Problem 49 - Beginning and Intermediate Algebra

Question

Three solutions of an equation are given. Use a system of three equations in three variables to find the constants and write the equation. $$\begin{array}{l} A x+B y+C z=12 ; \\ \left(1, \frac{3}{4}, 3\right),\left(\frac{4}{3}, 1,2\right), \text { and }(2,1,1) \end{array}$$

Step-by-Step Solution

Answer

The given solutions are plugged into the equation Ax + By + Cz = 12 to form a system of linear equations, which is then solved to find the constants A, B, and C. The resulting equation is \(\frac{7}{3}x + \frac{16}{3}y + 2z = 12\).

Step 1: Write down the system of equations with given solutions

We are given three solutions of the equation Ax + By + Cz = 12. We will plug in these solutions (1, 3/4, 3), (4/3, 1, 2), and (2, 1, 1) as values for x, y, and z in the equation to obtain a system of three linear equations: 1. A(1) + B(3/4) + C(3) = 12 2. A(4/3) + B(1) + C(2) = 12 3. A(2) + B(1) + C(1) = 12

Step 2: Solve the system of equations

To find the values of A, B, and C, we will solve the system of equations: 1. A + (3/4)B + 3C = 12 2. (4/3)A + B + 2C = 12 3. 2A + B + C = 12 First, let's multiply Eq. 1 by 4 and Eq. 2 by 3 so that all three equations have the same coefficients for B: 1. 4A + 3B + 12C = 48 2. 4A + 3B + 6C = 36 3. 2A + B + C = 12 Now, subtract equation 2 from equation 1 to get a new equation: new equation 4: 6C = 12 Divide by 6: C = 2 Next, let's substitute C = 2 back into equation 3: 2A + B + 2 = 12 Subtract 2 from both sides: 2A + B = 10 Now, multiply equation 3 by 2 and subtract the result from equation 1 to get a new equation: new equation 5: 3B + 8C = 32 Substitute C = 2 into equation 5: 3B + 8(2) = 32 Subtract 16 from both sides: 3B = 16 Divide by 3: B = 16/3 Finally, substitute C = 2 and B = 16/3 into equation 3: 2A + (16/3) = 10 Subtract 16/3 from both sides: 2A = (30 - 16)/3 Divide by 2: A = 14/6 Simplify: A = 7/3 Now we found A=7/3, B=16/3, and C=2.

Step 3: Write down the equation with the found constants

Substitute the values of A, B, and C we found into the Ax + By + Cz = 12 equation: (7/3)x + (16/3)y + 2z = 12 And this is the equation we were looking for.

Key Concepts

Linear Equations

Linear equations form the foundation of algebraic problem-solving. A linear equation is an equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can be written in the standard form, which is usually expressed as:

• Ax + By + Cz = D

where A, B, C, and D are constants, and x, y, and z are variables.
These variables typically stand for unknown values that we need to discover. If you visualize a linear equation with two variables in a two-dimensional plane, it forms a straight line, hence the name "linear". In three-dimensional space, using three variables, the representation becomes a plane.

One key aspect of linear equations is that they imply a linear relationship between the variables involved. This means that changes in one variable result in proportional changes in another, without any curves or higher-degree powers of variables. Understanding the representation and characteristics of linear equations can help demystify more complex algebraic expressions later on.

Solutions of Equations

When dealing with systems of equations, a solution represents the set of values that satisfies all equations simultaneously. In a system with linear equations, each equation is a constraint on the values of the variables. The objective is to find a point (x, y, z) that lies on the plane represented by each equation in the system.
Let's break this process down:

• **Substitution**: You replace one of the variables with an expression from another equation.
• **Elimination**: This involves adding or subtracting equations to eliminate one variable, thereby reducing the system's complexity.
• **Graphical Representation**: Visualizing solutions can entail plotting each equation in a common space and finding their intersection point.

In the case of linear systems, there can be one solution (the lines or planes intersect at a single point), infinitely many solutions (the lines or planes overlap), or no solution (the lines or planes do not intersect at all).
Understanding solutions in systems of equations helps develop a broader competence in solving varying types of algebraic problems.

Algebraic Equations

Algebraic equations encompass a broad category of equations involving variables and constants. Within this realm, algebraic equations can be classified by degree and structure. A linear equation is a first-degree algebraic equation, meaning the highest power of the variable is one.
Taking a close look at what algebraic equations involve:

• **Variables**: Represent unknown values we intend to solve for.
• **Constants**: Known values that provide constraints or conditions in the equation.
• **Operators**: Such as addition, multiplication, and sometimes division.

Algebraic equations can also take many forms, from simple linear equations to more complex quadratic or cubic ones. Linear examples are less complicated and often serve as an introduction to understanding the principles behind more elaborate equations.
With practice, interpreting and working with algebraic equations not only enhances problem-solving skills but also equips learners with techniques applicable in various scientific and engineering fields. The familiarity with manipulating these equations lays the groundwork for more advanced studies in mathematics and other analytical disciplines.