Question

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Answer 1

if you call the amount of the account invested at \(6\%\) \(x\) then since the total is \(\$24,000\) the amount invested at \(11\%\) must be
\[24,000-x\]
the total will be
\[.06x+.11(24,000-x)\] which you know is equal to \(\$1940\) set
\[.06x+.11(24,000-x)=1940\] and solve for \(x\)

Answer 2

probably easiest to multiply both sides by \(100\) to get rid of annoying decimals and solve
\[6x+11(24,000-x)=194,000\]

Answer 3

What happens with the negative x?
I got 6x+264000=194000

Answer 4

Is that right?

Answer 5

@satellite73

Answer 6

hold on let me check
should not be negative

Answer 7

\[6x+11(24,000-x)=194,000\]
\[6x+264,000-11x=194,000\]
\[-5x+264,000=194,000\]

Answer 8

think you might have forgotten to distribute the \(11\) properly