Question

Solve for \(x, y,\) and \(z\) in the system of equations. Explain each of your solution steps. \(3 x+2 y+z=42\) \(2 y+z+12=3 x\) \(x-3 y=0\)

Short answer

The solutions to the system of equations are \(x = 8.1\), \(y = 2.7\) and \(z = 12.3\).

Step-by-step solution

Re-arrange the third equation

The third equation can be re-arranged to \(x = 3y\) in order to make it easier to substitute \(x\) in the other equations.

Substitute value of x in first and second equations

Now, substitute \(x = 3y\) in first and second equations, you get \[3(3y) + 2y + z = 42\] and \[2y + z + 12 = 3(3y)\], which simplifies to \[9y + 2y + z = 42\] and \[2y + z + 12 = 9y\]. Further simplification gives \[11y + z = 42\] and \[z = 9y - 12\].

Substitute value of z from second to first equation

Next substitute the value of \(z\) from \[z = 9y - 12\] into \[11y + z = 42\], gives \[11y + 9y - 12 = 42\].

Solve for y

By simplifying the equation \[11y + 9y - 12 = 42\], you get \[20y = 54\] or \(y = 54 / 20 = 2.7\).

Find x and z using values of y

Substitute \(y = 2.7\) in \[x = 3y\] and \[z = 9y - 12\] to find the values of \(x\) and \(z\). After substitution, you get \(x = 8.1\) and \(z = 12.3\) respectively.

Key concepts

Substitution Method

The substitution method is a cornerstone of algebra, especially when solving systems of linear equations. It is particularly useful when one of the equations in the system can be solved for one variable in terms of the others. In the example provided, the third equation was rearranged to express one variable, in this case, x, purely in terms of another, y. This new equation, x = 3y, is then used as a substitution in the other equations.

Once the substitution is made, the system is effectively reduced from three to two equations with two unknowns, which simplifies the problem. By replacing x with 3y in the first two equations, new equations that only contain y and z are created. This technique systematically reduces the number of variables, thus making it possible to solve for the unknowns one at a time. In practice, finding opportunities for substitution can streamline solving complex systems and can often lead to quicker solutions.

Algebraic Manipulation

Algebraic manipulation involves the use of algebraic operations to simplify or rearrange equations and expressions. In solving our system of equations, after substitution, we engage in algebraic manipulation by combining like terms and isolating variables. For instance, in step 3, 11y + 9y - 12 = 42 is simplified to 20y = 54 by adding the y terms together and moving the constant to the other side.

Such manipulations adhere to the fundamental rules of algebra; performing the same operation on both sides of an equation to maintain equality. Mastering this skill is essential because it can significantly affect the complexity of solving a problem. Other examples of algebraic manipulation include factoring, expanding, and using the distributive property. These techniques transform the appearance of an equation to reveal a clearer path to the solution.

Linear Equations

Linear equations are the simplest form of equations you'll tackle in algebra. They represent relationships between variables that plot as straight lines when graphed on a coordinate plane. The equations in our set, 3x + 2y + z = 42, 2y + z + 12 = 3x, and x - 3y = 0, are all linear because each term is either a constant or the product of a constant and a single variable.

Recognizing a linear equation is about looking for the degree of the variables, which should always be one. The standard form of a linear equation is Ax + By + Cz = D, with A, B, and C being coefficients, and D a constant. Solving a system of linear equations, like in our example, involves finding the values for x, y, and z that make all the equations true simultaneously. Ideally, the intersection points of the graphs of the equations correspond to the solution of the system.