Problem 20 - Algebra 1: Concepts and Skills

Question

Choose a solution method to solve the linear system. Explain your choice, and then solve the system. $$ \begin{array}{c} {x-2 y=4} \\ {6 x+2 y=10} \end{array} $$

Step-by-Step Solution

Answer

The solutions to the linear system of equations are x = 2 and y = -1

Step 1: Identifying the System of Equations

Identify the system of equations in hand. \[ \begin{array}{c} {x-2 y=4} \ {6 x+2 y=10} \end{array}\]

Step 2: Applying the Elimination Method

In the given system, you can notice that the coefficients of the variable y in both equations have the same magnitude but opposite signs. Therefore, you can apply the elimination method directly by adding the two equations together. This serves to eliminate the variable y from the system. \(x - 2y + 6x + 2y = 4 + 10\) simplifies to \(7x = 14\)

Step 3: Solving for the First Variable

Divide each side by 7 to find the value of x = 2.

Step 4: Substituting \[x\] into one of the original equations

Substitute \(x = 2\) into the first original equation, \[x - 2y = 4\], to find the solution for y. The equation becomes \[2-2y=4\]

Step 5: Solving for the Second Variable

Simplify and solve for y. Firstly, subtract 2 from both sides to get \[-2y = 2\]. Then divide each side by -2 to find that \(y = -1\)

Key Concepts

System of Equations

A system of equations is a set of two or more equations that have the same variables and are solved simultaneously. The goal is to find a solution that satisfies all equations in the system. In the given problem, we have two linear equations:

\[\begin{array}{c}{x-2y=4} \{6x+2y=10}\end{array}\]

This pair of equations must be solved together to determine the values of x and y that make both equations true. There are three common methods to solve systems of equations: graphing, substitution, and elimination. The best method to choose depends on the structure of the equations and coefficients involved.

Elimination Method

The elimination method is a technique for solving systems of equations where one variable is eliminated by adding or subtracting the equations. This method is particularly useful when the coefficients of one of the variables are the same or opposites. In this problem, the coefficients of y are opposites (-2 and 2), making it a perfect candidate for the elimination method.

By adding the two equations:

\(x - 2y + 6x + 2y = 4 + 10\)

you can eliminate the y terms:

\(7x = 14\)

After elimination, you're left with an equation that has a single variable, x, which is then easy to solve for. Once x is found, it can be substituted into one of the original equations to find the value of y. The elimination method streamlines the process and provides a clear path to find the solution.

Algebraic Solutions

Algebraic solutions refer to finding the exact values of variables in equations through algebraic manipulations. This is in contrast to graphical solutions, which may only give approximate values based on visual interpretations of graphs. In our system of equations:

\(7x = 14\)

we obtain the algebraic solution for x by dividing both sides by 7:

\(x = 2\)

And then, by substituting x back into one of the original equations, we can solve for y algebraically as well:

\(2 - 2y = 4\)

With further manipulations:

\(-2y = 2\)

And finally:

\(y = -1\)

This step-by-step algebraic approach ensures we find the precise values of x and y that satisfy both original equations simultaneously.