Problem 15 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}y=\frac{2}{3} x-1 \\ 5 x-7 y=9\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \( x = 6 \) and \( y = 3 \).

Step 1: Substitute Expression for y

We start with the first equation: \( y = \frac{2}{3}x - 1 \). We substitute this expression for \( y \) in the second equation \( 5x - 7y = 9 \). This gives us:\[ 5x - 7 \left( \frac{2}{3}x - 1 \right) = 9 \]

Step 2: Simplify the Equation

Distribute \(-7\) in the equation:\[ 5x - \frac{14}{3}x + 7 = 9 \]

Step 3: Combine Like Terms

Combine \( 5x \) and \( -\frac{14}{3}x \) by finding a common denominator:\[ \frac{15}{3}x - \frac{14}{3}x + 7 = 9 \]\[ \frac{1}{3}x + 7 = 9 \]

Step 4: Isolate x

Subtract 7 from both sides to isolate terms with \( x \):\[ \frac{1}{3}x = 2 \]

Step 5: Solve for x

Multiply both sides by 3 to solve for \( x \):\[ x = 6 \]

Step 6: Substitute x back to find y

Substitute \( x = 6 \) back into the equation \( y = \frac{2}{3}x - 1 \):\[ y = \frac{2}{3} \cdot 6 - 1 \]\[ y = 4 - 1 = 3 \]

Step 7: Write the Solution

The solution to the system of equations is \( x = 6 \) and \( y = 3 \).

Key Concepts

Substitution Method

The substitution method is a handy tool in algebra for solving systems of equations, especially linear equations. Here's why it's effective: when you have two equations, you express one of the variables in terms of the other using one of the equations. This substitution turns the second equation into a single-variable equation, which is easier to solve.

• You start by solving one of the equations for either variable. In our exercise, we solved for y in the first equation, finding that y = \( \frac{2}{3}x - 1 \).
• Next, substitute this expression into the other equation. This helps in reducing a system of equations into a single one, which you'll solve like any basic equation.

This method is particularly useful when one of the equations is already solved for a variable, as it was here with y. The main aim is to simplify and solve step-by-step to reach the values of the unknown variables. This structured approach ensures no step is skipped, leading you efficiently to the solution of the system of equations.

Linear Equations

Linear equations are foundational in algebra, representing straight lines when graphed on a coordinate plane. Understanding these is crucial because they serve as building blocks for more complex concepts.

• Linear equations can often be recognized by their standard form, which is Ax + By = C, where A, B, and C are constants.
• In a system of linear equations like ours, each equation represents a line. The solution to the system is the point where these lines intersect, or meet.

In our particular problem, we transformed our second equation 5x - 7y = 9 using substitution. The steps of substitution and simplification help in determining the point of intersection. Once we found x and y, we verified this solution as the intersection point. This is because linear equations in a system can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (same line).

Algebraic Expressions

Algebraic expressions are made up of variables, constants, and operational symbols. They're central to forming the equations we solve using algebra. Understanding how to manipulate these expressions is key to mastering algebra.

• An algebraic expression consists of terms, which can include numbers, variables, and coefficients. For instance, in our problem \( \frac{2}{3}x - 1 \) is an expression with a variable term and a constant.
• The process of substitution involves replacing variables with expressions, or numerical values. When we substituted \( y = \frac{2}{3}x - 1 \) into the second equation, we essentially worked with a new expression to solve for x.

Manipulating algebraic expressions requires careful attention to mathematical operations like addition, subtraction, multiplication, and division. Each operation has rules that ensure expressions are correctly simplified. This is crucial for both understanding the expressions you're working with and executing the steps needed to solve them correctly in systems of equations.