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Chapter 4: Problem 54

Solve each system by substitution. If a system has no solution or infinitely many solutions, so state. $$ \left\\{\begin{array}{l} {y=2 x-9} \\ {x+3 y=8} \end{array}\right. $$

Short Answer

Expert verified
The solution is \( x = 5 \), \( y = 1 \).

Step by step solution

01

Identify the Equations

The system of equations given is: \( y = 2x - 9 \) and \( x + 3y = 8 \). Since the first equation is already solved for \( y \), we can use substitution.
02

Substitute for y in the Second Equation

Replace \( y \) in the second equation with the expression from the first equation: \( x + 3(2x - 9) = 8 \).
03

Simplify the Equation

Distribute the 3 in the equation: \( x + 6x - 27 = 8 \). Combine like terms to get: \( 7x - 27 = 8 \).
04

Solve for x

Add 27 to both sides of the equation: \( 7x = 35 \). Then divide both sides by 7: \( x = 5 \).
05

Substitute Back to Find y

Substitute \( x = 5 \) back into the first equation: \( y = 2(5) - 9 \). Simplify to find \( y = 1 \).
06

Verify the Solution

Substitute \( x = 5 \) and \( y = 1 \) back into the second equation to verify: \( 5 + 3(1) = 8 \). The equation holds true, confirming the solution.

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Key Concepts

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

Understanding the Substitution Method
The substitution method is a straightforward technique to solve a system of linear equations. It's especially useful when one of the equations is already solved for one variable. Here's how it works:
  • Identify the Equation: Find an equation where one variable is isolated. In the exercise, the equation is given as \( y = 2x - 9 \).
  • Substitute: Replace this isolated variable in the other equation with its expression. This helps transform the system into a single variable equation.
  • Solve for the Remaining Variable: You now have an equation in one variable, solve it as usual.
Once the value of one variable is known, use it to find the other variable by substituting back into either equation. This method ensures clarity and often simplifies calculations by reducing the number of variables in the equation.
Solving Linear Equations
To solve linear equations, follow some key steps that make the process organized:
  • Combine Like Terms: If your equation has similar variable terms, combine them. For instance, \( x + 6x \) becomes \( 7x \). Doing this helps simplify your equation.
  • Isolate the Variable: Move constants to the other side of the equation to start isolating your variable term. For example, adding or subtracting values so that one side contains only the variable terms.
  • Perform Operations: Finally, solve the equation by using arithmetic operations. Divide or multiply both sides of the equation to solve for the variable. In our example, we solved \( 7x = 35 \) by dividing both sides by 7, finding \( x = 5 \).
Taking these steps makes solving linear equations methodical and avoids errors, allowing one to find the correct solution efficiently.
Importance of Algebraic Verification
Verification is the final and crucial step in solving any system of equations. After finding a solution for the variables, it's key to ensure that these values satisfy both original equations. Here's how you can verify:
  • Substitute Back: Use the found values (in our example, \( x = 5 \) and \( y = 1 \)) and plug them back into the original equations. Substitute these values into both equations.
  • Check Compatibility: Ensure both equations hold true after substitution. For instance, substituting back into the equation \( x + 3y = 8 \): \( 5 + 3(1) = 8 \), we see the equation is true.
  • Confirm: If both equations are satisfied, your solutions are correct. If not, re-evaluate your steps.
This step confirms the reliability of your solution, ensuring you've accurately solved the problem. Verification builds confidence in your answers and highlights any potential arithmetic errors that might have occurred.

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Most popular questions from this chapter

Write a system of two linear equations such that \((2,3)\) is a solution of the first equation but is not a solution of the second equation.

Explain the error in the following work Solve: $$\left\\{\begin{array}{rlrl}{x+y} & {=1} & {x+y} & {=1} \\ {x-y} & {=5} & {\frac{+x-y}{2 x}} & {=6}\end{array}\right. \frac{2 x}{2}=\frac{6}{2}$$ $$x=3$$ The solution is 3

What concept does this diagram illustrate? $$ \left\\{\begin{array}{l} {6 x+5 y=11} \\ {y=(3 x-2)} \end{array}\right. $$

Gourmet Foods. A New York delicatessen sells marinated mushrooms for \(\$ 12\) a pint and stuffed Kalamata olives for \(\$ 9\) a pint. How many pints of each should be used to get 20 pints of a mixture that will sell for \(\$ 10\) a pint?

When solving a system, what advantages are there with the substitution method compared with the graphing method?
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