Problem 22
Question
In Exercises 21-26, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} x-5 y=5 \\ 3 x-15 y=15 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The system of equations is dependent, and therefore, it has infinitely many solutions.
1Step 1: Write the First Equation in Terms of x
The first equation is \(x - 5y = 5\). Solving this equation for \(x\), we obtain \(x = 5y +5\).
2Step 2: Substitute \(x\) into the Second Equation
Substitute \(5y + 5\) for \(x\) in the second equation \(3x - 15y = 15\), giving \(3(5y + 5) - 15y = 15\). This simplifies to \(15y + 15 - 15y = 15\), and then further simplifies to \(15 = 15\).
3Step 3: Analyze the Result
The result \(15 = 15\) is a true statement, which indicates that the equations are dependent and have infinitely many solutions. That is, any value of \(y\) will suffice for the variable \(y\), and replacing \(y\) in the equation for \(x\) will give the corresponding \(x\) value.
Key Concepts
Substitution MethodSystem of EquationsAlgebraic SolutionDependent System
Substitution Method
The substitution method for solving systems of equations involves replacing one variable with an expression from another equation that relates the two variables. This way, we reduce a two-variable system to a single equation with one variable. Here's how you can apply it:
This method is particularly useful when one of the equations is easily solvable for one of the variables, as was the case in our example where we solved for x in terms of y.
- Isolate one variable in one of the equations.
- Substitute the expression for this variable into the other equation.
- Solve the resulting single-variable equation.
- Substitute the solution back into one of the original equations to find the value of the other variable.
This method is particularly useful when one of the equations is easily solvable for one of the variables, as was the case in our example where we solved for x in terms of y.
System of Equations
A system of equations is a set of two or more equations with the same set of variables. In algebra, our goal is often to find the values of these variables that satisfy all equations simultaneously. Systems can be solved using various methods, including graphing, substitution, and elimination. When using the substitution method, it becomes important to look at the structure of the equations and determine the most efficient way to isolate and substitute the variables. The given system x - 5y = 5 and 3x - 15y = 15, showcases two linear equations where multicollinearity (a multiple of one equation is the other) indicates a special case within systems of equations.
Algebraic Solution
An algebraic solution to a system of equations provides a set of values for the variables that satisfy all of the equations. The primary steps involve manipulation of algebraic expressions and following systematic procedures, like substitution. When we worked through the substitution steps in the example, the algebra did not lead us to a specific solution for y. Instead, we ended up with the tautology 15 = 15. This indicates something unique about the system, leading us to further investigation rather than a straightforward single solution for each variable.
Dependent System
A dependent system of equations has an infinite number of solutions. This happens when the two equations are essentially the same line on a graph; one equation is just a multiple of the other. When substituting one equation into the other results in a true statement, like 15 = 15, it's a clear sign that we are dealing with a dependent system. Therefore, any value for y will yield a corresponding x when replaced in the equation x = 5y + 5. Recognising a dependent system is crucial since the method of solution will differ from the more common scenarios where there are distinct solutions.
Other exercises in this chapter
Problem 22
Compare the one-variable method with the two-variable method for modeling real-life problems.
View solution Problem 22
In Exercises \(19-24\), solve the system by the method of elimination. $$ \left\\{\begin{aligned} x-2 y &=0 \\ 0.2 x+0.8 y &=2.4 \end{aligned}\right. $$
View solution Problem 22
In Exercises 19-26, solve the system by graphing. $$ \left\\{\begin{array}{r} -x+10 y=30 \\ x+10 y=10 \end{array}\right. $$
View solution Problem 23
In Exercises 23-28, sketch the graph of the system of linear inequalities. $$ \left\\{\begin{array}{r} -3 x+2 y-2 \\ 2 x+y
View solution