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Solve each system by substitution. Determine whether each system is independent, inconsistent, or dependent. $$ \begin{array}{l} \frac{1}{2} x-\frac{1}{3} y=12 \\ \frac{1}{4} x-\frac{1}{2} y=1 \end{array} $$

Short Answer

Expert verified
(34, 15), independent

Step by step solution

01

- Solve the first equation for x

Start with the first equation: \[ \frac{1}{2} x - \frac{1}{3} y = 12 \]Multiply everything by 6 to clear the fractions: \[ 6 \left( \frac{1}{2} x - \frac{1}{3} y \right) = 6 (12) \]This simplifies to: \[ 3x - 2y = 72 \]Solve for x: \[ 3x = 2y + 72 \]\[ x = \frac{2}{3} y + 24 \]
02

- Substitute expression for x into the second equation

Take the expression for x from Step 1: \[ x = \frac{2}{3} y + 24 \]Substitute it into the second equation: \[ \frac{1}{4} \left( \frac{2}{3} y + 24 \right) - \frac{1}{2} y = 1 \]Multiply everything by 12 to clear fractions: \[ 12 \left( \frac{1}{4} \left( \frac{2}{3} y + 24 \right) \right) - 12 \left( \frac{1}{2} y \right) = 12 (1) \]Simplify the equation: \[ 3 \left( \frac{2}{3} y + 24 \right) - 6y = 12 \]\[ 2y + 72 - 6y = 12 \]
03

- Solve the resulting equation for y

Combine like terms: \[ 2y - 6y + 72 = 12 \]This simplifies to: \[ -4y + 72 = 12 \]Move 72 to the right side of the equation: \[ -4y = 12 - 72 \]\[ -4y = -60 \]Divide by -4: \[ y = 15 \]
04

- Solve for x using the expression from Step 1

Now that y has been found, substitute it back into the expression for x: \[ x = \frac{2}{3} y + 24 \]\[ x = \frac{2}{3} (15) + 24 \]\[ x = 10 + 24 \]\[ x = 34 \]
05

- Determine the type of system

The solution to the system is \((34, 15)\). Since there is exactly one solution, the system is independent.

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Key Concepts

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

substitution method
The substitution method is a popular technique for solving systems of linear equations. Here’s how it works:

1. Choose one of the equations and solve it for one variable in terms of the other variable.
2. Substitute this expression into the other equation. This step replaces the original variable with an equivalent expression involving the other variable.
3. Solve the new equation for the remaining variable.
4. Once you have one variable's value, substitute it back into the expression found in step 1 to find the value of the other variable.

This method effectively reduces the problem from two equations and two unknowns to a single equation with one unknown. It's especially useful when one of the equations is already solved for a variable. In the example provided, first, you solve one of the equations for x, then substitute this expression into the other equation to find y.
independent system
An independent system of equations has exactly one solution pair \((x, y)\). This means that the two lines representing the equations intersect at exactly one point.

In our example, after solving the system we found a single solution \( (34, 15) \). This indicates that the system is independent.

In contrast, a dependent system has infinitely many solutions (the same line), while an inconsistent system has no solutions (parallel lines that never intersect). The nature of the system can often be determined by analyzing the slopes and intercepts of the lines or by using algebraic techniques such as substitution or elimination.

So, if you find exactly one pair that satisfies both equations, your system is independent.
solving linear equations
Solving linear equations involves finding the values of the variables that satisfy the equations. Here are some steps to follow:

1. Simplify each equation if necessary (remove any fractions by multiplying by the least common multiple or rearranging terms).
2. Use algebraic methods such as substitution, elimination, or matrix techniques to combine the equations and reduce the number of variables.
3. Isolate the variable on one side of the equation to solve for its value.
4. Substitute the found value back into the original or modified equation to find the other variable.

Linear equations are foundational in algebra. Mastery of techniques for solving them is crucial for more complex mathematical topics.
fraction simplification
Fraction simplification is an important step in solving equations, especially when dealing with systems of linear equations. Here’s how to handle fractions effectively:

1. Identify the fractions in the equation and find the least common denominator (LCD).
2. Multiply every term in the equation by the LCD to eliminate the fractions.

For instance, in our example, the first equation was multiplied by 6 and the second by 12. This step clears the fractions and simplifies the equation to a more familiar form, making it easier to solve.

Simplifying fractions can prevent mistakes and make the algebraic manipulation more straightforward.
algebraic manipulation
Algebraic manipulation refers to the skills and operations used to rearrange and simplify equations. Here's a guide to effective manipulation:

1. Combining like terms: Ensure terms on one side of the equation are combined to simplify it.
2. Moving terms: Use addition or subtraction to move terms from one side of the equation to the other.
3. Factoring and distributing: Apply these techniques for simplification or to solve quadratic equations.
4. Isolating variables: Rearrange the equation to get the variable alone on one side.

In our example, algebraic manipulation was used at every step to simplify the equations and isolate the variables. Mastery of these techniques ensures you can handle increasingly complex problems.

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Most popular questions from this chapter

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