy$ term: $bcxy - adxy$



Therefore, all I need to find is $bc - ad$, right? I did this:



$bc - ad = 0$



$bc - (frac{18}{c})(frac{50}{b}) = 0$



$bc - frac{900}{bc} = 0$



$b^2c^2 - 900 = 0$



$b^2c^2 = 900$



$bc = 30$



I did the same with $ad$, and got the result $ad = 30$ as well.
That would mean $bc - ad = 0$. It doesn't feel right, but my answer key does confirm the answer of $0$. If I use numbers other than $18$ and $50$, the answer still seems to come out to $0$. How is this possible?






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 Jul 19 '18 at 18:11









RayDansh

841215




841215










asked Oct 18 '16 at 22:33









tarunbodtarunbod

111




111












  • $begingroup$
    You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
    $endgroup$
    – Jacky Chong
    Oct 18 '16 at 22:38










  • $begingroup$
    One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
    $endgroup$
    – Callus
    Oct 18 '16 at 22:39






  • 1




    $begingroup$
    I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
    $endgroup$
    – la flaca
    Oct 18 '16 at 22:40










  • $begingroup$
    I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
    $endgroup$
    – tarunbod
    Nov 24 '16 at 23:36


















  • $begingroup$
    You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
    $endgroup$
    – Jacky Chong
    Oct 18 '16 at 22:38










  • $begingroup$
    One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
    $endgroup$
    – Callus
    Oct 18 '16 at 22:39






  • 1




    $begingroup$
    I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
    $endgroup$
    – la flaca
    Oct 18 '16 at 22:40










  • $begingroup$
    I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
    $endgroup$
    – tarunbod
    Nov 24 '16 at 23:36
















$begingroup$
You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
$endgroup$
– Jacky Chong
Oct 18 '16 at 22:38




$begingroup$
You question is unclear. If you assume $ad=bc$ then of course $ad-bc=0$.
$endgroup$
– Jacky Chong
Oct 18 '16 at 22:38












$begingroup$
One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
$endgroup$
– Callus
Oct 18 '16 at 22:39




$begingroup$
One of the things you wrote in your derivation is $bc-ad=0$. This follows immediately from the assumption that $ad=bc$
$endgroup$
– Callus
Oct 18 '16 at 22:39




1




1




$begingroup$
I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
$endgroup$
– la flaca
Oct 18 '16 at 22:40




$begingroup$
I'm not sure I understand what the task is, so correct me if I'm wrong. You want to find the coefficient of the $xy$ term when the expresion $(ax+by)(cx-dy)$is expanded and you foun that coefficient is $bc-ad$, but you already know $bc=ad$, so what is the problem?
$endgroup$
– la flaca
Oct 18 '16 at 22:40












$begingroup$
I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
$endgroup$
– tarunbod
Nov 24 '16 at 23:36




$begingroup$
I'm so sorry guys, I didn't see that the question gave that ad = bc. I don't know what happened to me when I was solving this problem.
$endgroup$
– tarunbod
Nov 24 '16 at 23:36










2 Answers
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$begingroup$

Your answer is $0$. Since you know $bc = ad$ and you also know that $xy = bc - ad$. Then it must be $0$. Since you're basically subtracting a number with itself.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    I was working on the same problem. If you assume that $x$ and $y$ are both $1$, it really clears things up and it becomes obvious that the coefficient is $bc - ad$, and $bc = ad$, so like @dennis21469 said, you are subtracting a number from it's self. The $18$ and $50$ are just there to confuse you.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
      2






      active

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      0












      $begingroup$

      Your answer is $0$. Since you know $bc = ad$ and you also know that $xy = bc - ad$. Then it must be $0$. Since you're basically subtracting a number with itself.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Your answer is $0$. Since you know $bc = ad$ and you also know that $xy = bc - ad$. Then it must be $0$. Since you're basically subtracting a number with itself.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Your answer is $0$. Since you know $bc = ad$ and you also know that $xy = bc - ad$. Then it must be $0$. Since you're basically subtracting a number with itself.






          share|cite|improve this answer











          $endgroup$



          Your answer is $0$. Since you know $bc = ad$ and you also know that $xy = bc - ad$. Then it must be $0$. Since you're basically subtracting a number with itself.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 19 '18 at 17:52









          RayDansh

          841215




          841215










          answered Oct 19 '16 at 2:37









          Dennis21468Dennis21468

          1




          1























              0












              $begingroup$

              I was working on the same problem. If you assume that $x$ and $y$ are both $1$, it really clears things up and it becomes obvious that the coefficient is $bc - ad$, and $bc = ad$, so like @dennis21469 said, you are subtracting a number from it's self. The $18$ and $50$ are just there to confuse you.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I was working on the same problem. If you assume that $x$ and $y$ are both $1$, it really clears things up and it becomes obvious that the coefficient is $bc - ad$, and $bc = ad$, so like @dennis21469 said, you are subtracting a number from it's self. The $18$ and $50$ are just there to confuse you.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I was working on the same problem. If you assume that $x$ and $y$ are both $1$, it really clears things up and it becomes obvious that the coefficient is $bc - ad$, and $bc = ad$, so like @dennis21469 said, you are subtracting a number from it's self. The $18$ and $50$ are just there to confuse you.






                  share|cite|improve this answer











                  $endgroup$



                  I was working on the same problem. If you assume that $x$ and $y$ are both $1$, it really clears things up and it becomes obvious that the coefficient is $bc - ad$, and $bc = ad$, so like @dennis21469 said, you are subtracting a number from it's self. The $18$ and $50$ are just there to confuse you.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 19 '18 at 18:11









                  RayDansh

                  841215




                  841215










                  answered Oct 19 '16 at 3:51









                  BatmanBatman

                  1




                  1






























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