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Chapter 5: Problem 43

Solve the system using any method. $$ \begin{array}{l} 2 x=\frac{y}{2}+1 \\ 0.04 x-0.01 y=0.02 \end{array} $$

Short Answer

Expert verified
Infinitely many solutions: \( (x, 4x - 2) \) for any \(x\).

Step by step solution

01

Simplify the First Equation

Start with the equation: \(2x = \frac{y}{2} + 1\). Multiply both sides by 2 to clear the fraction: \(4x = y + 2\). Subtract 2 from both sides to isolate y: \(y = 4x - 2\).
02

Substitute into the Second Equation

Use the expression for y from Step 1 in the second equation: \(y = 4x - 2\). Replace y in \(0.04x - 0.01y = 0.02\): \(0.04x - 0.01(4x - 2) = 0.02\).
03

Solve for x

Expand the equation: \(0.04x - 0.04x + 0.02 = 0.02\). Simplify the equation: \(0.02 = 0.02\), which is always true, therefore this system has infinitely many solutions.
04

Verify the Solutions

Since the system has infinitely many solutions, any value of \(x\) works. Given \(y = 4x - 2\), the solutions can be written as \( (x, 4x - 2) \) for any \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving linear equations
Solving a system of linear equations involves finding the values of the variables that satisfy all the given equations. Typically, these equations are straight lines, and their solutions are the points where these lines intersect. There are various methods to solve linear equations, such as graphing, substitution, and elimination. In this exercise, we utilized the substitution method.
infinite solutions
Sometimes, systems of linear equations have more than one solution. If the equations represent the same line, they intersect at infinitely many points. This means there are countless pairs of values that satisfy both equations. In this exercise, after simplifying and substituting, we ended up with the equation \(0.02 = 0.02\). Since this is always true, it indicates that the system has infinite solutions. Therefore, any value of \(x\) paired with \(y = 4x - 2\) is a valid solution.
substitution method
The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. The general idea is to solve one of the equations for one of the variables and then substitute this expression into the other equation. This reduces the system to a single equation with one variable, which can be solved more straightforwardly. In our problem, we isolated \(y\) in the first equation, and then substituted this value into the second equation to find the solutions.
algebraic manipulation
Algebraic manipulation involves using basic algebraic operations to transform equations into simpler forms. This often involves isolating variables, simplifying expressions, and combining like terms. In this exercise, we performed operations such as multiplying both sides of an equation by the same number, distributing multiplication over addition, and isolating variables. These steps are essential in solving systems of equations accurately and efficiently.

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Most popular questions from this chapter

Jonas performed an experiment for his science fair project. He learned that rinsing lettuce in vinegar kills more bacteria than rinsing with water or with a popular commercial product. As a follow-up to his project, he wants to determine the percentage of bacteria killed by rinsing with a diluted solution of vinegar. a. How much water and how much vinegar should be mixed to produce 10 cups of a mixture that is \(40 \%\) vinegar? b. How much pure vinegar and how much \(40 \%\) vinegar solution should be mixed to produce 10 cups of a mixture that is \(60 \%\) vinegar?

Solve the system of equations by using the addition method. (See Examples \(3-4)\) $$ \begin{array}{l} 0.25 x-0.04 y=0.24 \\ 0.15 x-0.12 y=0.12 \end{array} $$

Solve the system by using any method. If a system does not have one unique solution, state whether the system is inconsistent or whether the equations are dependent. (See Examples \(5-6\) ) $$ \begin{array}{l} 3 x-4 y=6 \\ 9 x=12 y+4 \end{array} $$

A patient undergoing a heart scan is given a sample of fluorine- \(18\left({ }^{18} \mathrm{~F}\right)\). After \(4 \mathrm{hr}\), the radioactivity level in the patient is \(44.1 \mathrm{MBq}\) (megabecquerel). After \(5 \mathrm{hr}\), the radioactivity level drops to \(30.2 \mathrm{MBq}\). The radioactivity level \(Q(t)\) can be approximated by \(Q(t)=Q_{0} e^{-k t},\) where \(t\) is the time in hours after the initial dose \(Q_{0}\) is administered. a. Determine the value of \(k\). Round to 4 decimal places. b. Determine the initial dose, \(Q_{0}\). Round to the nearest whole unit. c. Determine the radioactivity level after \(12 \mathrm{hr}\). Round to 1 decimal place.

Use a graphing utility to approximate the solution(s) to the system of equations. Round the coordinates to 3 decimal places. $$ \begin{array}{l} x^{2}+y^{2}=32 \\ y=0.8 x^{2}-9.2 \end{array} $$
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