Problem 36 - Algebra 1: Concepts and Skills

Question

Which linear system has been correctly solved for one of the variables from the following system? $$ \begin{aligned} &2 x-y=-1\\\ &2 x+y=-7 \end{aligned} $$ $$ f. \quad \begin{aligned} &2 x-y=-1\\\ &y=2 x-7 \end{aligned} $$ $$ g.\quad \begin{aligned} &2 x-y=-1\\\ &y=-2 x+7 \end{aligned} $$ $$ h. \quad \begin{aligned} &y=2 x+1\\\ &2 x+y=-7 \end{aligned} $$ $$ j. \quad \begin{aligned} &y=-2 x-1\\\ &2 x+y=-7 \end{aligned} $$

Step-by-Step Solution

Answer

The linear system 'f' is the one that has been correctly solved for one of the variables from the original system.

Step 1 - Identify the Original System

The original system of linear equations is: \[ \begin{aligned} &2 x-y=-1\ &2 x+y=-7 \end{aligned} \]

Step 2 - Identify Candidate Systems

The candidate systems for comparison are: \[ \begin{aligned} &2 x-y=-1\ &y=2 x-7 \end{aligned} \] \[ \begin{aligned} &2 x-y=-1\ &y=-2 x+7 \end{aligned} \] \[ \begin{aligned} &y=2 x+1\ &2 x+y=-7 \end{aligned} \] \[ \begin{aligned} &y=-2 x-1\ &2 x+y=-7 \end{aligned} \]

Step 3 - Rearrange Original System

Rearrange the original system for y in terms of x by adding \(2x\) to both sides of the equation to get \( y = 2x + 1 \) for the first equation and subtract \(2x\) from both sides of the equation to get \( y = -2x - 7 \) for the second equation.

Step 4 - Identify Correct System

The system \(f\) \[ \begin{aligned} &2 x-y=-1\ &y=2 x-7 \end{aligned} \] fits the rearranged original system where y is expressed in terms of x from the second equation. Systems \(g\), \(h\), and \(j\) don't match our rearranged original equations, hence they are incorrect.

Step 5 - Verify Correct System

In order to verify the defined system \(f\), one can substitute \(y = 2x - 7\) into the first equation of the system and check that the equation is fulfilled.

Key Concepts

Linear Systems

A linear system is a collection of two or more linear equations involving the same set of variables. In the example from the exercise, the linear system is composed of two equations:

• \(2x - y = -1\)
• \(2x + y = -7\)

The goal when dealing with a linear system is to find values for the variables that satisfy all the equations simultaneously. This means we want to find the point(s) where all the lines represented by these equations intersect, which translates to finding the solution(s) for the variables \(x\) and \(y\) in this case.
A solved linear system indicates one or more variables have been isolated and expressed in terms of others. This action directly leads to finding specific values or relationships consistent across all equations.

Algebraic Equations

Algebraic equations involve one or more variables and a set of mathematical operations. When dealing with linear algebraic equations, which are first-degree equations, the general form is \(ax + by = c\) where \(a\), \(b\), and \(c\) are constants.
The primary task is to transform these equations to make them easier to interpret or solve. In our context, we needed to express one of the variables in terms of the others to find a common solution that satisfies all involved equations. This transformation process often involves rearranging equations by adding, subtracting, or even substituting values among the equations in the system.
Understanding these transformations is key to identifying the correct solution among a set of candidate systems. Seeing if one equation has been correctly isolated can help confirm if the solutions are consistent with the original system.

Variable Isolation

Variable isolation involves rewriting an equation so that one of the variables stands alone on one side of the equation, usually to find a specific solution. It's a fundamental method used to solve systems of equations, as seen in the original exercise.
In our example, isolating \(y\) from the equation \(2x - y = -1\) by rearranging terms gives us \(y = 2x + 1\). This means we've explicitly expressed \(y\) in terms of \(x\), simplifying our problem to one variable consideration in each equation.
The process often involves using basic algebraic operations:

• Adding or subtracting terms to both sides.
• Multiplying or dividing each side by a constant.

By mastering these techniques, students can shift complex systems into more manageable forms, facilitating easier problem-solving and analysis. In our example, the correct candidate system \(f\) was identified through precise variable isolation, confirming that the system aligns with the derived equations.