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# 5.9: Chapter Test

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1. Determine the point of intersection of the following pairs of lines by graphing both lines on the same set of axes.

(a) $\begin{Bmatrix} 3x+5y = -15\\ 2x-3y = 12\\\end{Bmatrix}$

(b) Prove this point algebraically.

1. If the following graph represents a linear programming problem with these constraints:

$\begin{Bmatrix} 6x+5y \le 450\\ 3x+5y \le 300\\ x \ge 0\\ y \ge 0\end{Bmatrix}$

and the profit statement is represented by the equation $P=12x+15y$ determine the maximum profit and show all the necessary work.

1. A college student opened a swimsuit business for the summer months. She is making tankini suits and bikini suits. Each tankini is processed for 4 minutes on the cutting machine and for 3 minutes on the sewing machine. Each bikini is processed for 3 minutes on the cutting machine and for 1 minute on the sewing machine. The cutting machine is available for a maximum of 120 minutes each day and the sewing machine for 60 minutes each day. If each tankini yields a profit of $45.00 and each bikini a profit of$30.00, how many of each swimsuit should she make each day to maximize a profit?
2. Solve the following system of linear equations.

$\begin{Bmatrix} \frac{1}{2}x+\frac{3}{4}y = -2\\ \frac{5}{3}x-\frac{2}{3}y = -\frac{1}{3}\\\end{Bmatrix}$

1. Solve the following system of linear equations by elimination.

$\begin{Bmatrix} 2(1-2x)-3(1-y)=-16\\ 4(2y+1)-3(x+3)=-22\\\end{Bmatrix}$

1. A vending machine contains 694 coins made up of quarters and nickels. The total value of the coins is \$131.90. How many of each coin are in the vending machine?
2. Determine the inequality that is modeled by the following graph.
3. Draw $\Delta ABC$. Place the following equations on the correct sides of the triangle:

$& BC: 3x-8y=-39\\& AC:4x+y=18\\& AB:x+2y=1$

Determine the coordinates of the vertices of the triangle.

(a) $\begin{Bmatrix} 3x+5y = -15\\ 2x-3y = 12\\\end{Bmatrix}$

$& 3x+5y=-15 && 2x-3y=12\\& 3x-3x+5y=-3x-15 && 2x-2x-3y=-2x+12\\& 5y=-3x-15 && -3y=-2x+12\\& \frac{\cancel{5}y}{\cancel{5}}=-\frac{3}{5}x-\frac{\overset{{\color{red}3}}{\cancel{15}}}{\cancel{5}} && \frac{\cancel{-3}y}{\cancel{-3}}=\frac{2}{3}x-\frac{\overset{{\color{red}-4}}{\cancel{12}}}{\cancel{-3}}\\& {\color{red}y=-\frac{3}{5}x-3} && {\color{red}y=\frac{2}{3}x-4}$

The point of intersection is:

$\begin{Bmatrix} 3x+5y=-15\\ 2x-3y=12\\\end{Bmatrix}$

$& 3x+5y=-15 \quad \rightarrow && -2(3x+5y=-15) \quad \rightarrow && -\cancel{6x}-10y=30\\& 2x-3y=12 && \quad \ 3(2x-3y=12) \quad \ \rightarrow && \ \ \underline{\;\;\; \cancel{6x}-9y=36 \;}\\& && && \qquad -19y=66\\& && && \qquad \ \frac{\cancel{-19}y}{\cancel{-19}}=-\frac{66}{19} \qquad \rightarrow\\& && && \qquad \qquad \ {\color{red}y=-\frac{66}{19}}$

$2x-3y&=12\\2x-3\left({\color{red}-\frac{66}{19}}\right)&=12\\2x{\color{red}+\frac{198}{19}}&=12\\2x+\frac{198}{19}{\color{red}-\frac{198}{19}}& =12{\color{red}-\frac{198}{19}}\\2x &={\color{red}\frac{228}{19}}-\frac{198}{19}\\2x &= {\color{red}\frac{30}{19}}\\\frac{\cancel{2}x}{\cancel{2}}&=\frac{30}{19}\left({\color{red}\frac{1}{2}}\right)\\x &= {\color{red}\frac{30}{38}}\\{\color{red}x}&{\color{red}\ =\frac{15}{19}}$

$l_1 \cap l_2 @ \left(\frac{15}{19},-\frac{66}{19}\right)$

$\begin{Bmatrix} 6x+5y \le 450\\ 3x+5y \le 300\\ x \ge 0\\ y \ge 0\end{Bmatrix}$

The $x-$intercept of $6x+5y \le 450$ is:

$6x+5y&=450\\6x+5({\color{red}0})&=450\\6x &= 450\\\frac{\cancel{6}x}{\cancel{6}}&=\frac{\overset{{\color{red}75}}{\cancel{450}}}{\cancel{6}}\\x&=75$

The $y-$intercept of $3x+5y \le 300$ is (0, 60)

The point of intersection of

$6x+5y &\le 450\\3x+5y &\le 300$

is:

$& 6x+5y=450 \quad \rightarrow && \qquad 6x+5y=450 \qquad \rightarrow && \ \ 6x + \cancel{5y}=450 && 6x+5y=450\\& 3x+5y=300 && -1(3x+5y=300) \quad \rightarrow && \underline{-3x{-\cancel{5y}}=-300 \;} && 6({\color{red}50})+5y=450\\& && && \qquad \ \ 3x=150 && {\color{red}300}+5y=450\\& && && \qquad \ \ \frac{\cancel{3}x}{\cancel{3}}=\frac{\overset{{\color{red}50}}{\cancel{150}}}{\cancel{3}} \quad \rightarrow && 300{\color{red}-300}+5y=450{\color{red}-300}\\& && && \qquad \quad \ x=50 && 5y={\color{red}150}\\& && && && \frac{\cancel{5}y}{\cancel{5}}=\frac{\overset{{\color{red}30}}{\cancel{150}}}{\cancel{5}}\\& && && && y=30$

$l_1 \cap l_2 @ (50, 30)$

$& (0,0) && P=12x+15y \rightarrow P=12(0)+15(0) \rightarrow P=0+0 \rightarrow P=0\\& && \text{Therefore} \ 12x+15y=\ 0\\\\& (0,60) && P=12x+15y \rightarrow P=12(0)+15(60) \rightarrow P=0+900 \rightarrow P=900\\& && \text{Therefore} \ 12x+15y=\ 900\\\\ & (50,30) && P=12x+15y \rightarrow P=12(50)+15(30) \rightarrow P=600+450 \rightarrow P=1050\\& && \text{Therefore} \ 12x+15y=\ 1050\\\\& (75,0) && P=12x+15y \rightarrow P=12(75)+15(0) \rightarrow P=900+0 \rightarrow P=900\\& && \text{Therefore} \ 12x+15y=\ 900$

The maximum profit occurred at the vertex (50, 30).

1. Let ‘$x$’ represent the number of tankinis to be made. Let ‘$y$’ represent the number to bikinis to be made. Table:
Tankinis Bikinis Time Available Min/hr
Cutting Machine 4min 3min 120min/hr
Sewing Machine 3 min 1min 60min/hr

Constraints:

$\begin{Bmatrix} 4x+3y \le 120\\ 3x+y \le 60\\ x \ge 0\\ y \ge 0\end{Bmatrix}$

Profit equation:$\boxed{p=45x+30y}$

Graph:

The feasible region is the area shaded in yellow.

Vertices:

The $y-$intercept of $4x+3y \le 120$ is (0, 40).

The $x-$intercept of $3x+y \le 60$ is:

$3x+y &= 60\\3x+({\color{red}0})&=60\\3x &= 60\\\frac{\cancel{3}x}{\cancel{3}}&=\frac{\overset{{\color{red}20}}{\cancel{60}}}{\cancel{3}}\\x &= 20$

The $x-$intercept is (20, 0).

$& 4x+3y=120 \quad \rightarrow && 3x+y=60 && 4x+3({\color{red}-3x+60})=120 && y=-3x+60\\& 3x+y=60 && 3x{\color{red}-3x}+y={\color{red}-3x}+60 \quad \rightarrow && 4x{\color{red}-9x+180}=120 && y =-3({\color{red}12})+60\\& && y=-3x+60 && {\color{red}-5x}+180=120 \qquad \qquad \qquad \rightarrow && y={\color{red}-36}+60\\& && && -5x+180{\color{red}-180}=120{\color{red}-180} && y ={\color{red}24}\\& && && -5x={\color{red}-60} && y=24\\& && && \frac{\cancel{-5}x}{\cancel{-5}}=\frac{\overset{{\color{red}12}}{\cancel{-60}}}{\cancel{-5}}\\& && && \quad \ \ x=12$

$l_1 \cap l_2 @ (12,24)$

Profit:

$& (0, 0) && P=45x+30y \rightarrow P=45(0)+30(0) \rightarrow P=0+0 \rightarrow P=0\\& && \text{Therefore} \ 12x+15y= \ 0\\\\& (0, 40) && P=45x+30y \rightarrow P=45(0)+30(40) \rightarrow P=0+1200 \rightarrow P=1200\\& && \text{Therefore} \ 12x+15y= \ 1200\\\\ & (20, 0) && P=45x+30y \rightarrow P=45(20)+30(0) \rightarrow P=900+0 \rightarrow P=900\\& && \text{Therefore} \ 12x+15y= \ 900\\\\& (12, 24) && P=45x+30y \rightarrow P=45(12)+30(24) \rightarrow P=540+720 \rightarrow P=1260\\& && \text{Therefore} \ 12x+15y= \ 1260$

The maximum profit occurred at the vertex (12, 24). She should make 12 tankinis and 24 bikinis daily to maximize her profit.

$\begin{Bmatrix} \frac{1}{2}x+\frac{3}{4}y = -2\\ \frac{5}{3}-\frac{2}{3}y=-\frac{1}{3}\\\end{Bmatrix}$

$4\left(\frac{1}{2}x+\frac{3}{4}y=-2\right) & \rightarrow 2x+3y=-8\\3\left(\frac{5}{3}x-\frac{2}{3}y=-\frac{1}{3}\right) & \rightarrow 5x-2y=-1$

$2(2x+3y=-8) & \rightarrow 4x + {\cancel{6y}}=-16 && 2x+3y=-8\\3(5x-2y=-1) & \rightarrow \underline{15x - \cancel{6y}=-3\;\;} && 2({\color{red}-1})+3y=-8\\& \qquad \quad \ 19x=-19 && {\color{red}-2}+3y=-8\\& \qquad \quad \frac{\cancel{19}x}{\cancel{19}}=\frac{\overset{{\color{red}-1}}{\cancel{-19}}}{\cancel{19}} \quad \rightarrow && -2{\color{red}+2}+3y=-8{\color{red}+2}\\& \qquad \qquad x=-1 && \frac{\cancel{3}y}{\cancel{3}}=\frac{\overset{{\color{red}-2}}{\cancel{-6}}}{\cancel{3}}\\& && y=-2$

$I_1 \cap I_2 @ (-1,-2)$

$\begin{Bmatrix} 2(1-2x)-3(1-y)=-16\\ 4(2y+1)-3(x+3)=-22\\\end{Bmatrix}$

$& \begin{Bmatrix} 2(1-2x)-3(1-y)=-16\\ 4(2y+1)-3(x+3)=-22\\\end{Bmatrix}\\& {\color{red}2-4x-3+3y=-16} \rightarrow -4x+3y{\color{red}-1}=-16 \rightarrow -4x+3y-1{\color{red}+1}=-16{\color{red}+1} \rightarrow -4x+3y=-15\\& {\color{red}8y+4-3x-9=-22} \rightarrow -3x+8y{\color{red}-5}=-22 \rightarrow -3x+8y-5{\color{red}+5}=-22{\color{red}+5} \rightarrow -3x+8y=-17\\$

$& -3(-4x+3y=-15) \quad \rightarrow \qquad \cancel{12x}-9y=45\\& \quad 4(-3x+8y=-17) \quad \rightarrow \quad \underline{-\cancel{12x}+32y=-68}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ 23y=-23\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \frac{\cancel{23}y}{\cancel{23}}=\frac{\overset{{\color{red}-1}}{-23}}{23}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ y=-1$

$-4x+3y &=-15\\-4x+3({\color{red}-1})&=-15\\-4x{\color{red}-3}&=-15\\-4x-3{\color{red}+3}&=-15{\color{red}+3}\\-4x&={\color{red}-12}\\\frac{\cancel{-4}x}{\cancel{-4}}&=\frac{\overset{{\color{red}3}}{\cancel{-12}}}{\cancel{-4}}\\x&=3$

1. Let ‘$q$’ represent the number of quarters in the vending machine. Let ‘$n$’ represent the number of nickels in the vending machine.

$\begin{Bmatrix} q+n=694\\ .25q+.05n=131.90\\\end{Bmatrix}$

$& q+n =694\\& q+n{\color{red}-n}=694{\color{red}-n}\\& {\color{red}q-694-n}$

$.25q+0.5n&=131.90\\.25({\color{red}694-n})+.05n&=131.90\\{\color{red}173.50-.25n}+.05n&=131.90\\173.50{\color{red}-.20n}&=131.90\\173.50{\color{red}-173.50}-.20n&=131.90{\color{red}-173.50}\\-.20n&={\color{red}-41.60}\\\frac{\cancel{-.20}n}{\cancel{-.20}}&=\frac{\overset{{\color{red}208}}{\cancel{-41.60}}}{\cancel{-.20}}\\n&=208$

$q&=694-n\\q&=694-208\\q&={\color{red}486}$

There are 486 quarters and 208 nickels in the vending machine.

The slope of the graphed line is: $m=\frac{rise}{run} \quad m=\frac{-3}{4}={\color{red}-\frac{3}{4}}$

The $y-$intercept is (0, 4).

The line is a solid line and the shading is below the line. Therefore the inequality symbol is $\le$.

The inequality is $\boxed{y \le -\frac{3}{4}x+4}$ (Slope-intercept form) or $\boxed{3x+4y \le 16}$ (Standard Form).

1. Sketch the triangle. Label the vertices $ABC$. Place the equation on the correct side of the triangle. To determine the coordinates of the vertices, solve the two equations (a $2\times 2$ system) of the sides that form the vertex.

$A (5, -2) \qquad B (-5, 3) \qquad C (3, 6)$

$& \begin{Bmatrix} 4x+y=18\\ x+2y=1\\ \end{Bmatrix} && \begin{Bmatrix} x+2y=1\\ 3x-8y=-39\\ \end{Bmatrix} && \begin{Bmatrix} 4x+y=18\\ 3x-8y=-39\\ \end{Bmatrix}$

$& \begin{Bmatrix} 4x+y=18\\ x+2y=1\\ \end{Bmatrix}$

$& -2(4x+y=18) \quad \rightarrow \quad {\color{red}-8x-\cancel{2y}=-36} && 4x+y=18\\& \qquad x+2y=1 \qquad \rightarrow \qquad \ \underline{x+\cancel{2y}=1 \;\;\;\;} && 4({\color{red}5})+y=18\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad -7x=-35 \quad \rightarrow && {\color{red}20}+y=18\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \frac{\cancel{-7}x}{\cancel{-7}}=\frac{\overset{{\color{red}5}}{\cancel{-35}}}{\cancel{-7}} && 20{\color{red}-20}+y=18{\color{red}-20}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x=5 && y={\color{red}-2}\\& && y=-2$

$\begin{Bmatrix} x+2y=1\\ 3x-8y=-39\\ \end{Bmatrix}$

$& 4(x+2y=1) \qquad \rightarrow \qquad \ {\color{red}4x+\cancel{8y}=4} && x+2y=1\\& \ 3x-8y=-39 \quad \rightarrow \qquad \ \underline{3x-\cancel{8y}=-39} && ({\color{red}-5})+2y=1\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad 7x=-35 \quad \rightarrow && -5{\color{red}+5}+2y=1{\color{red}+5}\\& \qquad \qquad \qquad \qquad \qquad \qquad \ \ \frac{\cancel{7}x}{\cancel{7}}=\frac{\overset{{\color{red}-5}}{\cancel{-35}}}{\cancel{7}} && 2y={\color{red}6}\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ x=-5 && \frac{\cancel{2}y}{\cancel{2}}=\frac{\overset{{\color{red}3}}{\cancel{6}}}{\cancel{2}}\\& && y=3$

$\begin{Bmatrix} 4x+y=18\\ 3x-8y=-39\\ \end{Bmatrix}$

$& 8(4x+y=18) \quad \ \rightarrow \ \quad {\color{red}32x+\cancel{8y}=144} && 4x+y=18\\& \ 3x-8y=-39 \quad \rightarrow \ \ \quad \underline{3x-\cancel{8y}=-39} && 4({\color{red}3})+y=18\\& \qquad \qquad \qquad \qquad \qquad \qquad \quad 35x={\color{red}105} \quad \rightarrow && 12{\color{red}-12}+y=18{\color{red}-12}\\& \qquad \qquad \qquad \qquad \qquad \qquad \ \ \frac{\cancel{35}x}{\cancel{35}}=\frac{\overset{{\color{red}3}}{\cancel{105}}}{\cancel{35}} && y={\color{red}6}\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad x=3 && y=6$ --User:John.Kim 15:26, 5 December 2012 (PST)

Jan 16, 2013

Nov 12, 2014