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Find all values \(a\) for which there exists an ordered pair \((a, b)\) satisfying the following system of equations:

\(\begin{align*}a + ab^2 & = 40b, \\a - ab^2 & = -32b.\end{align*}\)

 Jul 20, 2019
 #1
avatar+8756 
+5

a + ab2   =   40b

a - ab2   =   -32b

 

The purple values are equal to each other and the blue values are equal to each other.

 

a - ab2   =   -32b                     Add  40b  to both sides of the equation.

 

a - ab2 + 40b   =   -32b + 40b        Since  a + ab2  =  40b   we can substitute  a + ab2  in for  40b

 

a - ab2 + a + ab2   =   -32b + 40b      The elimination method is really just like substitution  smiley

 

a - ab2 + a + ab2   =   -32b + 40b      Simplify both sides by combining like terms.

 

2a   =   8b          Divide both sides of the equation by  8

 

\(\frac14\)a  =  b

 

Now we can substitute this value for  b  into one of the original equations.

 

a + ab2  =  40b

                                    Substitute   \(\frac14\)a   in for   b

a + a(\(\frac14\)a)2  =  40(\(\frac14\)a)

                                    Simplify both sides of the equation.

a + \(\frac{1}{16}\)a3   =   10a

                                    Multiply through by  16

16a + a3  =  160a

                                    Subtract  16a  from both sides and subtract  a3  from both sides

0  =  144a - a3

                                    Factor  a  out of both terms on the right side

0  =  a( 144 - a2 )

                                           Factor   144 - a2   as a difference of squares

0  =  a( 12 - a )( 12 + a )

                                           Set each factor equal to  0  and solve for  a

0  =  a___ or ___12 - a  =  0___ or ___12 + a  =  0

 

 

a  =  0 a  =  12 a  =  -12 
 Jul 20, 2019
 #3
avatar+1544 
+3

Hectitar is correct

CalculatorUser  Jul 20, 2019
 #2
avatar+1544 
+4

EDIT: The following is wrong, but why?

 

1. Divide both sides by b

 

\(\frac{a}{b}+ab=40\)

 

\(\frac{a}{b}-ab=-32\)

 

2. Substitute out a/b so we get:
 

\(40-ab=-32+ab\)

 

3. Solving for ab we get

 

\(ab=36\)

 

So the values of a can be \((-36, -18, -12 -9, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 9, 12, 18, 36)\)

 

 

 

Assuming if the problem asks for integer solutions

 Jul 20, 2019
edited by CalculatorUser  Jul 20, 2019
 #4
avatar+8756 
+5

Here is something taken from what I wrote in my notes on my phone a long time ago... smileylaughlaugh

 

----------

 

a + b  =  10a + b   Divide both sides by  a

 

a/a + b/a   =   10a/a + b/a

 

1 + b/a  =  10 + b/a    Subtract  b/a  from both sides

 

1  =  10

 

So you'd be like, "NO SOLUTION!!!" right??? WRONG!! because  a  =  0  IS a solution to the original equation, but when you divided both sides by  a  it made it look like there was no solution. So that is why you gotta always say   a ≠ 0  whenever you divide by an unknown number. So then the final answer would be basically saying "No solution when  a ≠  0.   So  a = 0  might be a solution might not be I dunno." Then you gotta check if  a = 0  is a solution.

 

----------

 

So that is why   0  is missing from your list.  laugh

 

When  you divided both sides by  b,  you didn't note that the following equation holds only when  b ≠ 0

 

Otherwise, it is true that  ab = 36

 

There are three solutions to this system of equations (see here) and in both cases where  b ≠ 0,  ab = 36

 

But  ab = 36  is not the full story! There are more restrictions...

 

Notice that  (1, 36)  can't be a solution for  (a, b)  because

 

a + ab2 must equal  40b       but       1 + 362  ≠  40(36)

 

So  (1, 36)  can't be a solution.

 

There are more restrictions on  a  and  b  besides just  ab = 36  ( when  b ≠ 0 )  smiley

hectictar  Jul 20, 2019

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