\(\displaystyle \mbox{1. The following system of equations has more than}\)
\(\displaystyle \mbox{one solution }\, (x,\, y).\)
. . . .\(\displaystyle 2ax\, +\, 6y\, =\, 5\)
. . . .\(\displaystyle 4x\, +\, 3ay\, =\, b\)
\(\displaystyle \mbox{The value of }a\, +\, b\mbox{ where }a,\, b\, >\, 0,\, \mbox{ is....}\)
Your small and sideways work is too hard for me to follow. Sorry.i tried to answer this question but i dont know if i did was right, and im still really confused, could someone help me
[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]
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?
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.could someone please show me how to work this out