Question

Use the given statements to write a system of equations. Solve the system by elimination. The sum of a number \(x\) and a number \(y\) is \(13 .\) The difference of \(x\) and \(y\) is 3 .

Short answer

The solution to the system of equations is \(x = 8\) and \(y = 5\).

Step-by-step solution

Form the System of Equations

The problem provides two pieces of information that can be established as two equations. The first statement is 'The sum of a number \(x\) and a number \(y\) is \(13\)', which translates to: \(x + y = 13 \). The second statement is 'The difference of \(x\) and \(y\) is \(3\)' that translates to: \(x - y = 3 \)

Solve by Elimination

To solve the system by elimination, we will manipulate the equations such that adding them together will eliminate one of the variables. In this case, if we add the two equations together: \( (x + y) + (x - y) = 13 + 3 \), the \(y\) variables will cancel out, yielding \(2x = 16\)

Solve for x

To solve for \(x\), we divide both sides of the equation by 2: \(2x/2 = 16/2 \), which yields \(x = 8\)

Solve for y

With the value of \(x\) now known, we can substitute \(x=8\) into either of the original equations to find \(y\). Using the equation \(x + y = 13\), replace \(x\) with 8: \(8 + y = 13\). To find the value of \(y\), we will subtract 8 from both sides of the equation: \(8 - 8 + y = 13 - 8 \), which yields \(y = 5\)

Key concepts

Algebra

Algebra is a branch of mathematics that utilizes mathematical statements to describe relationships between variables, typically denoted by letters, and constants, which are numeric values. In algebra, we are often tasked with solving equations that contain one or more unknowns. A fundamental skill in algebra is the ability to formulate and manipulate these equations to find the values of the unknown variables.

For example, in the given exercise, numbers x and y are variables, representing quantities that we need to determine. Algebraic operations, like addition and subtraction, as well as more complex processes such as factoring, are used to isolate these variables and solve for their values. Understanding algebra is crucial since it provides a basis for more advanced fields of mathematics and applications in real-world problem-solving scenarios.

Elimination Method

The elimination method is a strategy used in algebra to solve systems of linear equations. This method involves adding or subtracting equations to eliminate one of the variables, making it possible to solve for the other variable.

In our exercise, to apply the elimination method, we must align the equations and perform operations that will cancel out one of the variables. As shown in the steps, by adding the equations \(x + y = 13\) and \(x - y = 3\), we are able to eliminate the variable y, because \(y - y = 0\). This leaves us with an equation in one variable, \(2x = 16\), which is straightforward to solve. The elimination method is particularly useful because it can be applied to any system of linear equations, regardless of the number of variables involved, as long as the system has an equivalent number of equations.

Linear Equations

Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable. A linear equation looks like \(ax + b = 0\), where a and b are constants. They are called 'linear' because the graph of these equations is always a straight line. When we deal with two variables, these lines are plotted in a two-dimensional space and display how one variable is related to another.

The system of equations given in the exercise comprises two linear equations. When we graph such equations, their point of intersection represents the solution set for the system. If the lines intersect at a single point, the system has one solution; if they are parallel, there is no solution; if they are the same line, there are infinitely many solutions. Solving them algebraically, as in the exercise, gives us the precise point of intersection, that is, the values of x and y that satisfy both equations simultaneously.