Problem 14 - College Algebra Essentials

Question

In Exercises \(5-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} 2 x+5 y=1 \\ -x+6 y=8 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution for the system of equations is \(x = -28/17\) and \(y = 16/17\).

Step 1: Isolate a Variable in One Equation

In the second equation \(-x + 6y = 8\), isolate for x by adding x to both sides which simplifies the equation to \(x = 6y - 8\).

Step 2: Substitute Isolated Variable into Second Equation

Sub next the isolated x from the second equation into the first equation, replacing x with \(6y - 8\). This gives the equation \(2*(6y - 8) + 5y = 1\). Simplifying results in \(17y = 16\). Dividing by 17 on both sides gives us \(y = 16/17\).

Step 3: Find the Other Variable

The last step is to substitute \(y = 16/17\) into the rearranged equation of x that we got earlier which is \(x = 6y - 8\)this gives us \(x = 6*(16/17) - 8\). Simplifying results in \(x = -28/17\).

Key Concepts

Linear Equations

Linear equations are algebraic expressions representing straight lines, often in the form of \(ax + by = c\). In our exercise, both equations \(2x + 5y = 1\) and \(-x + 6y = 8\) are linear equations. These equations involve two variables, \(x\) and \(y\). Each equation can be graphed as a straight line on a coordinate plane.

Understanding linear equations is crucial because they represent relationships with constant rates of change. This means that any change in one variable results in a proportional change in the other variable. In the context of solving system of equations, the solution to these equations represents the point where the lines intersect. This is because the intersection point satisfies both equations simultaneously.

When solving linear equations, especially in systems, it's helpful to rewrite or manipulate them to reveal solutions more easily. You may rearrange terms, combine like terms, or use substitution methods, as demonstrated in this exercise.

System of Equations

A system of equations consists of two or more equations with the same variables. In this particular exercise, we are dealing with a system of two linear equations:

• \(2x + 5y = 1\)
• \(-x + 6y = 8\)

The main aim when dealing with a system of equations is to find the value of the variables that satisfies all given equations simultaneously. Solving such systems helps in understanding how two or more algebraic relationships interact with each other.

There are various methods to solve systems of equations, such as graphing, substitution, and elimination. Each method has its own advantages and is suitable for different types of problems. In this exercise, we specifically use the substitution method, where one equation is solved for one variable in terms of the other, and then substituted into the second equation to find a solution. This method is handy when one of the equations is easily rearranged to isolate a variable.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of the variables that make the equation true. In this exercise, we use the substitution method to solve the algebraic equations provided.

Here’s a breakdown of how we solve it:

• Isolate one variable: Start by isolating one variable in one of the equations. From \(-x + 6y = 8\), rearranging gives us \(x = 6y - 8\).
• Substitute the isolated variable: Replace the isolated variable in the other equation. Substitute \(x = 6y - 8\) into \(2x + 5y = 1\), resulting in \(2(6y - 8) + 5y = 1\).
• Solve for the variable: Simplify and solve the new equation for one variable. For example, simplifying \(2(6y - 8) + 5y = 1\) results in \(17y = 16\), giving \(y = \frac{16}{17}\).
• Find the other variable: Substitute \(y = \frac{16}{17}\) back into the original equation for \(x\), producing \(x = -\frac{28}{17}\).

Solving algebraic equations is a foundational skill in algebra, providing crucial practice in manipulating equations and understanding variable relationships.