Question

MULTIPLE CHOICE Find the \(x\) -value of the solution for the linear system below. \(y=2 x-2\) \(y=3 x+1\) \((\mathbf{A})-8\) \((\mathbf{B})-3\) (c) 3 (D) 8

Short answer

The \(x\) - value of the solution for this linear system is -3. So, the correct option is \(\mathbf{(B)}-3\).

Step-by-step solution

Set equations equal to each other

Start the solution by setting the two provided equations equal to each other: \[2x - 2 = 3x + 1\]. This will find the \(x\) value at which the two lines intersect.

Rearrange to isolate \(x\)

Next, rearrange the previous equation by moving terms via subtraction or addition to isolate \(x\) on one side and have a numerical value on the other side: \[2x - 3x = 2 + 1\]. This simplifies the equation to \[-x = 3\]

Solve

At the moment, \(x\) is equal to \(-3\), so the final step is to multiply both sides of the equation by \(-1\) to get \(x = -3\).

Key concepts

Linear Equations

Linear equations form the foundation of algebra and are equations of the first degree. This means they contain only variables raised to the power of one. They graph as straight lines and are characterized by their slope and y-intercept, determining their steepness and position on the y-axis, respectively.

In the context of our exercise, we encounter two linear equations, \( y = 2x - 2 \) and \( y = 3x + 1 \). Each equation can be visualized as a straight line on a graph. The intersection of these lines corresponds to the solution of the system, which is the coordinate pair \( (x, y) \) shared by both equations. The intersection represents the values of \( x \) and \( y \) that satisfy both equations simultaneously. Our initial task was to find the \( x \) value at this intersection point.

Algebraic Manipulation

Algebraic manipulation is a suite of techniques used to solve for unknown variables in equations. In practical terms, it involves adding, subtracting, multiplying, and dividing both sides of the equation by the same number, or manipulating the equation in such a way to isolate the variable of interest.

When we set \( 2x - 2 = 3x + 1 \) to solve our linear system, we're utilizing algebraic manipulation. To find the value of \( x \) that satisfies both equations, we isolate \( x \) on one side. We do this by, for example, subtracting \( 2x \) from both sides, resulting in \( -2 = x + 1 \). Following this, we subtract \( 1 \) from both sides, hence isolating \( x \) as \( x = -3 \). This is the cornerstone of algebraic manipulation—using such operations in a strategic manner to solve for unknown variables.

Systems of Equations

A system of equations is a set of two or more equations that have the same set of variables. The solutions to these systems are the values that satisfy all equations simultaneously. There are various methods to solve these systems, such as graphing, substitution, elimination, and matrix methods.

In this exercise, we approach solving a linear system by substitution. This method involves finding an expression for one variable in terms of the other in one equation and then substituting this expression into the other equation. However, our exercise simplifies the substitution by setting the two expressions for \( y \) equal to each other since they are both equal to \( y \) for the same value of \( x \) at the intersection. Systematic substitution as seen here allows us to combine the equations and solve for one variable, paving the way to find the solution for the entire system.