Problem 20 - Precalculus with Limits

Question

Solving a System by Substitution In Exercises \(15-24\) , solve the system by the method of substitution. $$\left\\{\begin{aligned} 0.5 x+3.2 y &=9.0 \\ 0.2 x-1.6 y &=-3.6 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 18\) and \(y = 0\).

Step 1: Isolate \(x\) in the second equation

We will start by isolating \(x\) in the second equation: \(0.2x = 1.6y + 3.6\). Dividing both sides by \(0.2\), it results in \(x = 8y + 18\).

Step 2: Substitute \(x\) in the first equation

Now, replace \(x\) in the first equation (\(0.5x + 3.2y = 9\)) by the expression we found in Step 1: \(0.5(8y + 18) + 3.2y = 9\). Simplifying gives \(4y + 9 + 3.2y = 9\), which simplifies to \(7.2y = 0\), and thus \(y = 0\).

Step 3: Substitute \(y\) into the isolated equation from Step 1 to find \(x\)

Now replace \(y\) with \(0\) in the formula derived from Step 1: \(x = 8(0) + 18\), thus \(x = 18\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of equations. This approach involves isolating one variable in one of the equations, and then substituting that expression into the other equation. It effectively reduces the system from two equations with two variables to a single equation with one variable.

For example, in our original problem, we used the second equation to solve for \(x\). This turned the second equation into an expression solely in terms of \(y\): \(x = 8y + 18\). This substitution simplifies the system, allowing us to solve for \(y\) in the first equation directly.

The benefits of the substitution method are its straightforwardness and its applicability to various types of systems, especially when one equation is simple to solve for one variable. When using this method, it's essential to ensure all calculations and algebraic manipulations are carefully executed to avoid errors.

Linear Equations

Linear equations are equations of the first order, meaning they involve only linear terms (terms to the power of one). They can be written in the form of \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

In the given exercise, both equations are linear: \(0.5x + 3.2y = 9.0\) and \(0.2x - 1.6y = -3.6\). The variables \(x\) and \(y\) have coefficients (0.5, 3.2, 0.2, and -1.6) and no exponents other than one, indicative of linear equations. This property makes systems of linear equations well-suited for the substitution method.

Linear equations have straightforward graphical representations: they are lines on a Cartesian plane. The intersection of these lines represents the solution to the system of equations, giving the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Algebraic Manipulation

Algebraic manipulation involves the rearrangement of algebraic expressions and equations to simplify, solve, or transform them into a more convenient form. This is a foundational skill necessary for executing the substitution method effectively.

In the solution process, algebraic manipulation was used to isolate \(x\) in the second equation: from \(0.2x - 1.6y = -3.6\) to \(x = 8y + 18\). This was achieved by first adding \(1.6y + 3.6\) to both sides and then dividing everything by \(0.2\), ensuring that the equation remained balanced.

Additionally, algebraic manipulation was key in substituting the expression for \(x\) into the first equation and simplifying it to solve for \(y\). Simplification allowed us to reduce \(0.5(8y + 18) + 3.2y = 9\) to \(7.2y = 0\), leading directly to \(y = 0\). Mastery of manipulation techniques is crucial for problem-solving in algebra.