Problem 49 - Introductory Algebra for College Students

Question

When using substitution to solve $$\left\\{\begin{array}{rr}5 x-4 y= & 9 \\\x-2 y= & -3\end{array}\right.$$ I find it easiest to solve for \(x\) in the first equation.

Step-by-Step Solution

Answer

The solutions of the given system of equations are \(x = 5\) and \(y = 4\).

Step 1: Solve the equation for x

The second equation in the system is \(x = 2y - 3\), which has already solved for x.

Step 2: Substitute for x in the first equation

Replace x in the first equation with 2y - 3: \[5(2y - 3) - 4y = 9\]. This simplifies to \(10y - 15 - 4y = 9\) or \(6y = 24\). Solving this gives \(y = 4\).

Step 3: Substitute y in the original second equation to find x

Replace y in the second equation with 4: \(x = 2*4 - 3 = 5\). Thus, the solution of the system is x = 5 and y = 4.

Step 4: Check the solution in both of the initial equations

Replace x by 5 and y by 4 in the initial equations: \[5*5 - 4*4 = 9\] and \[5 - 2*4 = -3\]. Both of these equations are valid, thus the solutions for x and y are correct.

Key Concepts

Understanding the Substitution Method

The substitution method is a powerful tool for solving systems of equations. In essence, this method involves replacing one variable with another by expressing it in terms of the second variable. This is particularly useful when one of the equations in the system already solves for one variable, as it simplifies the process of finding the solution.

For instance, in the system of equations given in the exercise, the second equation explicitly solves for x. By expressing x in terms of y, or vice versa, you can substitute this expression into the other equation. This turns the system of equations into a single equation with one variable, which is typically much easier to solve.

Once you solve for one variable, you replace this value back into your substituted expression to find the value of the other variable. The substitution method is not only a foundational strategy in algebra but also an essential skill for solving more complex mathematical problems.

Finding Algebraic Solutions

Algebraic solutions require a methodical approach to manipulating equations. The goal is to isolate one variable, making it easy to solve the equation step by step. It's crucial to maintain equal balance by performing the same operations on both sides of an equation. This ensures that the equation's properties remain unchanged.

In the exercise provided, we started by isolating x in the second equation, resulting in an expression that allows substitution into the first equation.When simplifying, ensure to combine like terms and pay careful attention to the arithmetic operations involved. During the process, equation balancing is still imperative. Once you've solved for y, you substitute that value back into the equation used for the initial substitution to find the value for x. Despite seeming simple, algebraic solutions can sometimes involve complex manipulations, so keeping your steps clear and logical is key for success.

Working with Linear Equations

Linear equations form the basis for much of algebra. They are equations of the first degree, meaning they contain variables that are to the power of one. The general form of a linear equation in two variables is given by Ax + By = C, where A, B, and C are coefficients. Linear equations graph as straight lines on the Cartesian plane.

The system provided in the textbook consists of two linear equations. When solving such a system, the goal is to find a common solution that satisfies both equations simultaneously. This common solution represents a point of intersection between the two lines if graphed on the Cartesian plane. The combination of these two core techniques—graphical representation and algebraic manipulation—provides a complete understanding of linear equations and their solutions.