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Chapter 7: Problem 15

Solve the system by the method of elimination and check any solutions algebraically. $$\left\\{\begin{array}{l} 5 x+3 y=6 \\ 3 x-y=5 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 1.5\) and \(y = -0.5\)

Step by step solution

01

Match Coefficients

We want to eliminate one variable by making the coefficients of that variable the same in both equations. Multiplying the second equation by 3 yields \(9x - 3y = 15\). Now the system of equations is \[ \begin{cases} 5x + 3y = 6 \ 9x - 3y = 15 \end{cases} \]
02

Eliminate Variable

We can add the two equations together to eliminate y. This gives \(14x = 21\). After simplifying, we find that \(x = 1.5\).
03

Substitute and Solve

Substitute \(x = 1.5\) into the first equation. This results in \(5(1.5) + 3y = 6\), simplifying to \(7.5 + 3y = 6\), and finally \(y = -0.5\).
04

Check the Solution

Substitute \(x = 1.5\) and \(y = -0.5\) into both original equations. Both equations hold true, meaning the solution to this system of equations is \(x = 1.5\) and \(y = -0.5\)

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Key Concepts

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

Elimination Method
The elimination method is a popular technique used to solve systems of linear equations. It works by eliminating one of the variables, allowing you to solve for the remaining variable. Here's how it generally works:
  • First, you manipulate the equations so that one variable has the same coefficient in both equations, but with opposite signs.
  • By adding or subtracting these equations, the chosen variable cancels out, letting you solve for the other variable with ease.
  • Once you have one variable, you can substitute it back into one of the original equations to find the other variable.
In our exercise, we focused on adjusting the second equation so that the coefficients of y are equal and opposite. We multiplied the second equation by 3, which gave us a new system with aligned coefficients for y. Now, by adding these equations, the y terms cancel each other out, leaving us with a single equation in terms of x.
Algebraic Solution
An algebraic solution involves resolving equations using algebra rules rather than graphical or numerical methods. The process relies on manipulating equations to isolate variables.
  • Equations are manipulated based on standard arithmetic operations: addition, subtraction, multiplication, and division.
  • We aim to isolate variables to easily solve for them one at a time.
  • A thorough check is important to ensure the solution satisfies the original equations.
In this example, after employing the elimination method to resolve one equation with only x, dividing both sides by 14 allows us to find that the value of x is 1.5. It鈥檚 essential to recalculate the equations with these variables to check the accuracy of the solution. Hence, by substituting x back into either equation, the value of y is found, confirming the solution is correct.
Substitution
The substitution method is another fundamental technique to solve systems of equations, where you solve one of the equations for one variable and then substitute this expression into the other equation. Here鈥檚 a concise way to understand it:
  • First, solve one of the equations for one variable in terms of the other.
  • Then, replace this variable in the other equation with the expression found in the first step.
  • This method gives a single equation with one variable, which you can solve directly.
In our case, after determining x using the elimination method, we employed substitution by plugging x back into the first equation. Solving for y needed simple arithmetic to find its value. Substitution, thus, acts as an effective way to verify solutions and ensure consistency in results, providing an excellent cross-check alongside the elimination method.

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