Question

solve the system. if there are infinitely many solutions, write each answer as a function of ( t ) like ( x = x(t), y = y(t) ) where ( z = t ). if there are no solutions, type dne in each answer box.
\begin{cases} x + y + z = 4 \\ y - 3z = 5 \\ 2x + y + 5z = 4 end{cases}
( x= )
( y= )
( z= )
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question 19
a chemist has three different acid solutions. the first acid solution contains ( 25 % ) acid, the second contains ( 35 % ) and the third contains ( 80 % ). he wants to use all three solutions to obtain a mixture of 60 liters containing ( 55 % ) acid, using 2 times as much of the ( 80 % ) solution as the ( 35 % ) solution. how many liters of each solution should be used?
the chemist should use:
liters of ( 25 % ) solution
liters of ( 35 % ) solution
liters of ( 80 % ) solution
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question 20
country days scholarship fund receives a gift of ( $ 215,000 ). the money is invested in stocks, bonds, and cds. cds pay ( 5.2 % ) interest, bonds pay ( 5.1 % ) interest, and stocks pay ( 6.8 % ) interest. country day invests ( $ 30,000 ) more in bonds than in cds. if the annual income from the investments is ( $ 12,130 ), how much was invested in each vehicle?
country day invested ( $ ) in stocks.

Explanation:

Step1: Solve the first system of equations

Given the system:

$$\begin{cases} x + y+z = 4&(1)\\ y-3z = 5&(2)\\ 2x + y+5z = 4&(3) \end{cases}$$

From equation \((2)\), we can express \(y\) as \(y=3z + 5\).
Substitute \(y = 3z+5\) into equation \((1)\):
\(x+(3z + 5)+z=4\), which simplifies to \(x+4z=-1\), so \(x=-4z - 1\).
Substitute \(x=-4z - 1\) and \(y = 3z+5\) into equation \((3)\):
\(2(-4z - 1)+(3z + 5)+5z=4\)
\(-8z-2 + 3z+5+5z=4\)
\((-8z+3z + 5z)+(-2 + 5)=4\)
\(0z+3 = 4\), which is a contradiction. So this system has no solutions.

Step2: Solve the acid - solution problem

Let \(x\) be the liters of \(25\%\) acid solution, \(y\) be the liters of \(35\%\) acid solution, and \(z\) be the liters of \(80\%\) acid solution.
We have the following system of equations:
\(

$$\begin{cases}x + y+z=60&(a)\\z = 2y&(b)\\0.25x+0.35y + 0.8z=0.55\times60&(c)\end{cases}$$

\)
Substitute \(z = 2y\) into equation \((a)\): \(x+y + 2y=60\), so \(x=60 - 3y\).
Substitute \(x = 60 - 3y\) and \(z = 2y\) into equation \((c)\):
\(0.25(60 - 3y)+0.35y+0.8\times(2y)=33\)
\(15-0.75y+0.35y + 1.6y=33\)
\((-0.75y+0.35y+1.6y)=33 - 15\)
\(1.2y=18\), so \(y = 15\).
If \(y = 15\), then \(z=2y = 30\) and \(x=60-(15 + 30)=15\).

Step3: Solve the investment problem

Let \(x\) be the amount invested in CDs, \(y\) be the amount invested in bonds, and \(z\) be the amount invested in stocks.
We know that \(x + y+z=215000\), \(y=x + 30000\), and \(0.052x+0.051y+0.068z=12130\).
Substitute \(y=x + 30000\) into \(x + y+z=215000\), we get \(x+(x + 30000)+z=215000\), so \(z=185000 - 2x\).
Substitute \(y=x + 30000\) and \(z=185000 - 2x\) into \(0.052x+0.051y+0.068z=12130\):
\(0.052x+0.051(x + 30000)+0.068(185000 - 2x)=12130\)
\(0.052x+0.051x+1530+12580-0.136x=12130\)
\((0.052x+0.051x-0.136x)+(1530 + 12580)=12130\)
\(-0.033x+14110=12130\)
\(-0.033x=12130 - 14110=-1980\)
\(x = 60000\)
If \(x = 60000\), then \(y=x + 30000=90000\) and \(z=185000-2\times60000 = 65000\)

Answer:

For the first system: \(x=\text{DNE}\), \(y=\text{DNE}\), \(z=\text{DNE}\)
For the acid - solution problem: \(15\) liters of \(25\%\) solution, \(15\) liters of \(35\%\) solution, \(30\) liters of \(80\%\) solution
For the investment problem: \(\$65000\) was invested in stocks.