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Chapter 5: Problem 1

Solve the system of equations. If the system does not have one unique solution, state whether the system is inconsistent or the equations are dependent. $$ \begin{array}{l} x=5 y+12 \\ \frac{1}{2} x=\frac{1}{3}-\frac{1}{3} y \end{array} $$

Short Answer

Expert verified
The solution is \( x = 2 \) and \( y = -2 \).

Step by step solution

01

- Rewrite the second equation

Rewrite the second equation to make it easier to manipulate. Start with: \( \frac{1}{2} x = \frac{1}{3} - \frac{1}{3} y \). Multiply every term by 6 to clear the fractions: \( 6 \times \frac{1}{2} x = 6 \times \frac{1}{3} - 6 \times \frac{1}{3} y \) Simplifying: \( 3x = 2 - 2y \)
02

- Solve one equation for one variable

We already have \( x \) in terms of \( y \) from the first equation: \( x = 5y + 12 \).
03

- Substitute and simplify

Substitute \( x = 5y + 12 \) into the simplified second equation: \( 3(5y + 12) = 2 - 2y \). Expand and simplify: \( 15y + 36 = 2 - 2y \) Combining like terms: \( 15y + 2y = 2 - 36 \) \( 17y = -34 \) Thus, \( y = -2 \).
04

- Solve for the other variable

Now substitute \( y = -2 \) back into the first equation: \( x = 5(-2) + 12 \) Simplify: \( x = -10 + 12 \) \( x = 2 \).
05

- Verify the solution

Substitute \( x = 2 \) and \( y = -2 \) into the original equations to verify: First equation: \( 2 = 5(-2) + 12 \) \( 2 = -10 + 12 \) \( 2 = 2 \) (True) Second equation: \( \frac{1}{2}(2) = \frac{1}{3} - \frac{1}{3}(-2) \) \( 1 = \frac{1}{3} + \frac{2}{3} \) \( 1 = 1 \) (True)

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Key Concepts

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

Linear Equations
Linear equations are mathematical statements that describe a straight line when graphed on a coordinate plane. These are equations of the first degree, meaning they involve variables (like x and y) raised only to the power of one. For example, in our exercise, we have two equations:

\( x = 5y + 12 \)
and
\( \frac{1}{2} x = \frac{1}{3} - \frac{1}{3} y \).

The goal in solving linear equations is to find the specific values of the variables that satisfy both equations simultaneously. This brings us to the methods used to find these solutions, such as the substitution method.
Substitution Method
One efficient way to solve a system of linear equations is the substitution method. This involves solving one of the equations for one variable and then substituting that expression into the other equation.

In our exercise, we took the first equation
\( x = 5y + 12 \)
and solved it for x. We then substituted this expression into the second equation:
\( \frac{1}{2} x = \frac{1}{3} - \frac{1}{3} y \).

By substituting the first equation into the second, we effectively reduced the problem to a single equation with one variable, which is easier to solve. This method is particularly useful when one equation is easy to solve for a particular variable.
Algebraic Manipulation
Algebraic manipulation involves performing operations on equations to isolate variables and solve the system. This can include adding, subtracting, multiplying, or dividing both sides of an equation to simplify it.

In the given problem, we started by simplifying the second equation:
\( \frac{1}{2} x = \frac{1}{3} - \frac{1}{3} y \).
We multiplied every term by 6 to clear the fractions:
\( 3x = 2 - 2y \).

Next, we substituted
\( x = 5y + 12 \)
into this equation and solved for y. After finding y, we substituted it back into the first equation to find x. Such manipulations make complex equations more manageable and lead to straightforward solutions.
Consistent and Dependent Systems
A system of equations can be classified based on the number of solutions it has. If a system has at least one solution, it is called consistent. If it has infinitely many solutions, it is both consistent and dependent.

In our example, we discovered that
\( x = 2 \)
and
\( y = -2 \)
solve both linear equations. This means our system is consistent because a solution exists.
If we had found the equations were multiples of each other, indicating identical lines, the system would be dependent and have infinitely many solutions.

Consistent systems have intersecting lines at one or more points, while dependent systems have coincident lines, sharing all points on the same line.

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Most popular questions from this chapter

Solve the system. $$ \begin{array}{l} (x+3)^{2}+(y-2)^{2}=4 \\ (x-1)^{2}+y^{2}=8 \end{array} $$

The average of an electrician's hourly wage and a plumber's hourly wage is \(\$ 33 .\) One day a contractor hires the electrician for \(8 \mathrm{hr}\) of work and the plumber for \(5 \mathrm{hr}\) of work and pays a total of \(\$ 438\) in wages. Find the hourly wage for the electrician and for the plumber.

A sporting goods store sells two types of exercise bikes. The deluxe model costs the store $$\$ 540$$ from the manufacturer and the standard model costs the store $$\$ 420$$ from the manufacturer. The profit that the store makes on the deluxe model is $$\$ 180$$ and the profit on the standard model is $$\$ 120$$. The monthly demand for exercise bikes is at most \(30 .\) Furthermore, the store manager does not want to spend more than $$\$ 14,040$$ on inventory for exercise bikes. a. Determine the number of deluxe models and the number of standard models that the store should have in its inventory each month to maximize profit. (Assume that all exercise bikes in inventory are sold.) b. What is the maximum profit? c. If the profit on the deluxe bikes were $$\$ 150$$ and the profit on the standard bikes remained the same, how many of each should the store have to maximize profit?

Use a graphing utility to graph the solution set to the system of inequalities. \(y<\frac{4}{x^{2}+1}\) \(y>\frac{-2}{x^{2}+0.5}\)

A college theater has a seating capacity of 2000 . It reserves \(x\) tickets for students and \(y\) tickets for general admission. For parts (a)-(d) write an inequality to represent the given statement. a. The total number of seats available is at most 2000 . b. The college wants to reserve at least 3 times as many student tickets as general admission tickets. c. The number of student tickets cannot be negative. d. The number of general admission tickets cannot be negative. e. Graph the solution set to the system of inequalities from parts (a)-(d).
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