solve system w/ matrices; 13% investment; boat speed

[bethmet]

1: Solve by Matrix Methods-Use Row Operations

7x+5y-z=94
x+5y+2z=52
2x+y+z=29

[7 5 -1 94
1 5 2 52
2 1 1 29]

Multiply Row 1 by 1/7 [1 .71 -1 13.4
1 5 2 52
2 1 1 29]
2: Helen Weller invested $12,000 in an account that pays 10% simple interest. How much additional money must be invested in an account that pays 13% simple interest so that the total interest is equal to the interest on th two investments at the rate of 11%?

3: The speed of a stream is 5 mph. If a boat travels 82 miles downstream in the same time that it takes to travel 41 miles upstream, what is the speed of the boat in still water?

Distance Rate Time
82 x+y 82(x+y)
41 x-y 41(x-y)

82(x+y)=41(x-y)

 

[Denis]

Sooooo....you typed 3 problems...what's your question(s)?

 

[soroban]

Hello, bethmet!

\(\displaystyle \text{1) Solve by Matrix Methods: }\;\begin{array}{ccc}7x+5y-z&=& 94 \\x+5y+2z&=&52 \\2x+y+z&=& 29\end{array}\)

Some advice: avoid fractions!


\(\displaystyle \text{Rearrange the equations: }\;\begin{array}{ccc}x + 5y + 2z &=& 52 \\ 2x + y + z &=& 29 \\ 7x + 5y - z &=& 94 \end{array}\)


\(\displaystyle \text{We have: }\;\left[\begin{array}{ccc|c}1 &5&2&52 \\2&1&1&29\\ 7&5&\text{-}1&94 \end{array}\right]\)


\(\displaystyle \begin{array}{c}\\R_2-2R_1 \\ R_3-7R_1\end{array}\left[\begin{array}{ccc|c}1 & 5 & 2 & 52 \\ 0 & \text{-}9 & \text{-}3 & \text{-}75 \\ 0 & \text{-}30 & \text{-}15 & \text{-}270 \end{array}\right]\)


. . \(\displaystyle \begin{array}{c}\\ \text{-}\frac{1}{3}R_2 \\ \text{-}\frac{1}{15}R_3 \end{array}\left[\begin{array}{ccc|c}1 &5&2&52\\ 0&3&1&25\\0&2&1&18\end{array}\right]\)


\(\displaystyle \begin{array}{c}\\ R_2-R_3 \\ \\ \end{array} \left[\begin{array}{ccc|c}1&5&2&52\\0&1&0&7 \\ 0&2&1&18 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-5R_2 \\ \\ R_3-2R_2\end{array} \left[\begin{array}{ccc|c}1&0&2&17 \\ 0&1&0&7\\ 0&0&1&4 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-2R_3 \\ \\ \\ \end{array} \left[\begin{array}{ccc|c}1&0&0&9 \\ 0&1&0&7 \\ 0&0&1 & 4 \end{array}\right]\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix}\;x\;=\;9\; \\ \;y\;=\;7\;\\\;z\;=\;4\; \end{Bmatrix}\)

 

[fasteddie65]

2: Helen Weller invested $12,000 in an account that pays 10% simple interest. How much additional money must be invested in an account that pays 13% simple interest so that the total interest is equal to the interest on th (sic) two investments at the rate of 11%?

Let a = the amount invested at 13%

0.10(12000) = 0.13a = 0.11(12000 + a)

This should give the amount invested at 13%.

 

[fasteddie65]

3: The speed of a stream is 5 mph. If a boat travels 82 miles downstream in the same time that it takes to travel 41 miles upstream, what is the speed of the boat in still water?

Your apparent set-up makes no sense. d = rt, so t = d/r
Let b = speed of the boat in still water

82/(b + 5) = 41(b - 5)

Upstream, the current's speed is added to the boat's speed; upstream, it is subtracted from the boat's speed.