Problem 20 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{l} 2 x-y=9 \\ 7 x+4 y=1 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(\left( \frac{37}{15}, -\frac{61}{15} \right)\).

Step 1: Solve for y in one equation

Let's take the first equation: \(2x - y = 9\). To isolate \(y\), start by moving \(2x\) to the other side which gives: \(-y = -2x + 9\). Divide everything by -1 to solve for \(y\): \(y = 2x - 9\).

Step 2: Substitute the expression into the other equation

Now that we have \(y = 2x - 9\), substitute this expression into the second equation \(7x + 4y = 1\). Replace \(y\) with \(2x - 9\): \(7x + 4(2x - 9) = 1\).

Step 3: Simplify and solve for x

Expand and simplify the equation: \(7x + 8x - 36 = 1\). Combine like terms: \(15x - 36 = 1\). Add 36 to both sides: \(15x = 37\). Divide by 15 to solve for \(x\): \(x = \frac{37}{15}\).

Step 4: Substitute back to find y

Now that we know \(x = \frac{37}{15}\), substitute it back into the expression \(y = 2x - 9\). Calculate: \(y = 2 \times \frac{37}{15} - 9\). Simplify: \(y = \frac{74}{15} - \frac{135}{15}\). Thus, \(y = -\frac{61}{15}\).

Step 5: Write the solution as an ordered pair

The solution to the system of equations is the ordered pair \(\left( \frac{37}{15}, -\frac{61}{15} \right)\), representing the \(x\) and \(y\) values we found.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. The primary aim is to find the values of these unknowns that satisfy all the given equations simultaneously. For instance, in the problem presented, we have two equations:

• \(2x - y = 9\)
• \(7x + 4y = 1\)

Both equations involve the variables \(x\) and \(y\). Solving a system of equations can show where the equations intersect when plotted as lines on a graph. In this exercise, the substitution method is used, which involves solving one of the equations for one variable and then substituting this expression into the other equation. This method is particularly useful when one of the equations is easily rearranged, allowing for straightforward substitution into the second equation. The final goal is to find a solution expressed as an ordered pair \((x, y)\) that satisfies both equations.

Linear Equations

Linear equations are equations of the first degree, meaning each term is either a constant or the product of a constant and a single variable. They are called linear because their graph is a straight line. The general form of a linear equation in two variables is \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. In our current example, the equations \(2x - y = 9\) and \(7x + 4y = 1\) are both linear equations. Demonstrating how these equations translate into straight lines, each line corresponds to an infinite set of \((x, y)\) pairs that satisfy the equation. When solving a system of linear equations, we are looking for the point(s) where these lines intersect, which represents the solution(s) to the system. Linear equations provide a foundational concept in algebra and serve as a building block for more complex concepts.

Algebra Problem Solving

Algebra problem solving often involves using various methods to find unknown values. Among these methods is the substitution method, which is efficiently applied to systems of linear equations. In our example, we start by isolating one variable in one of the equations. Once isolated, we substitute it into the other equation to find the second variable. Breaking down this process:

• Solve one equation for one of the variables (e.g., solve for \(y\) in \(2x - y = 9\)).
• Substitute this expression into the other equation to find the other variable (substitute \(y = 2x - 9\) into \(7x + 4y = 1\)).
• Solve for the remaining variable (simplify and solve \(15x - 36 = 1\) for \(x\)).
• Substitute the solution back into one of the original equations to find the first variable.
• Combine these solutions to express the final result as an ordered pair.

Understanding these steps strengthens problem-solving skills, offering a systematic approach to tackling algebra problems. Developing proficiency in these methods is essential for delving into more advanced mathematical concepts.