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avatar+42 

For what values of $y$ does the following equation have no solutions for $x$ ?  

 

$$xy + 5y + 4x + 20=1.$$

 Mar 27, 2020
 #1
avatar+111330 
+2

Not totally sure about this  but....here's my best attempt

 

xy    +  5y  + 4x +  20   = 1      rearrange as   

 

xy   + 4x     =   19 - 5y

 

x [ y + 4]  =  [  19  -  5y  ]        divide both sides  by  y + 4

 

x  =   [  19 - 5y ]  /   [  y +  4 ]

 

Note that if    y   =  -4  , then x is undefined  because  we will be dividing the right side  by  0

 

So.....there  are  no solutions for   x  if y   =  -4

 

 

cool cool cool

 Mar 27, 2020
 #2
avatar+42 
+1

Thanks

Tervvy  Mar 27, 2020
 #3
avatar+111330 
+1

Here's  the  proof

 

x(-4)  + 5(-4) + 4x  + 20  = 1

 

-4x   - 20  + 4x  + 20  =  1

 

0  =  1

 

Impossible

 

So if y -4, we  have a contradiction which no "x"will resolve...!!!!

 

cool cool cool

 Mar 27, 2020

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