Problem 39 - Algebra 2

Question

Solve each system of equations. $$ \left\\{\begin{array}{l}{-x+y+z=5} \\ {2 x+y-z=2} \\ {3 x+2 y+4 z=0}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is $$x = -\frac{13}{8},$$ $$y = \frac{69}{16},$$ $$z = -\frac{15}{16}.$$

Step 1 - Write down the system of equations

Start by writing down the given system of linear equations:1. $$-x + y + z = 5$$ (Equation 1)2. $$2x + y - z = 2$$ (Equation 2)3. $$3x + 2y + 4z = 0$$ (Equation 3)

Step 2 - Eliminate variable z from Equations 1 and 2

Add Equation 1 and Equation 2 together to eliminate variable z. This yields:$$(-x + y + z) + (2x + y - z) = 5 + 2$$Simplify to get:$$x + 2y = 7$$ (Equation 4)

Step 3 - Express x in terms of y from Equation 4

Isolate variable x in Equation 4:$$x = 7 - 2y$$ (Equation 5)

Step 4 - Substitute Equation 5 into Equation 3

Substitute x from Equation 5 into Equation 3 to find z in terms of y:$$3(7 - 2y) + 2y + 4z = 0$$Simplify to obtain a relationship between y and z:$$21 - 6y + 2y + 4z = 0$$$$21 - 4y + 4z = 0$$$$4z = 4y - 21$$Divide both sides by 4 to solve for z:$$z = y - \frac{21}{4}$$ (Equation 6)

Step 5 - Substitute Equation 6 into Equation 2 to find y

Replace z in Equation 2 with the expression from Equation 6:$$2x + y - (y - \frac{21}{4}) = 2$$Now substitute x from Equation 5 into the above equation:$$2(7 - 2y) + y - y + \frac{21}{4} = 2$$Simplify to solve for y:$$14 - 4y + \frac{21}{4} = 2$$Multiply through by 4 to eliminate the fraction:$$56 - 16y + 21 = 8$$Combine like terms:$$77 - 16y = 8$$$$-16y = 8 - 77$$$$-16y = -69$$Divide both sides by -16:$$y = \frac{69}{16}$$ (Equation 7)

Step 6 - Find z using Equation 6 and y from Equation 7

Substitute the value of y from Equation 7 into Equation 6 to find z:$$z = \frac{69}{16} - \frac{21}{4}$$To combine the terms, convert \(\frac{21}{4}\) to have the same denominator as \(\frac{69}{16}\):$$z = \frac{69}{16} - \frac{84}{16}$$Now subtract to find the value of z:$$z = \frac{69 - 84}{16}$$$$z = -\frac{15}{16}$$ (Equation 8)

Step 7 - Find x using Equation 5 and y from Equation 7

Plug the value of y from Equation 7 into Equation 5 to determine x:$$x = 7 - 2\times\frac{69}{16}$$$$x = 7 - \frac{138}{16}$$Convert 7 to have the same denominator as \(\frac{138}{16}\):$$x = \frac{112}{16} - \frac{138}{16}$$Now subtract to find the value of x:$$x = \frac{112 - 138}{16}$$$$x = -\frac{26}{16}$$$$x = -\frac{13}{8}$$ (Equation 9)

Key Concepts

System of Equations

When we speak of a 'system of equations', we refer to a collection of two or more linear equations with the same set of variables. The goal here is to find the variable values that will satisfy all equations in the system simultaneously.

A system of linear equations can have one solution, infinitely many solutions, or no solution at all. In our exercise, we are working with a system of three linear equations with three variables, which generally matches the conditions for a unique solution if the equations are independent and consistent.

Elimination Method

One of the techniques to solve a system of linear equations is the elimination method. This approach involves manipulating the equations at hand to eliminate one or more variables. Once a variable is eliminated, the system is simplified, making it easier to solve.

The step-by-step solution provided for the exercise demonstrates the elimination method effectively by first adding Equation 1 and Equation 2 to eradicate the variable z, simplifying the process of solving for the remaining variables, x and y.

Substitution Method

The substitution method, another strategy for solving systems of equations, entails expressing one variable in terms of another and then substituting this expression into the other equations. This method turns a system of equations into simpler and more manageable forms, often leading to a solution through consecutive substitutions.

The solution to the exercise exemplifies this by expressing x in terms of y, and then substituting this expression into other equations to solve for y and z ultimately.

Linear Algebra

Linear algebra is the mathematical discipline that deals with vectors, vector spaces, linear transformations, and systems of linear equations. It provides the vocabulary and the formal tools to describe and solve equations involving linear functions, which is why the problems like the one in the exercise belong to the realm of linear algebra.

The techniques used in our exercise, including the elimination and substitution methods, are fundamental concepts in linear algebra that are applied to find solutions for linear systems.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like x, y, or z), and operators (such as add, subtract, multiply, and divide). The solution steps in our exercise evolved around manipulating these algebraic expressions to simplify and eventually resolve the system of equations, showcasing the importance of understanding how to work with algebraic expressions effectively.