Problem 8
Question
In Exercises 5-14, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} 2 x-y=0 \\ 3 y=6 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution to the system of equations is \(x = 1\) and \(y = 2\)
1Step 1: Isolate Y from the second equation
According to the second equation, \(3y = 6\). divide both sides by 3 to isolate y, which gives us \(y = 6 / 3 = 2\)
2Step 2: Substitute Y value into the first equation
Substitute \(y = 2\) into the first equation. This gives us \(2x - 2 = 0 \). Add 2 to both sides of equation to isolate \(x\), which gives us \(2x = 2\)
3Step 3: Solve for X
Now, divide both sides by 2 to solve for \(x\). This results in \(x = 2 / 2 = 1\)
Key Concepts
Algebraic SubstitutionIsolating VariablesSystem of Equations
Algebraic Substitution
Algebraic substitution is a powerful problem-solving method used in algebra to simplify solving equations or systems of equations. In substitution, a value or an expression is replaced by another that holds the same value. When working with systems of equations, this technique allows us to find a common ground where one equation can be rewritten to include a variable from the other equation.
For instance, if we have two equations like in the provided exercise, we can use this method to replace one variable with an equivalent expression. Once substituted, the resulting single-variable equation is much simpler to solve. The beauty of algebraic substitution is that it turns a potentially complicated situation into a sequence of more straightforward problems that lead progressively to the solution.
For instance, if we have two equations like in the provided exercise, we can use this method to replace one variable with an equivalent expression. Once substituted, the resulting single-variable equation is much simpler to solve. The beauty of algebraic substitution is that it turns a potentially complicated situation into a sequence of more straightforward problems that lead progressively to the solution.
Isolating Variables
Isolating variables is a fundamental step in the process of solving equations and systems of equations. It involves moving all terms containing the variable of interest to one side of the equation and everything else to the other side. The goal is to get the variable alone, or in other words, 'isolated' so that it becomes easy to determine its value.
It is crucial for students to practice this method as it builds a strong foundation for more complex algebraic procedures, such as factoring or solving quadratic equations.
Steps to Isolate Variables:
- Perform arithmetic operations such as addition, subtraction, multiplication, or division to both sides of the equation as needed.
- Use inverse operations to undo complex expressions or functions surrounding the variable.
- Keep track of the operations, as you'll need to apply them to the other equation in a system for consistency.
It is crucial for students to practice this method as it builds a strong foundation for more complex algebraic procedures, such as factoring or solving quadratic equations.
System of Equations
A system of equations is a set of two or more equations with the same set of variables. The goal when solving such a system is to find the values of these variables that will satisfy all the equations simultaneously. Systems can be solved using various methods including graphing, substitution, and elimination.
Solving a system of equations by the substitution method, often preferred for its straightforward approach, involves finding a solution for one variable and then 'substituting' this solution back into another equation to find the remaining variable(s). Exercise improvement advice often highlights the importance of clearly presenting each step to avoid confusion. Proper explanation, as seen in the solved exercise, must address how and why each operation is performed and how it leads us closer to the final solution.
Solving a system of equations by the substitution method, often preferred for its straightforward approach, involves finding a solution for one variable and then 'substituting' this solution back into another equation to find the remaining variable(s). Exercise improvement advice often highlights the importance of clearly presenting each step to avoid confusion. Proper explanation, as seen in the solved exercise, must address how and why each operation is performed and how it leads us closer to the final solution.
Other exercises in this chapter
Problem 8
In Exercises \(7-10\), use a system of linear equations to solve the problem. The selling price of a cellular phone is \(\$ 149.92\). The markup rate is \(60 \%
View solution Problem 8
In Exercises 7-12, solve the system by the method of elimination. $$ \left\\{\begin{array}{r} 4 a+5 b=9 \\ 2 a+5 b=7 \end{array}\right. $$
View solution Problem 8
In Exercises \(5-10\), solve the system by graphing. $$ \left\\{\begin{array}{l} y=\frac{1}{2} x+2 \\ y=-x+8 \end{array}\right. $$
View solution Problem 9
In Exercises \(7-16\), sketch the graph of the system of linear inequalities. $$ \left\\{\begin{array}{l} y
View solution