Solving equations

Last updated: 18 Nov 2008

We have already done much of the work in solving this problem by changing it from the word problem

'Carl has twice as much money invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per year. If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?'

to the algebraic problem

\[ \begin{split} S &= 2B \\ ( 10\% \times S ) + ( 5\% \times B ) &= 800 \\ S &= ? \end{split} \]

To solve the system of equations you want to reduce the problem from two variables in two equations to one variable in one equation. Usually the easiest way to do this is by substitution i.e. replacing one of the variables by the other.

\[ \begin{split} (10\% \times S) + (5\% \times B) &= 800 \qquad \text{We know that S = 2B so replace in the equation} \\ (10\% \times 2B) + (5\% \times B) &= 800 \qquad \text{Multiply out } 10\% \times 2B \\ (20\% \times B) + (5\% \times B) &= 800 \qquad \text{20\% of B and 5\% of B are 25\% of B} \\ 25\% \times B &= 800 \qquad \text{We know that } 25\% = \frac{1}{4} \\ \frac{1}{4} \times B &= 800 \qquad \text{Multiply both sides by 4} \\ B &= 800 \times 4 \\ B &= 3200 \end{split} \]

Careful at this point not to assume that you have finished. You have found the amount of money invested in bonds, now you need to use the equation \(S = 2B\) and calculate the amount invested in stocks.

\[S = 2B = 2 \times 3200 = 6400\]

Returning to the question.

Carl has twice as much money invested in stocks as in bonds. Stocks earn 10% interest per year and bonds 5% per year.

If Carl earned a total of $800 dollars from his stocks and bonds last year how much money did he have invested in stocks?

  1. $9,600
  2. $8,000
  3. $6,400
  4. $4,000
  5. $3,200

The amount invested in stocks was $6,400 and the answer is C.

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