Equations #22 - Questions and Answers

Question 1 Directions: Define 2 variables, write 2 equations, and solve the system of equations. The North Ridgeville Football team played a game against Avon last week. A total of 125 points were scored by both teams. The teams scored a total of 22 six-point touchdowns and three-point field goals. In addition, there were 4 two-point safeties and 12 one-point kicks scored. How many touchdowns and field goals were scored by the teams during the game? Answer Step 1: Let 𝑥 x and 𝑦 y denote the number of touchdowns and field goals scored by the teams during the game. Given that: Total touchdowns and field goals scored = 22 𝑥 + 𝑦 = 22 x+y=22 Total score = Total scores of touchdowns, field goals, and additional points, that is, 125 = 6 𝑥 + 3 𝑦 + 4 ⋅ 2 + 12 ⋅ 1 125=6x+3y+4⋅2+12⋅1 125 = 6 𝑥 + 3 𝑦 + 8 + 12 125=6x+3y+8+12 6 𝑥 + 3 𝑦 = 105 6x+3y=105 2 𝑥 + 𝑦 = 35 2x+y=35 Step 2: The system of equations to solve is: 𝑥 + 𝑦 = 222 𝑥 + 𝑦 = 35 x+y=222x+y=35 First, eliminate 𝑦 y: 2 𝑥 + 𝑦 − ( 𝑥 + 𝑦 ) = 35 − 22 2x+y−(x+y)=35−22 𝑥 = 13 x=13 Step 3: Solve for 𝑦 y by substituting the value for 𝑥 x in either of the two equations: 13 + 𝑦 = 22    ⟹    𝑦 = 9 13+y=22⟹y=9 Thus, 13 touchdowns and 9 field goals were scored by the teams. Question 2 Solve the system of equations: 6 𝑥 − 2 𝑦 = 89 𝑥 − 3 𝑦 = 12 6x−2y=89x−3y=12 Answer Step 1: We'll use elimination to resolve this. We'll work to get rid of 𝑦 y. We multiply the first equation by 3 and the second equation by -2: 3 ( 6 𝑥 − 2 𝑦 = 8 )    ⟹    18 𝑥 − 6 𝑦 = 24 3(6x−2y=8)⟹18x−6y=24 − 2 ( 9 𝑥 − 3 𝑦 = 12 )    ⟹    − 18 𝑥 + 6 𝑦 = − 24 −2(9x−3y=12)⟹−18x+6y=−24 Step 2: The result of adding the two equations is: 18 𝑥 − 6 𝑦 + ( − 18 𝑥 + 6 𝑦 ) = 24 − 24 18x−6y+(−18x+6y)=24−24 0 = 0 0=0 That is always accurate. So, there are infinitely many solutions to the system of equations. We solve for 𝑦 y or 𝑥 x: 6 𝑥 − 2 𝑦 = 8    ⟹    𝑦 = 3 𝑥 − 4 6x−2y=8⟹y=3x−4 Answer: There are infinitely many solutions of the form 𝑦 = 3 𝑥 − 4 y=3x−4, where 𝑥 = any real number. x=any real number. Question 3 Find the equilibrium price and quantity for the supply and demand equations: 𝑑 ( 𝑝 ) = 14 , 400 − 5 𝑝 d(p)=14,400−5p 𝑠 ( 𝑝 ) = 1 , 500 + 2 𝑝 s(p)=1,500+2p Answer Step 1: Consider the given equations: 𝑑 ( 𝑝 ) = 14 , 400 − 5 𝑝 d(p)=14,400−5p 𝑠 ( 𝑝 ) = 1 , 500 + 2 𝑝 s(p)=1,500+2p Step 2: For equilibrium, demand equals supply: 𝑑 ( 𝑝 ) = 𝑠 ( 𝑝 ) d(p)=s(p) 14 , 400 − 5 𝑝 = 1 , 500 + 2 𝑝 14,400−5p=1,500+2p 7 𝑝 = 12 , 900 7p=12,900 𝑝 = 1 , 842.86 p=1,842.86 Step 3: Quantity for supply is determined as follows: 𝑠 ( 𝑝 ) = 1 , 500 + 2 𝑝 s(p)=1,500+2p 𝑠 ( 1 , 842.86 ) = 1 , 500 + 2 ⋅ 1 , 842.86 s(1,842.86)=1,500+2⋅1,842.86 𝑠 ( 𝑝 ) = 5 , 185.72 s(p)=5,185.72 Question 4 Find a least-squares solution of 𝐴 𝑥 = 𝑏 Ax=b by constructing and solving the normal equations: 𝐴 = [ 3 1 1 1 1 4 ] ,   𝑏 = [ 1 1 1 ] A= ​ 3 1 1 ​ 1 1 4 ​ ​ ,b= ​ 1 1 1 ​ ​ Answer Constructing the normal equations: 𝐴 𝑇 𝐴 = [ 3 1 1 1 1 4 ] [ 3 1 1 1 1 4 ] = [ 11 8 8 18 ] A T A=[ 3 1 ​ 1 1 ​ 1 4 ​ ] ​ 3 1 1 ​ 1 1 4 ​ ​ =[ 11 8 ​ 8 18 ​ ] 𝐴 𝑇 𝑏 = [ 3 1 1 1 1 4 ] [ 1 1 1 ] = [ 5 6 ] A T b=[ 3 1 ​ 1 1 ​ 1 4 ​ ] ​ 1 1 1 ​ ​ =[ 5 6 ​ ] Solving 𝐴 𝑇 𝐴 𝑥 = 𝐴 𝑇 𝑏 A T Ax=A T b: [ 11 8 8 18 ] [ 𝑥 1 𝑥 2 ] = [ 5 6 ] [ 11 8 ​ 8 18 ​ ][ x 1 ​ x 2 ​ ​ ]=[ 5 6 ​ ] Step-by-step solution: 𝑥 1 + 10 𝑥 2 = − 3 x 1 ​ +10x 2 ​ =−3 21 𝑥 2 = 5    ⟹    𝑥 2 = − 113 21 , 𝑥 1 = − 21 21 21x 2 ​ =5⟹x 2 ​ = 21 −113 ​ ,x 1 ​ = 21 −21