Question

Solve the linear system. $$ \begin{aligned} &3 x-5 y=8\\\ &-2 x+3 y=3 \end{aligned} $$

Short answer

The solution for the linear system is \(x = -27\) and \(y = - \frac{89}{5}\)

Step-by-step solution

Isolation of Variable in First Equation

Isolate the variable y in the first equation:\[y = \frac{1}{5}(3x - 8)\]

Substitution in Second Equation

Find x by substituting the above value of y into the second equation:\[-2x + 3(\frac{1}{5}(3x - 8)) = 3\]This simplifies to:\[-2x + \frac{9}{5}x - \frac{24}{5} = 3\]Combine like terms and solve for x:\[-\frac{1}{5}x = \frac{27}{5}\]Multiply by -5 to isolate x:\[x = -27\]

Back Substitution

Substitute x back into the first equation:\[3(-27) - 5y = 8\]This simplifies to:\[-81 - 5y = 8\]Solve for y:\[-5y = 89\]Thus, \[y = -\frac{89}{5}\]

Key concepts

System of Equations

A system of equations is a collection of two or more equations with a set of variables. The objective is to find the values of these variables that satisfy all the given equations simultaneously. In the context of linear algebra, we often deal with linear systems, which are made up of linear equations like the ones shown in the provided exercise.

Linear systems can have either one solution, no solution, or infinitely many solutions. When we have exactly one solution, the system is termed 'consistent and independent'. The solution to a system of linear equations is the point or set of points where the graphs of the equations intersect. In the problem given, we're looking to find the specific values for x and y that satisfy both equations.

Substitution Method

The substitution method is a technique used to solve systems of equations. This method involves solving one of the equations for one variable in terms of the others, then substituting this expression into the other equations. This effectively reduces the number of equations by one. The substitution method is particularly useful when one equation can be easily solved for one variable.

For instance, in our exercise, we solved for y in the first equation and substituted it into the second, allowing us to find the value of x. Once x is known, y can be found by plugging the value of x back into one of the original equations, a process known as back substitution.

Isolating Variables

When isolating variables, you manipulate an equation to express one variable in terms of the other variables. This involves performing algebraic operations such as adding, subtracting, multiplying, dividing, or factoring, while keeping the equation balanced.

The goal is to get the target variable alone on one side of the equation, usually with a coefficient of 1. In the challenge we faced with our example, the first step was isolating variable y in the first equation. By isolating variables systematically, you simplify the process of solving equations, as seen when we isolated the variable x to find its value.

Linear Algebra

Linear algebra is one of the foundations of mathematics, particularly useful for solving systems of equations. It deals with vectors, matrices, and linear transformations. It allows us to use matrix operations to efficiently solve systems of equations, amongst other things.

In linear algebra, a system of linear equations can be represented as a matrix equation. The solution to these systems, when they exist, can be found using techniques like graphing, substitution, elimination, or more advanced methods such as matrix operations or determinants. Our example had us solve a system of linear equations, which is a cornerstone application of linear algebra.