Problem 40 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(6 x+2 y=5\) \(8 x+2 y=3\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = -1\) and \(y = 5.5\)

Step 1: Harmonizing the Equations

First, we will align the equations according to their variables and constants for easier calculation. Equations rewritten in simplified form are: \(6x + 2y = 5\) and \(8x + 2y = 3\)

Step 2: Perform Linear Combination

Next, we subtract the second equation from the first. The goal is to eliminate one variable (in this case, y). The process looks as follows: \((6x + 2y) - (8x + 2y) = 5 - 3\) which simplifies to \(-2x = 2\) .

Step 3: Solving for x

Now, we can isolate x by dividing both sides of the equation by -2 which gives us \(x = -1\) .

Step 4: Solving for y using x value

We now insert the x value we calculated into the first original equation to get the value of y. The calculation looks like: \(6(-1) + 2y = 5\) which simplifies to \(-6 + 2y = 5\) and on further simplification we get: \(2y = 11\) and finally, divide by 2 across to get \(y = 5.5\) .

Key Concepts

Understanding Linear Systems

Linear systems are sets of linear equations that share common variables. These equations are essential tools in mathematics, particularly when solving real-life problems that involve finding unknowns with specific constraints. Most commonly, linear systems involve two or more equations, each with two or more unknowns, such as the equations given in the original exercise:

• \(6x + 2y = 5\)
• \(8x + 2y = 3\)

In a graphical context, each equation in a linear system represents a line, and the solution to the system is the point at which these lines intersect. This point represents the values of the variables that satisfy all equations simultaneously.
It's crucial to manipulate these equations methodically to find this intersection point, often through methods like substitution or elimination.

Solving Linear Equations

Solving linear equations involves finding the values of the variables that make the equations true. The process usually starts with simplifying the equations, if necessary, using algebraic techniques such as:

• Adding or subtracting terms from both sides
• Multiplying or dividing all terms by a constant

In the exercise, the method involved performing linear operations to simplify and solve the equations efficiently. First, the equations were aligned and simplified. Then, through subtraction, we isolate one variable at a time.
For instance, extracting the variable \(x\) was achieved by ensuring the terms involving \(y\) canceled each other out, resulting in a straightforward equation to solve: \(-2x = 2\), which simplifies to \(x = -1\). Once \(x\) was known, it could be substituted back to find \(y\).

Elimination Method in Action

The elimination method is a powerful technique in solving systems of linear equations, especially when the goal is to remove one of the variables by adding or subtracting equations. This method hinges on aligning terms so that one variable can cancel out, simplifying the overall solution process.
In the provided example, the equations:

• \(6x + 2y = 5\)
• \(8x + 2y = 3\)

were processed by subtracting the second equation from the first. The subtraction effectively canceled out \(y\), as both equations had the same coefficient for \(y\) (which is \(2\)). Thus, "eliminating" \(y\) reduced the equations to a single variable form: \(-2x = 2\), allowing us to quickly solve for \(x\).
Once \(x\) was found, it was substituted back into one of the original equations to solve for \(y\), demonstrating the efficiency of the elimination method in solving linear systems.