McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
9. Perfect Squares
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Exercise 63 Page 71

Create one variable to represent lawn seats and one variable to represent pavilion seats, then use the Substitution Method.

H

Practice makes perfect
To write this situation as a system of equations, we need to first assign variables to each item. We can say that lawn seats are l and pavilion seats are p. We are told that the price of 2 lawn seats and 2 pavilion seats is $120. This can be illustrated using an equation. 2l+ 2p=120 The price of 3 lawn seats and 4 pavilion seats is $225. This can also be illustrated using an equation. 3l+ 4p=225 The two equations above form a system of equations. 2l+2p=120 3l+4p=225 Let's solve this system of equations using the Substitution Method.
2l+2p=120 & (I) 3l+4p=225 & (II)
l+p=60 & (I) 3l+4p=225 & (II)
l=60-p 3l+4p=225
l=60-p 3( 60-p)+4p=225
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(II): Solve for p
l=60-p 180-3p+4p=225
l=60-p 180+p=225
l=60-p p=45
Great! Now, to find the value of l, we need to substitute p=45 into either one of the equations in the given system. Let's use the first equation.
l=60-p p=45
l=60- 45 p=45

(I): Simplify product and terms

l=15 p=45
The solution to the system of equations is l=15 and p=45. Therefore, the correct answer is H.

Alternative Solution

Solving a System of Equations Using a Matrix
To solve the given system of equations using a matrix, we should first write the matrix for the system. Let l be number of lawn seats and p number of pavilion seats. 2 l + 2p= 120 3 l+ 4p= 225 ⇓ [ cc|c2 & 2 & 120 3 & 4 & 225 ] In order to solve the matrix, we will use row operations to obtain a matrix in the following form. [ cc|c 1 & 0 & a 0 & 1 & b ] This final matrix represents the solution of the system of equations, where l= a, p= b.Let's solve the matrix!
[ cc|c2 & 2 & 120 3 & 4 & 225 ]
[ cc|c-3 & -3 & -180 3 & 4 & 225 ]
[ cc|c-3 & -3 & -180 3+( -3) & 4+( -3) & 225+( -180) ]
[ cc|c-3 & -3 & -180 0 & 1 & 45 ]
[ cc|c1 & 1 & 60 0 & 1 & 45 ]
[ cc|c1 & 1- 1 & 60- 45 0 & 1 & 45 ]
[ cc|c1 & 0 & 15 0 & 1 & 45 ]
Looking at the right-hand column, we can see that the solution to the system is the unique point (15,45). To help visualize this answer, we have also written the matrix that resulted from using row operations in system notation. 1l+0p=15 0l+1p=45 ⇒ l=15 p=45 In the end, we can say that the correct answer is H.