Question

Use the substitution method to solve the linear system. $$\begin{aligned} &y=x-4\\\ &4 x+y=26 \end{aligned}$$

Short answer

The solution to the given system of linear equations using substitution method is \(x = 6, y = 2\).

Step-by-step solution

Substitute Equation 1 in Equation 2

Replace y in the second equation with the equal value from the first equation, which is \(x - 4\). So we get: \(4x + (x - 4) = 26\)

Solve the resulting equation

Simplify the equation to single variable equation: \[5x - 4 = 26\] This will simplify further to: \[5x = 30\] When divided by 5, the value of x: \[x = 6\]

Substitute x into Equation 1

Now use this x-value in the first equation to find the corresponding value of y: \[y = 6 - 4\]

Calculate the value of y

By solving the equation: \[y = 2\]

Key concepts

Substitution Method

The substitution method is a technique used in algebra to solve systems of equations. It involves replacing one variable with an expression obtained from another equation involving the same variable. To apply this method, one must first solve one of the equations for one variable in terms of the other variables. Once this is done, the resulting expression is substituted into the other equations. This process transforms the system of equations into an equivalent system where one of the equations has only one variable. By solving for that variable, we can then go back and find the value of the original variable.

For example, in the given exercise, the substitution method begins by isolating the variable y in the first equation, which gives us the expression \(y = x - 4\). This expression is then substituted into the second equation in place of y, leading us to an equation in one variable. The simplicity of the substitution method makes it a valuable tool for solving linear systems, especially when they are suited for such an approach.

Linear Equations

A linear equation is an algebraic expression that represents a straight line when graphed on a coordinate plane. These equations typically appear in the form \(ax + by = c\), where x and y are variables and a, b, and c are coefficients. A key feature of a linear equation is that it has at most one solution for each variable and does not contain any powers higher than one for its variables.

Understanding the structure of linear equations is essential as they lay the groundwork for more complex algebraic concepts. In the context of systems of equations, linear equations interact with one another to form a system that can be solved for a set of values that satisfy all equations simultaneously. In the exercise, both equations presented are linear, which indicates that a unique solution likely exists provided the equations are independent.

Systems of Equations

A system of equations consists of two or more equations with a common set of unknowns. The goal is to find values for the variables that satisfy all equations in the system. There are several ways to solve such systems, including graphing, substitution, elimination, and matrix methods.

In a system of linear equations, like the one presented in the exercise, the graphical representation would show each equation as a line on the coordinate plane. The solution to the system is the point or points where the lines intersect. Algebraically, this translates to finding a set of values for the variables that make all the equations true when substituted back into the original equations. The given set of linear equations is a simple example of a system that can be solved using algebraic methods such as substitution.

Algebraic Solutions

Algebraic solutions refer to the process and the answers obtained by solving equations or systems of equations using algebraic techniques. These methods include manipulation and simplification of expressions, substituting variables, and applying properties of equality to isolate and solve for unknowns. The aim is to find the precise values of variables that satisfy the given conditions.

When we talk about the algebraic solution of a system of equations, we pin down the exact values of the variables, as opposed to a graphical solution that might only give us an approximation. In the provided exercise, an algebraic solution is achieved by first using the substitution method to find the value of x, and then using that value to solve for y, yielding a precise and exact solution to the system of equations.