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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.

Answers to Test

(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)

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CK.MAT.ENG.SE.1.Algebra-I---Honors.5.9

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