confusing system of equation question

m1a3

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i tried to answer this question but i dont know if i did was right, and im still really confused, could someone help meCONFUSING question.jpgattempted.jpg
 
\(\displaystyle \mbox{1. The following system of equations has more than}\)
\(\displaystyle \mbox{one solution }\, (x,\, y).\)

. . . .\(\displaystyle 2ax\, +\, 6y\, =\, 5\)
. . . .\(\displaystyle 4x\, +\, 3ay\, =\, b\)

\(\displaystyle \mbox{The value of }a\, +\, b\mbox{ where }a,\, b\, >\, 0,\, \mbox{ is....}\)
i tried to answer this question but i dont know if i did was right, and im still really confused, could someone help me
Your small and sideways work is too hard for me to follow. Sorry.

Starting fresh: The only way to have "more than one solution" is to have infinitely-many solutions. In such a situation, the equations are the same, other than some multiple; in particular, one can make them the same via clever multiplications.

So multiply each of the rows in order to make at least one of the terms (say, the first one in each) identical. Then you know that the other terms must be equal. Set them equal, and solve for the values of a and b. Pick the pair of values that are positive, and add.

I do get one of the listed answers. ;)
 
i dont even know if what im doing right now is even close to it beign right...
what.jpg
could someone please show me how to work this out, a solution to this answer maybe. Also what should i be looking at to understand this question if my attempt is not even close. :\
 

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[FONT=MathJax_Main]1. The following system of equations has more than[/FONT]
[FONT=MathJax_Main]one solution [/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main].[/FONT]

. . . .[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]5[/FONT]
. . . .[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]b[/FONT]

[FONT=MathJax_Main]The value of [/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main] where [/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Main] is....[/FONT]

I would rewrite those equations as:

\(\displaystyle \displaystyle \frac{2a}{5}x \ + \ \frac{6}{5}y \ = \ 1\)

and

\(\displaystyle \displaystyle \frac{4}{b}x \ + \ \frac{3a}{b}y \ = \ 1\)

Now equate the coefficients of 'x' of the two equations.

do similarly for the coefficients of 'y'.

Now solve....
 
sorry i dont understand any of this now :\, could you just please tell me what the answer is? on my first attempt at the very top ^, i managed to get 5 (B)... is this correct? if not which is the answer?
 
sorry i dont understand any of this now :\, could you just please tell me what the answer is? on my first attempt at the very top ^, i managed to get 5 (B)... is this correct? if not which is the answer?

That is not an acceptable answer!

If that is true, you need to first take English class - so that you can understand instructions.

And your answer (as I understand) a + b = 5 is not correct.
 
could someone please show me how to work this out
Try following the specific step-by-step instructions I'd provided you earlier. In particular, instead of creating some sort of system of equations after the cancellation (due to one pair of terms obviously being equal), try doing what I said (and which I'd advised does lead to the correct answer): Set the other pairs of terms equal, too. Solve. ;)
 
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