Problem 39
Question
Find \(x\) and \(y\) in terms of \(a\) and \(b\). $$\left\\{\begin{array}{ll} x+y=0 & (a \neq 1) \\ x+a y=1 \end{array}\right.$$
Step-by-Step Solution
Verified Answer
\( x = \frac{1}{1-a} \), \( y = -\frac{1}{1-a} \)
1Step 1: Express y in terms of x
Start by manipulating the first equation: \( x + y = 0 \). We can express \( y \) in terms of \( x \) as \( y = -x \). This expression will help us substitute into the second equation.
2Step 2: Substitute y in the second equation
Substitute \( y = -x \) into the second equation \( x + ay = 1 \). This turns the equation into \( x + a(-x) = 1 \). Simplify this to \( x - ax = 1 \).
3Step 3: Factor and solve for x
Factor out \( x \) from the equation: \( x(1-a) = 1 \). Since \( a eq 1 \), divide both sides by \( (1-a) \) to solve for \( x \). Thus, \( x = \frac{1}{1-a} \).
4Step 4: Substitute x back to find y
Using \( y = -x \), substitute the expression for \( x \): \( y = -\frac{1}{1-a} \). Thus, \( y \) is expressed in terms of \( a \).
5Step 5: Final expressions for x and y
Now, we have \( x = \frac{1}{1-a} \) and \( y = -\frac{1}{1-a} \). These are the expressions for \( x \) and \( y \) in terms of \( a \).
Key Concepts
Substitution MethodSolving EquationsExpression of Variables
Substitution Method
The substitution method is a popular technique to solve systems of linear equations. By using substitution, you transform the system into a single equation in one variable. Let's break down this process with our example:
The first equation from the system is: - \( x + y = 0 \).
We can start by solving this equation for \( y \) in terms of \( x \). This results in \( y = -x \).
This expression is now ready for substitution.
Now, with \( y \) expressed as \( -x \), you substitute this expression back into the second original equation:
- \( x + ay = 1 \)
This process is the backbone of the substitution method. It reduces complexity by allowing us to work with a single equation.
In summary, the substitution method is helpful because it simplifies the system into manageable parts, making it easier to find solutions for complex equations.
The first equation from the system is: - \( x + y = 0 \).
We can start by solving this equation for \( y \) in terms of \( x \). This results in \( y = -x \).
This expression is now ready for substitution.
Now, with \( y \) expressed as \( -x \), you substitute this expression back into the second original equation:
- \( x + ay = 1 \)
This process is the backbone of the substitution method. It reduces complexity by allowing us to work with a single equation.
In summary, the substitution method is helpful because it simplifies the system into manageable parts, making it easier to find solutions for complex equations.
Solving Equations
Once we have substituted and simplified our system of equations, the next step involves solving the equation for a particular variable. Following the substitution, our equation becomes:
\( x + a(-x) = 1 \).
By simplifying, we get:
\( x - ax = 1 \).
Here, we are looking to solve for \( x \). Factor out \( x \) as follows:
\( x(1-a) = 1 \).
Solving equations at this step involves isolating the variable of interest. Since \( a eq 1 \), the equation can be rearranged to find \( x \):
\( x = \frac{1}{1-a} \).
Keep in mind, solving equations with factoring is efficient as it directly tackles the variable term, leading to a cleaner solution.
This process exemplifies careful manipulation of algebraic expressions to achieve a clear, single-variable equation.
\( x + a(-x) = 1 \).
By simplifying, we get:
\( x - ax = 1 \).
Here, we are looking to solve for \( x \). Factor out \( x \) as follows:
\( x(1-a) = 1 \).
Solving equations at this step involves isolating the variable of interest. Since \( a eq 1 \), the equation can be rearranged to find \( x \):
\( x = \frac{1}{1-a} \).
Keep in mind, solving equations with factoring is efficient as it directly tackles the variable term, leading to a cleaner solution.
This process exemplifies careful manipulation of algebraic expressions to achieve a clear, single-variable equation.
Expression of Variables
Expressing variables is a crucial part of solving systems of equations. Once you find one variable, you typically express the other using this newly found result.
In our case, by substituting back to find \( y \), we use the relation we had from the beginning: \( y = -x \).
Plug in the expression of \( x \) we found:
\( y = -\frac{1}{1-a} \).
Now both variables, \( x \) and \( y \), are expressed in terms of \( a \).
This demonstrates the importance of creating expressions for variables in terms of other known quantities. It simplifies solving systems of equations and understanding the relationships among variables.
Expressing variables helps in visualizing solutions and ensuring each solution satisfies all given conditions in the problems.
In our case, by substituting back to find \( y \), we use the relation we had from the beginning: \( y = -x \).
Plug in the expression of \( x \) we found:
\( y = -\frac{1}{1-a} \).
Now both variables, \( x \) and \( y \), are expressed in terms of \( a \).
This demonstrates the importance of creating expressions for variables in terms of other known quantities. It simplifies solving systems of equations and understanding the relationships among variables.
Expressing variables helps in visualizing solutions and ensuring each solution satisfies all given conditions in the problems.
Other exercises in this chapter
Problem 39
Solve for \(x\) and \(y\) $$\left[\begin{array}{cc} x & 2 y \\ 4 & 6 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\ 2 x & -6 y \end{array}\right]$$
View solution Problem 39
Solve the system of linear equations. $$\left\\{\begin{array}{l} -x+2 y+z-3 w=3 \\ 3 x-4 y+z+w=9 \\ -x-y+z+w=0 \\ 2 x+y+4 z-2 w=3 \end{array}\right.$$
View solution Problem 40
Find the partial fraction decomposition of the rational function. $$\frac{2 x^{2}-x+8}{\left(x^{2}+4\right)^{2}}$$
View solution Problem 40
Find the inverse of the matrix. $$\begin{aligned} &\left[\begin{array}{llll} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d \end{array}\right]
View solution