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Chapter 11: Problem 9

$$11 x+14 y=30 \text { and } 3 x+4 y=12$$ Quantity \(\mathbf{A}\) \(x+y\) Quantity \(\mathbf{B}\) \\[ (x+y)^{-2} \\] a. Quantity A is greater. b. Quantity B is greater. c. The two quantities are equal. d. The relationship cannot be determined from the information given.

Short Answer

Expert verified
a. Quantity A is greater.

Step by step solution

01

Identify and Solve the Given System of Linear Equations

We have the system of equations: \[11x+14y=30 \text{ and } 3x+4y=12\]. The system is solved easily by dividing the second equation by 3, getting: \[x + \frac{4y}{3} = 4\]. The first equation is then divided by 11, getting: \[x + \frac{14y}{11} = \frac{30}{11}\]
02

Solve for the Variables X and Y

We use the method of substitution to solve the system. Setting the two equations equal to each other will cancel out variable x: \[\frac{4y}{3} = \frac{14y}{11}+\frac{30}{11}\]. Solving for y will yield y = 2. Plugging y = 2 into either of the original equations (here, the second is used) will give the value of x = 2.
03

Substitute into Quantity A and B

Substituting y = 2 and x = 2 into Quantity A gives: \[x + y = 2 + 2 = 4\]. Substitute y = 2 and x = 2 into Quantity B gives: \[\frac{1} {(x + y)^2} = \frac{1} {4^2} = \frac{1}{16}\].
04

Compare Quantity A and Quantity B

At this point we can compare Quantity A and Quantity B. We have Quantity A = 4 and Quantity B = \frac{1}{16}. Obviously, 4 is greater than \frac{1}{16}

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Most popular questions from this chapter

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