\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Linear Systems - Substitution} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2010} \rfoot{\small Last revised: September 2, 2015} \textwidth 7.6in \textheight 9.65in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\LARGE Sample Problems}\bigskip \bigskip \end{center} \begin{enumerate} \item Solve each of the following system of linear equations. a) \ $\left\{ \begin{array}{c} 2x-y=16 \\ 3x+5y=11% \end{array}% \right. $ \ \ \ \ \ \ b) $\ \left\{ \begin{array}{c} x+3y=11 \\ 12y=-4x+7% \end{array}% \right. $ \ \ \ \ \ \ \ c) \ $\left\{ \begin{array}{c} 4y=6x+10 \\ 3x-2y=-5% \end{array}% \right. $ \item There is an animal farm where chickens and cows live. \ All together, there are $85$ heads and $238$ legs. \ How many chickens and how many cows are there on the farm? \item We have a jar of coins, all pennies and dimes. All together, we have $% 372$ coins, and the total value of all coins in the jar is $\$20.91$. How many pennies are there in the jar? \item We invested $\$7000$ into two bank accounts. One account earns $14\%$ per year, the other account earns $9\%$ per year. How much did we invest into each account if after the first year, the combined interest from the two accounts is $\$840$? \item How many gallons of each of a $4\%$ and an $11\%$ salt solutions should be mixed to obtain $35$ gallons of a $7\%$ solution?\bigskip \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Practice Problems}\bigskip \end{center} \begin{enumerate} \item Solve each of the following system of linear equations. a) \ $\left\{ \begin{array}{c} 3x+y=-4 \\ x-3y=-8% \end{array}% \right. $ \ \ \ \ \ \ \ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $d) \ $\left\{ \begin{array}{c} \dfrac{1}{2}x+\dfrac{1}{4}y=5 \\ \dfrac{1}{2}y-\dfrac{1}{3}x=-6% \end{array}% \right. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g) \ $\left\{ \begin{array}{c} 2x+3y=3 \\ 5x-2y=4% \end{array}% \right. $ b) $\ \left\{ \begin{array}{c} 5\left( p-1\right) -2\left( q-1\right) =22 \\ p-q=8% \end{array}% \right. $ \ \ \ \ \ \ e) \ $\left\{ \begin{array}{c} 2x-y=1 \\ 2\left( y-3\right) =6\left( x-1\right)% \end{array}% \right. $\ \ \ \ \ \ \ \ h) \ $\left\{ \begin{array}{c} 3x-2y=-8 \\ -2x+3y=12% \end{array}% \right. $ c) \ $\left\{ \begin{array}{c} a+3b=10 \\ b=\dfrac{-a-10}{3}% \end{array}% \right. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f) \ $\left\{ \begin{array}{c} 2a+3b=4 \\ 4a=-6b+8% \end{array}% \right. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i) \ $\left\{ \begin{array}{c} 2r-0.5s=-1.7 \\ 1.5r+s=0.65% \end{array}% \right. $ \item Given the equations of two straight lines, find both coordinates of all intersection points. a) \ $2x-5y=-41$ \ and \ $x+y=4$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $5x-y=-35$ \ and $y=-\dfrac{3}{4}x+\dfrac{1}{2}$ b) \ $x+y=-5$ \ and \ $2y=-2x-10$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e) \ $% y=-\dfrac{2}{3}x+2$ \ \ and \ $2x+3y=6$ c) \ $y=\dfrac{3}{4}x-2$ \ \ \ \ and \ \ $3x-4y=-24$ \item There is an animal farm where chickens and cows live. \ All together, there are $52$ heads and $134$ legs. \ How many chickens and how many cows are there on the farm? \item We invested $\$9700$ into two bank accounts. One account earns $7\%$ per year, the other account earns $12\%$ per year. How much did we invest into each account if after the first year, the combined interest from the two accounts is $\$1004$? \item We have $54$ coins, all dimes and quarters, in the total value of $% \$10.05.$ \ How many quarters and how many dimes are there? \item We invested $\$7800$ into two bank accounts. One account earns $9\%$ per year, the other account earns $10\%$ per year. How much did we invest into each account if after the first year we have a total of $\$8549$ in the accounts? \item How many gallons of each of a $22\%$ and a $10\%$ salt solutions should be mixed to obtain $72$ gallons of a $13\%$ solution? \item How many gallons of each of a $41\%$ and a $20\%$ sugar solutions should be mixed to obtain $147$ gallons of a $26\%$ sugar solution? \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Sample Problems \ -- \ Answers}\bigskip \bigskip \end{center} \begin{enumerate} \item a) \ $x=7,y=-2$ \ \ \ \ b) \ there is no solution \ \ \ \ \newline c) \ $x$ can be any number, and then $y=\dfrac{3x+5}{2}$ \item $51$ chickens and $34$ cows \item $181$ pennies \item $\$4200$ $\ $at $14\%$ $\ $and $\ \$2800$ $\ $at $\ 9\%$ \item $20$ gallons of $4\%$ solution with $15$ gallons of $11\%$ solution\bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Practice Problems \ -- \ Answers}\bigskip \bigskip \end{center} \begin{enumerate} \item a) \ $x=-2,y=2$ \ \ \ \ \ \ \ \ b) $\ p=3,q=-5$ \ \ \ \ \ \ \ c) \ there is no solution \ \ \ \ \ \ d) \ $x=12,y=-4$ e) \ $x=-1,y=-3$ \ \ \ \ \ \ f) \ there are infintely many solutions; $a$ can be any number and then $b=% \dfrac{4-2a}{3}$ g) \ $x=\dfrac{18}{19},y=\dfrac{7}{19}$ \ \ \ \ \ \ \ \ \ h) \ $x=0,y=4$ \ \ \ \ \ \ \ i) \ $r=-0.5,s=1.\,\allowbreak 4$ \item a) \ $\left( -3,7\right) $ \ \ \ \ \ b) \ all points on $y=-x-5$ are common; the two lines given are identical c) \ no common points; the two lines given are parallel \ \ \ \ \ d) \ $% \left( -6,5\right) $ e) \ all points on $y=-\dfrac{2}{3}x+2$ are common; the two lines given are identical. \item $37$ chickens, $15$ cows \item $\$3200$ at $7\%$ and \ $\$6500$ at $12\%$ \item $23$ dimes and $31$ quarters \item $\$3100$ at $9\%$ \ and $\$4700$ at $10\%$ \item $18$ gallons of $22\%$ and $54$ gallons of $10\%$ solution \item $42$ gallons of $41\%$ solution and $105$ gallons of $20\%$ solution\pagebreak \end{enumerate} \begin{center} {\LARGE Sample Problems \ -- \ Solutions}\bigskip \bigskip \end{center} \begin{enumerate} \item a) $\left\{ \begin{array}{c} 2x-y=16 \\ 3x+5y=11% \end{array}% \right. $ Solution: \ We will first solve for $y$ in terms of $x$ in the first equation% $.$% \begin{eqnarray*} 2x-y &=&16\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ add }y \\ 2x &=&y+16\text{ \ \ \ \ \ \ \ subtract }16 \\ 2x-16 &=&y \end{eqnarray*}% We substitute $y=2x-16$ into the second equation and solve for $x.$% \begin{eqnarray*} 3x+5y &=&11 \\ 3x+5\left( 2x-16\right) &=&11 \\ 3x+10x-80 &=&11 \\ 13x-80 &=&11\text{ \ \ \ \ \ \ \ \ \ \ add }80 \\ 13x &=&91\text{ \ \ \ \ \ \ \ \ \ \ divide by }13 \\ x &=&7 \end{eqnarray*}% Now that we know the value of $x,$ we can easily compute $y$ since $y=2x-16$% \begin{equation*} y=2x-16=2\cdot 7-16=-2 \end{equation*}% Thus the solution is $x=7,$ \ $y=-2.$ \ We check: the pair should be a solution for both equations.% \begin{equation*} \begin{array}{c} 2\cdot 7-\left( -2\right) =14+2=16~~\checkmark \\ 3\cdot 7+5\left( -2\right) =21-10=11~~\checkmark% \end{array}% \end{equation*}% Thus our solution is correct. b) $\left\{ \begin{array}{c} x+3y=11 \\ 12y=-4x+7% \end{array}% \right. $ Solution: \ We will first solve for $x$ in terms of $y$ in the first equation% $.$% \begin{eqnarray*} x+3y &=&11\text{ \ \ \ \ \ \ \ \ subtract }3y \\ x &=&-3y+11\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }3 \end{eqnarray*}% We substitute $x=-3y+11$ into the second equation and solve for $y.$% \begin{eqnarray*} 12y &=&-4x+7 \\ 12y &=&-4\left( -3y+11\right) +7\text{ \ \ \ \ \ \ \ \ \ distribute} \\ 12y &=&12y-44+7 \\ 12y &=&12y-37\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }12y \\ 0 &=&-37 \end{eqnarray*}% Since there is no value for $y$ that could make the statement $0=-37$ true, there is no solution for this system. \ A linear system like this is called an \textbf{inconsistent system}.\pagebreak c) \ $\left\{ \begin{array}{c} 4y=6x+10 \\ 3x-2y=-5% \end{array}% \right. $ Solution: \ We will first solve for $y$ in terms of $x$ in the second equation.% \begin{eqnarray*} 3x-2y &=&-5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ add }2y \\ 3x &=&2y-5\text{ \ \ \ \ \ \ \ \ add }5 \\ 3x+5 &=&2y\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ \dfrac{3x+5}{2} &=&y \end{eqnarray*}% We substitute $y=\dfrac{3x+5}{2}$ into the first equation and solve for $x.$% \begin{eqnarray*} 4y &=&6x+10 \\ 4\left( \dfrac{3x+5}{2}\right) &=&6x+10 \\ \dfrac{4\left( 3x+5\right) }{2} &=&6x+10\text{ \ \ \ \ \ \ \ \ \ simplify} \\ 2\left( 3x+5\right) &=&6x+10\text{ \ \ \ \ \ \ \ \ \ \ distribute} \\ 6x+10 &=&6x+10\text{ \ \ \ \ \ \ \ \ \ subtract }6x \\ 10 &=&10 \end{eqnarray*}% $10=10$ is in fact true for all values of $x.$ \ What happens here, one equation establishes a connection between $x$ and $y,$ namely, $y=\dfrac{3x+5% }{2}.$ \ The other equation does not contain any new information, it is just a disguised re-statement of the same connection. \ A system like this is called a \textbf{dependent system}. \ This system has infinitely many solutions. \ $x$ can take any value, and then $y$ must be $y=\dfrac{3x+5}{2}% . $ \ Thus, there are infinitely many solutions. \item There is an animal farm where chickens and cows live. \ All together, there are $85$ heads and $238$ legs. \ How many chickens and how many cows are there on the farm?\newline Solution: We will denote the number of chickens by $x$ and the number of cows by $y$. The first equation will express the number of heads, the second equation will express the number of legs.% \begin{eqnarray*} x+y &=&85 \\ 2x+4y &=&238 \end{eqnarray*}% To simplify our system, we divide the second equation by $2$. \begin{eqnarray*} x+y &=&85 \\ x+2y &=&119 \end{eqnarray*}% We solve for $y$ in terms of $x$ in the first equation: \ $y=85-x$ \ \ We substitute this into the second equation:% \begin{eqnarray*} x+2y &=&119 \\ x+2\left( 85-x\right) &=&119\text{ \ \ \ \ \ \ \ \ \ distribute }2 \\ x+170-2x &=&119\text{ \ \ \ \ \ \ \ \ \ combine like terms} \\ -x+170 &=&119\text{ \ \ \ \ \ \ \ \ \ subtract }170 \\ -x &=&-51\text{ \ \ \ \ \ \ \ \ \ multiply by }-1 \\ x &=&51 \end{eqnarray*}% Now that we know the value of $x$, we compute $y$.% \begin{equation*} y=85-x=85-51=34 \end{equation*}% Thus we have $51$ chickens and $34$ cows. We check: the number of heads is $% 51+34=85,$ and the number of legs is $2\left( 51\right) +4\left( 34\right) =102+136=238$. So our solution is correct. \item We have a jar of coins, all pennies and dimes. All together, we have $% 372$ coins, and the total value of all coins in the jar is $\$20.91$. How many pennies are there in the jar? Solution: \ Let us denote the number of pennies by $x$ and the number of dimes by $y$. \ The number of coins will give us one equation:% \begin{equation*} x+y=372 \end{equation*}% The second equation will express the total value of the coins. \ We can express it in dollars: each penny is worth $0.01$ dollars and each dime is worth $0.1$ dollars. \ So the equation is% \begin{equation*} 0.01x+0.1y=20.91 \end{equation*}% We would want to get rid of the decimals by multiplying both sides of this equation by $100$ and then get% \begin{equation*} x+10y=2091 \end{equation*}% Our second choice is to express the total value of the coins in pennies. \ That will immediately give us the equation $x+10y=2091$. \ So we need to solve the system% \begin{equation*} \left\{ \begin{array}{c} x+y=372 \\ x+10y=2091% \end{array}% \right. \end{equation*}% We will solve this using substitution. \ From the first equation, $y=372-x$. \ Then the second equation becomes% \begin{eqnarray*} x+10\left( 372-x\right) &=&2091\text{ \ \ \ \ \ \ \ \ \ distribute }10 \\ x+3720-10x &=&2091\text{ \ \ \ \ \ \ \ \ \ combine like terms} \\ -9x+3720 &=&2091\text{ \ \ \ \ \ \ \ \ \ subtract \ }3720\text{ } \\ -9x &=&-1629\text{ \ \ \ \ \ \ \ \ divide by }-9 \\ x &=&181 \end{eqnarray*}% So we have $181$ pennies. \ To check, we should also figure out the number of dimes. \ \newline It is $y=372-x=372-181=191$. \ So, if we have $181$ pennies and $191$ dimes. \ Now we check against the conditions stated in the problem:% \begin{eqnarray*} x+y &=&181+191=372\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ total number of coins is }372 \\ 0.01x+0.1y &=&0.01\left( 181\right) +0.1\left( 191\right) =1.81+19.1=20.91% \text{ \ \ \ \ \ \ \ \ total value of coins is }\$20.91 \end{eqnarray*}% and so our solution is correct. \pagebreak \item We invested $\$7000$ into two bank accounts. One account earns $14\%$ per year, the other account earns $9\%$ per year. How much did we invest into each account if after the first year, the combined interest from the two accounts is $\$840$?\newline Solution: \ Let us denote the amount invested at $14\%$ by $x$ and the amount invested at $9\%$ by $y$. The two equations express that% \begin{eqnarray*} x+y &=&7000\text{ \ \ \ \ \ \ the amounts add up to }\$7000 \\ 0.14x+0.09y &=&840\text{ \ \ \ \ \ \ the interests earned add up to }\$840 \end{eqnarray*}% We solve the system of equation by substitution. But let us first make the second equation simpler:% \begin{eqnarray*} 0.14x+0.09y &=&840\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiply by }100 \\ 14x+9y &=&84\,000 \end{eqnarray*}% We now have \begin{eqnarray*} x+y &=&7000 \\ 14x+9y &=&84\,000 \end{eqnarray*}% We will solve for $y$ in terms of $x$ in the first equation: \ $y=7000-x$ and substitute that into the second equation.% \begin{eqnarray*} 14x+9y &=&84\,000 \\ 14x+9\left( 7000-x\right) &=&84\,000\text{ \ \ \ \ \ \ \ \ \ \ distribute }9 \\ 14x+63\,000-9x &=&84\,000\text{ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 5x+63\,000 &=&84\,000\text{ \ \ \ \ \ \ \ \ \ \ subtract }63\,000 \\ 5x &=&21\,000\text{ \ \ \ \ \ \ \ \ \ \ divide by }5 \\ x &=&4200 \end{eqnarray*}% Then $y=7000-x=7000-4200=2800.$ \ Thus we invested $\$4200$ at $14\%$ and $% \$2800$ at $9\%$. \ We check: the amounts add up to $\$4200+\$2800=\$7000$. The interest from the accounts are:% \begin{equation*} 14\%\text{ \ of \ }4200\text{ is \ }0.14\left( 4200\right) =588\text{ \ and \ }9\%\text{ \ of \ }2800\text{ is \ }0.09\left( 2800\right) =252 \end{equation*}% Since $588+252=840$, our solution is correct. \pagebreak \item How many gallons of each of a $4\%$ and an $11\%$ salt solutions should be mixed to obtain $35$ gallons of a $7\%$ solution? Solution: Let us denote by $x$ the amount of $4\%$ solution and by $y$ the amount of $11\%$ solution. \ Clearly, the two amounts should add up to $35$ gallons, giving us the equation $x+y=35$. \begin{tabular}{llll} & Amount of Solution (gallons) & Percentage & Amount of Solvant (gallons) \\ \cline{2-4} Component 1 & \multicolumn{1}{|c}{$x$} & \multicolumn{1}{|c}{$0.04$} & \multicolumn{1}{|c|}{$0.04x$} \\ \cline{2-4} Component 2 & \multicolumn{1}{|c}{$y$} & \multicolumn{1}{|c}{$0.11$} & \multicolumn{1}{|c|}{$0.11y$} \\ \cline{2-4} Mixture & \multicolumn{1}{|c}{$35$} & \multicolumn{1}{|c}{$0.07$} & \multicolumn{1}{|c|}{} \\ \cline{2-4} \end{tabular} Since we have two unknown variables, we will need two equations. \ The first one is easy, the volume of the mixture should be $35$.% \begin{equation*} x+y=35 \end{equation*}% We obtain the second equation by stating that the amount of solvant in the components must add up to the amount of solvant. (In other words, the last entry in the third row can be written in two different ways: the product of $% 35$ and $7\%$; and the sum of $0.04x$ \ and $0.11$) \ \begin{equation*} 0.07\left( 35\right) =0.04x+0.11y \end{equation*}% And this equation can be immediately made much nicer by simply multiplying both sides by $100.$ \ Then we have:% \begin{eqnarray*} 7\left( 35\right) &=&4x+11y \\ 4x+11y &=&245 \end{eqnarray*}% So our system is now% \begin{equation*} \left\{ \begin{array}{c} x+y=35 \\ 4x+11y=245% \end{array}% \right. \end{equation*}% We solve this system using substitution: $y=35-x$ from the first equation. \ Then the second equation becomes% \begin{eqnarray*} 4x+11\left( 35-x\right) &=&245\text{ \ \ \ \ \ \ \ \ \ distribute }11 \\ 4x+385-11x &=&245\text{ \ \ \ \ \ \ \ \ \ combine like terms} \\ -7x+385 &=&245\text{ \ \ \ \ \ \ \ \ \ \ subtract }385 \\ -7x &=&-140\text{ \ \ \ \ \ \ \ \ divide by }-7 \\ x &=&20 \end{eqnarray*}% If $x=20,$ then the other amount, denoted by $y=35-x$ must be $% 35-20=\allowbreak 15$. Thus we need to mix $20$ gallons of $4\%$ solution with $15$ gallons of $11\% $ solution. We check our solution: suppose we mix the two solutions specified above. \ We need to find how much solution and how much solvant we have, hoping that the amount of solvant indeed will be $8\%$ of the amount of mixture. \begin{tabular}{llll} $\text{ }$ & Amount of Solution & & Amount of Solvant \\ \cline{2-4} Component 1 & \multicolumn{1}{|l}{$20$ gallons \ \ \ \ \ \ \ \ \ \ \ \ of $% 4\%$ solution} & \multicolumn{1}{|l}{$\Longrightarrow $} & \multicolumn{1}{|l|}{$0.04\left( 20\right) =\allowbreak 0.8$ gallons} \\ \cline{2-4} Component 2 & \multicolumn{1}{|l}{$15\text{ gallons \ \ \ \ \ \ \ \ \ \ \ \ of }11\%\text{ solution}$} & \multicolumn{1}{|l}{$\Longrightarrow $} & \multicolumn{1}{|l|}{$0.11\left( 15\right) =\allowbreak 1.\,\allowbreak 65$ gallons} \\ \cline{2-4} \multicolumn{1}{c}{} & \multicolumn{1}{|c}{$\Downarrow $} & \multicolumn{1}{|c}{} & \multicolumn{1}{|c|}{$\Downarrow $} \\ \cline{2-4} & \multicolumn{1}{|l}{$35$ gallons} & \multicolumn{1}{|l}{} & \multicolumn{1}{|l|}{$0.8+1.65=\allowbreak 2.\,\allowbreak 45$ gallons} \\ \cline{2-4} \end{tabular} $7\%$ of $35$ is $0.07\left( 35\right) =\allowbreak 2.\,\allowbreak 45$ Thus our solution is correct. \end{enumerate} \bigskip \bigskip \bigskip \bigskip \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}