How much is invested At %6 and at % 10

Let x be the amount invested in the 6% account and y be the amount invested in the 10%. then we can set the following system of equations:
[tex]\begin{gathered} 0.06x+0.10y=470, \\ x+y=7000. \end{gathered}[/tex]Substracting y from the second equation we get:
[tex]x=7000-y\text{.}[/tex]Substituting the above equation in the first one we get:
[tex]0.06(7000-y)+0.10y=470.[/tex]Simplifying the above result we get:
[tex]\begin{gathered} 420-0.06y+0.10y=470, \\ 420+0.04y=470. \end{gathered}[/tex]Subtracting 420 from the above equation we get:
[tex]\begin{gathered} 420+0.04y-420=470-420, \\ 0.04y=50. \end{gathered}[/tex]Dividing the above equation by 0.04 we get:
[tex]\begin{gathered} \frac{0.04y}{0.04}=\frac{50}{0.04}, \\ y=1250. \end{gathered}[/tex]Finally, substituting y=1250 at x=7000-y we get:
[tex]\begin{gathered} x=7000-1250, \\ =5750. \end{gathered}[/tex]Answer:
$5750 at 6%.
$1250 at 10%.