Problem 58 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{c} 8 x-2.8 y=4 \\ 0.3 x+y=11.2 \end{array}\right.$$

Step-by-Step Solution

Answer

x = 4, y = 10

Step 1 - Write down the equations separately

Extract the equations from the given system: 1) \(8x - 2.8y = 4\) 2) \(0.3x + y = 11.2\)

Step 2 - Solve the second equation for y

From equation (2) \(0.3x + y = 11.2\), solve for \(y\): \(y = 11.2 - 0.3x\)

Step 3 - Substitute y in the first equation

Substitute \(y\) from Step 2 into equation (1) \(8x - 2.8(11.2 - 0.3x) = 4\)

Step 4 - Simplify and solve for x

Performing the substitution and then simplifying: 8x - 2.8 \(11.2 - 0.3x\) = 4 8x - 31.36 + 0.84x = 4 8.84x = 35.36 \Rightarrow x = 4\

Step 5 - Solve for y using x

Substitute \(x = 4\) back into the expression for \(y\) obtained in Step 2: \(y = 11.2 - 0.3 \cdot 4 \Rightarrow y = 10\)

Step 6 - Write the solution as an ordered pair

The solution to the system is \(x, y = (4, 10)\)

Key Concepts

solving by substitution

Solving systems of equations by substitution is a method where we solve one equation for one variable and then substitute that expression into another equation. This allows us to find the values of both variables step-by-step. In our exercise, we had two equations: \(8x - 2.8y = 4\) and \(0.3x + y = 11.2\). We first solved the second equation for \(y\): \y = 11.2 - 0.3x\. Then, we substituted this expression into the first equation. By substituting, we transformed the system into a single equation with one variable, making it easier to solve. This process helps to isolate and find the values of the unknowns efficiently.

linear equations

Linear equations are equations of the first degree, meaning they have no exponents greater than one and form straight lines when graphed. In this exercise, we dealt with two linear equations: \(8x - 2.8y = 4\) and \(0.3x + y = 11.2\). Each equation represents a straight line on a coordinate plane. The solution to the system of linear equations is the point where these two lines intersect. Understanding that each linear equation represents a constraint, and their intersection represents a common solution, helps in visualizing and solving the system.

algebraic manipulation

Algebraic manipulation involves using basic algebraic rules to simplify equations and expressions. In the given exercise, we used it to substitute expressions and find solutions step-by-step. For instance, after expressing \(y\) from \(0.3x + y = 11.2\) as \(y = 11.2 - 0.3x\), we substituted it into the first equation: \(8x - 2.8(11.2 - 0.3x) = 4\). Through careful algebraic manipulation, we expanded, simplified, and isolated \(x\) to find its value. These steps demonstrate how algebraic manipulation helps in transforming and solving equations effectively.

ordered pairs

An ordered pair represents a point in the coordinate system, typically written as \( (x, y) \). In the context of systems of equations, an ordered pair is the solution where both equations intersect. For our exercise, the solution was found to be \( (4, 10) \). This means that plugging \( x = 4 \) and \( y = 10 \) into both of our original equations would satisfy them, confirming that \( (4, 10) \) is the intersection point. Understanding ordered pairs is crucial as it gives us a concrete solution to where the variables meet the conditions of both equations.