Chapter 5: Problem 23
Solve the system of equations. $$\left\\{\begin{aligned} x+4 z &=1 \\ x+y+10 z &=10 \\ 2 x-y+2 z &=-5 \end{aligned}\right.$$
Short Answer
Expert verified
The solutions to the system of equations are \( x = 1 \), \( y = 9 \), and \( z = 0 \).
Step by step solution
01
Substitute the first equation into the others
From the first equation \(x=1-4z\). Substitute this into the second and third equations to eliminate x. This results in: \(1-4z+y+10z=10\) and \(2(1-4z)-y+2z=-5\)
02
Simplify the equations
Simplify the equations obtained in Step 1 by performing operations like addition, subtraction, multiplication, or division. This results in: \(6z+y=9\), \(2-8z-y+2z=-5\).
03
Solve for y
From the equations in step 2, we can express both in terms of y, giving us: \(y=9-6z \) and \( y=7-6z \). equating the two, It's determined that z must be zero for these equations to hold true.
04
Substitute z into other equations
Substitute \( z=0 \) into the equations for x and y to get: \( x=1-4(0)=1 \) and \( y=9-6(0)=9 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Algebraic Substitution
One of the foundational techniques in solving systems of equations is algebraic substitution. This method involves expressing one variable in terms of another using one of the equations and then 'substituting' this expression into the other equations.
For example, given a system of equations like: \[x + 4z = 1\] A student could solve for x and get \[x = 1 - 4z\]This expression for x can then be used in place of x in the other equations of the system. This step simplifies the system, as it reduces the number of variables. Simplifying complex problems into simpler parts is a key strategy in effectively understanding and solving algebraic equations.
When applying substitution, it's essential to ensure that each substituted variable is isolated on one side of the equation. This involves being comfortable with basic algebraic operations such as addition, subtraction, multiplication, and division. Mastery of algebraic substitution not only helps with systems of equations but is a valuable skill across all fields of mathematics and science.
For example, given a system of equations like: \[x + 4z = 1\] A student could solve for x and get \[x = 1 - 4z\]This expression for x can then be used in place of x in the other equations of the system. This step simplifies the system, as it reduces the number of variables. Simplifying complex problems into simpler parts is a key strategy in effectively understanding and solving algebraic equations.
When applying substitution, it's essential to ensure that each substituted variable is isolated on one side of the equation. This involves being comfortable with basic algebraic operations such as addition, subtraction, multiplication, and division. Mastery of algebraic substitution not only helps with systems of equations but is a valuable skill across all fields of mathematics and science.
System of Linear Equations
A system of linear equations consists of two or more linear equations that share the same set of variables. Linear equations are called 'linear' because, when graphed, they produce a straight line. The point where these lines intersect is the solution to the system, representing the values of the variables that satisfy all the equations simultaneously.
For instance, the system given: \[\begin{aligned} x + 4z &= 1 \ x + y + 10z &= 10 \ 2x - y + 2z &= -5\end{aligned}\]represents three planes in a three-dimensional space. The point where all three planes meet is the solution (x, y, z). Solving such systems can be achieved through various methods, including graphing, algebraic substitution, elimination, and matrix operations. Among these, algebraic substitution and elimination are most often taught in early algebra courses due to their straightforward and systematic approaches.
For instance, the system given: \[\begin{aligned} x + 4z &= 1 \ x + y + 10z &= 10 \ 2x - y + 2z &= -5\end{aligned}\]represents three planes in a three-dimensional space. The point where all three planes meet is the solution (x, y, z). Solving such systems can be achieved through various methods, including graphing, algebraic substitution, elimination, and matrix operations. Among these, algebraic substitution and elimination are most often taught in early algebra courses due to their straightforward and systematic approaches.
Solving for Variables
The process of solving for variables requires manipulation of equations to find the values of unknowns. This entails operations that maintain the equality of the equations, such as adding the same number to both sides. In a system of equations, the goal is to determine the values of each variable that will satisfy all equations at once.
In complex systems, like the one in our example, you start by solving for one variable in terms of another, reducing the severity of the system. You continue doing so until you can isolate one variable completely. Our exercise started with isolating x to find \[x = 1 - 4z\]Then, this information was used to find y in terms of z, giving two equations with both y expressed as a function of z. Equating the two expressions for y provided the value of z. Once you have the value for one variable, in this case, z, it's just a matter of substituting back into the other equations to find the remaining variables, x, and y—bringing this meticulous process full circle. Solving for variables is a dance of substitution and simplification until all variables are accounted for.
In complex systems, like the one in our example, you start by solving for one variable in terms of another, reducing the severity of the system. You continue doing so until you can isolate one variable completely. Our exercise started with isolating x to find \[x = 1 - 4z\]Then, this information was used to find y in terms of z, giving two equations with both y expressed as a function of z. Equating the two expressions for y provided the value of z. Once you have the value for one variable, in this case, z, it's just a matter of substituting back into the other equations to find the remaining variables, x, and y—bringing this meticulous process full circle. Solving for variables is a dance of substitution and simplification until all variables are accounted for.
Equation Simplification
The process of equation simplification is a crucial skill in algebra. Simplifying an equation means to make it as straightforward as possible by performing all possible operations. This includes combining like terms, distributing multiplications over additions or subtractions, and reducing fractions to their lowest terms.
Take for instance the equations \[1 - 4z + y + 10z = 10 \]and\[2(1 - 4z) - y + 2z = -5\]In the solution steps provided, these were simplified to \[6z + y = 9\]and\[2 - 8z - y + 2z = -5\]Simplifying equations reduces the chance for error in subsequent steps and makes it easier to solve for variables. It is akin to organizing one's thoughts before addressing a problem, which helps in maintaining clarity throughout the solving process. Without simplification, it would be difficult to identify the relationships between variables and to perform operations that bring one closer to the solution.
Take for instance the equations \[1 - 4z + y + 10z = 10 \]and\[2(1 - 4z) - y + 2z = -5\]In the solution steps provided, these were simplified to \[6z + y = 9\]and\[2 - 8z - y + 2z = -5\]Simplifying equations reduces the chance for error in subsequent steps and makes it easier to solve for variables. It is akin to organizing one's thoughts before addressing a problem, which helps in maintaining clarity throughout the solving process. Without simplification, it would be difficult to identify the relationships between variables and to perform operations that bring one closer to the solution.
