# Thread: Hi, rather easy systems of equations problem that I would love help on

1. ## Hi, rather easy systems of equations problem that I would love help on

thanks for looking!

Please say what equations you used. Thanks a lot, I love this site!

2. Hello, lahoo!

Person X invested $10,000 in three accounts. . . Account A pays 4.5% interest. . . Account B pays 6.2% interest. . . Account C pays 3.5% interest. X decides to invest into account B double the sum of the other two accounts. When the first dividends came, they earned$549 interest.
How much invested into each account?

If we organize the information, everything is clearer . . .

. . $\displaystyle \begin{array}{|c|c|c||c|} \hline \text{Account} & \text{Amount} & \text{Rate} & \text{Interest} \\ \hline\hline A & x & 0.045 & 0.045x \\ \hline B & 2x+2y & 0.062 & 0.062(2x+2y) \\ \hline C & y & 0.035 & 0.035y \\ \hline\hline \text{Total} & 10,\!000 & & 549 \\ \hline \end{array}$

In the second column, we have: .$\displaystyle x + (2x + 2y) + y \:=\:10,\!000 \quad\Rightarrow\quad 3x + 3y \:=\:10,\!000\;\;{\color{blue}[1]}$

In the last column: .$\displaystyle 0.045x + 0.062(2x+2y) + 0.35y \:=\:549 \quad\Rightarrow\quad 0.169x + 0.159y \:=\:549$

Multiply by 1000: .$\displaystyle 169x + 159y \:=\:549,\!000\;\;{\color{blue}[2]}$

Solve the system of equations: .$\displaystyle \begin{array}{cccc}3x + 3y &=& 10,\!000 & {\color{blue}[1]} \\ 169x + 159y &=& 549,\!000 & {\color{blue}[2]} \end{array}$

. . $\displaystyle \begin{array}{cccccc}\text{Multiply }{\color{blue}[1]}\text{ by -53:} & \text{-}159x - 159y &=& \text{-}530,\!000 \\ \text{Add }{\color{blue}[2]}\!: & 169x + 159y &=& 549,\!000 \end{array}$

And we get: .$\displaystyle 10x \:=\:19,\!000 \quad\Rightarrow\quad x \:=\:\boxed{\$1900} $in account$\displaystyle A.$Substitute into [1]: .$\displaystyle 3(1900) + 3y \:=\:10,\!000 \quad\Rightarrow\quad y \:=\:\frac{4300}{3} \:=\:\boxed{\$1433.33}$ in account $\displaystyle C.$

Then: .$\displaystyle 2(1900 + 1433.33) \;=\;\boxed{\$6666.67}$in account$\displaystyle B.\$