system of eqns, 2ax+6y=5, 4x+3ay=b, has more than 1 soln; find value of a+b for a,b>0

[MethMath11]

1. The following system of equations has more than one solution, (x, y):

. . . . .[math]2ax\, +\, 6y\, =\, 5[/math]
. . . . .[math]4x\, +\, 3ay\, =\, 6[/math]
The value of a + b, where a, b >0, is....

(A) 3. . . . .(B) 5. . . . .(C) 7. . . . .(D) 14. . . . .(E) 24

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Please do help, english isnt my main language and i do not know how to solve it

 

[Deleted member 4993]

1. The following system of equations has more than one solution, (x, y):

. . . . .[math]2ax\, +\, 6y\, =\, 5[/math]
. . . . .[math]4x\, +\, 3ay\, =\, 6[/math]
The value of a + b, where a, b >0, is....

(A) 3. . . . .(B) 5. . . . .(C) 7. . . . .(D) 14. . . . .(E) 24

---

Please do help, english isnt my main language and i do not know how to solve it

There are several ways to solve this problem.

What method have you been taught in your class? Please share your work following one of those methods.

 

[HallsofIvy]

The first thing I would do it try to solve it an see what happens!
The equations are
2ax+ 6y= 5 and
4x+ 3ay= b.

Multiply the first equation by 2 to get 4ax+ 12y= 10.
Multiply the second equation by a to get 4ax+ 3a^2y= ab.

Subtract: (3a^2- 12)y= ab- 10.

Now, there are three possibilities:
1) If 3a^2- 12 is not 0 we can divide by it to get a unique solution.
2) if 3a^2- 12= 0 but ab- 10 is not, we have 0 equal to a non-zero number which is not true. There is NO solution.
3) if 3a^2- 12= 0 and ab- 10= 0 we have 0= 0 which is true for any value of y- there is more than one solution.

So we must have 3a^2- 12= 0, 3a^2= 12, a^2= 4, a= 2 or a= -2.
We must also have ab- 10= 0, ab= 10. If a= 2, b= 5. If a= -2, b= -5.
Since the problem says "a,b> 0" we must have a= 2, b= 5 so a+ b= 7.

 

[MethMath11]

The first thing I would do it try to solve it an see what happens!
The equations are
2ax+ 6y= 5 and
4x+ 3ay= b.

Multiply the first equation by 2 to get 4ax+ 12y= 10.
Multiply the second equation by a to get 4ax+ 3a^2y= ab.

Subtract: (3a^2- 12)y= ab- 10.

Now, there are three possibilities:
1) If 3a^2- 12 is not 0 we can divide by it to get a unique solution.
2) if 3a^2- 12= 0 but ab- 10 is not, we have 0 equal to a non-zero number which is not true. There is NO solution.
3) if 3a^2- 12= 0 and ab- 10= 0 we have 0= 0 which is true for any value of y- there is more than one solution.

So we must have 3a^2- 12= 0, 3a^2= 12, a^2= 4, a= 2 or a= -2.
We must also have ab- 10= 0, ab= 10. If a= 2, b= 5. If a= -2, b= -5.
Since the problem says "a,b> 0" we must have a= 2, b= 5 so a+ b= 7.

 

[MethMath11]

Thank you , you helped alot. but i still dont get it why you eliminate the y (probbaly due to the bad translation)

 

[HallsofIvy]

We have the equation Ay= B.
If A is not 0, divide both sides by A to get the unique solution y= B/A.
If A is 0 and B is not, "Ay= B" becomes "0= B" which is not true for any y. There is no solution.
If both A and B are 0, "Ay= B" becomes "0= 0" which is true for any y. There are infinitely many solutions.

In this problem we are told "The following system of equations has more than one solution (x, y)" so the third must be true. Once I had reduced the system of equations to "(3a^2- 12)y= ab- 10" I knew I must have both 3a^2- 12= 0 and ab- 10= 0.

 

[Deleted member 4993]

Thank you , you helped a lot. but i still don't get it why you eliminate the y (probably due to the bad translation)

You may choose to eliminate 'x' instead - you should end up with same conclusion.

 

[HallsofIvy]

You may choose to eliminate 'x' instead - you should end up with same conclusion.

I started to answer that way but I don't think that was what MethMath11 was referring to. I think he was referring to when I went from "(3a^2- 12)y= ab- 10" to "3a^2- 12= 0" and "ab- 10= 0".

(MethMath11, perhaps it would help if you stopped doing "Meth" and "Math" at the same time!)

 

[JeffM]

Thank you , you helped alot. but i still dont get it why you eliminate the y (probbaly due to the bad translation)

Halls did NOT eliminate the y. He SOLVED for y in a particular case. In that particular case, he got a unique solution for x and y, but the problem stipulates that there is not a unique solution. What can you conclude from that contradiction?

 

[Steven G]

I started to answer that way but I don't think that was what MethMath11 was referring to. I think he was referring to when I went from "(3a^2- 12)y= ab- 10" to "3a^2- 12= 0" and "ab- 10= 0".

(MethMath11, perhaps it would help if you stopped doing "Meth" and "Math" at the same time!)

No, 1st he does Meth, then he does Math which is a deadly combination.

 

[topsquark]

Actually I do better at Quantum Mechanics when I have a good buzz on.

-Dan