Problem 47 - Thinking Mathematically

Question

In Exercises 47-48, solve each system for \(x\) and \(y\), expressing either value in terms of a or b, if necessary. Assume that \(a \neq 0\) and \(b \neq 0\). \(\left\\{\begin{array}{l}5 a x+4 y=17 \\ a x+7 y=22\end{array}\right.\)

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1/a\) and \(y = 3\).

Step 1 Solve for \(x\) in the first equation

You will start by solving for \(x\) in the first equation: \(5ax + 4y = 17\) rearranged gives: \(x = (17 - 4y)/(5a)\)

Step 2 Substitute \(x\) in the second equation

Next we substitute \(x\) into the second equation: \(ax + 7y = 22\) becomes \((a(17 - 4y))/(5a) + 7y = 22\). Multiply through by \(5a\) to clear the fraction: \(a(17 - 4y) + 35ay = 110a\). Distribute \(a\), and group terms to solve for \(y\): \(17a - 4ay + 35ay = 110a\), \(31ay = 93a\), \(y = 93/31\).

Step 3 Substitute \(y\) in the first equation

Now we put \(y = 93/31\) into the first equation to solve for \(x\): \(5ax + 4(93/31) = 17\), \(x = (17 - (372/31))/(5a)\), \(x = (527 - 372)/(155a)\), \(x = 155/(155a)\)

Step 4 Simplify the results

Upon simplification we get: \(x = 1/a\) and \(y = 3\).

Key Concepts

Linear Equations

Linear equations are fundamental to algebra and represent relationships where the variables involved change in proportion to each other. These equations take the form of \( ax + by = c \), where \(x\) and \(y\) are variables, and \(a\), \(b\), and \(c\) are constants with \(a\) and \(b\) not both zero. The graph of a linear equation is always a straight line in a two-dimensional plane.

In the context of a system of linear equations, we are working with two or more equations which we aim to solve simultaneously. The goal is to find a common solution that satisfies all equations in the system. This particular exercise presents a system of two linear equations with two variables, which often intersect at a single point, \( (x, y) \), signifying the unique solution to the system.

Algebraic Methods

Algebraic methods encompass various techniques to solve equations and systems of equations. One powerful aspect of algebra is its ability to manipulate equations to isolate variables, which allows us to find their values systematically. When facing a system of linear equations, methods such as substitution, elimination, and matrix approaches (like the Gauss-Jordan method) are commonly used.

The exercise provided demonstrates an algebraic approach by strategically isolating one variable and then substituting its value into the other equation. This is a stepping stone for solving the entire system. By engaging in algebraic manipulations like distributing, combining like terms, and simplifying expressions, we establish a path to the solutions for \(x\) and \(y\) within the given constraints, namely \(a \eq 0\) and \(b \eq 0\).

Substitution Method

The substitution method is a technique for solving systems of equations where you solve one equation for one variable and then substitute that expression into the other equation. This process reduces the system from two equations in two variables to a single equation in one variable.

Let's delve into the provided exercise: In Step 1, the expression for \(x\) is derived from the first equation and then substituted into the second, as seen in Step 2. Following this substitution, the second equation now only contains the variable \(y\), making it solvable. After finding the value for \(y\), it's plugged back into any of the original equations to solve for \(x\), as shown in Step 3. This method is particularly useful when equations in the system can be easily manipulated to express one variable in terms of the others, as it was with this particular system where \(x\) was expressed in terms of \(y\) and \(a\).