Problem 84
Question
Solve by expressing \(x\) and \(y\) in terms of \(a\) and \(b\) : $$ \left\\{\begin{array}{l} x-y=a \\ y=2 x+b \end{array}\right. $$
Step-by-Step Solution
Verified Answer
Solutions are \(x = -a - b\) and \(y = -2a - b\)
1Step 1: Substitution
Substitute \(y\) from the second equation into the first equation, so it becomes \(x - (2x + b) = a\).
2Step 2: Simplify the Equation for 'x'
Combine similar terms and solve for 'x'. The equation simplifies to be: \(-x - b = a\) or \(x = -a - b\).
3Step 3: Substitute 'x' into the Second Equation
Substitute \(x\) from the simplified equation into the second equation to find 'y'. Value of 'y' becomes: \(y = 2(-a - b) + b\).
4Step 4: Simplify the Equation for 'y'
Combine similar terms to solve for 'y'. The equation simplifies to be: \(y = -2a - b\).
Key Concepts
Linear EquationsSubstitution MethodAlgebraic Expressions
Linear Equations
When we talk about linear equations, we're referring to algebraic equations where each term is either a constant or the product of a constant and a single variable. These equations form straight lines when graphed on a coordinate plane. The general form of a linear equation in two variables, like x and y, is ax + by = c, where a, b, and c are constants.
The system given in the exercise is a classic example of a system of linear equations with two unknowns, x and y. The uniqueness of linear equations is that they have a single solution (where the lines intersect), no solution (if the lines are parallel and don't intersect), or infinitely many solutions (if the two equations describe the same line). Learning to solve these systems is foundational in algebra, as it applies to a multitude of mathematical and real-life problems.
The system given in the exercise is a classic example of a system of linear equations with two unknowns, x and y. The uniqueness of linear equations is that they have a single solution (where the lines intersect), no solution (if the lines are parallel and don't intersect), or infinitely many solutions (if the two equations describe the same line). Learning to solve these systems is foundational in algebra, as it applies to a multitude of mathematical and real-life problems.
Substitution Method
The substitution method is a widely utilized technique for solving a system of equations. This method involves solving one of the equations for one variable and then substituting the resulting expression into the other equation. This substitution provides an equation in only one variable, which can usually be solved quite easily. Once one variable is found, it can be substituted back into one of the original equations to solve for the other variable.
The steps provided in the exercise demonstrate the substitution method succinctly—starting with solving the second equation for y and then substituting the expression for y in terms of x and b into the first equation. By following this process, we systematically reduce a system of equations down to one equation with one variable, then solve that single equation. This method is particularly effective when one of the equations is already solved for one of the variables.
The steps provided in the exercise demonstrate the substitution method succinctly—starting with solving the second equation for y and then substituting the expression for y in terms of x and b into the first equation. By following this process, we systematically reduce a system of equations down to one equation with one variable, then solve that single equation. This method is particularly effective when one of the equations is already solved for one of the variables.
Algebraic Expressions
An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. Expressions are the building blocks of algebra and are used to describe relationships and changes between quantities. They don't have an equal sign as equations do; therefore, they cannot be solved but can be simplified. In the context of the provided exercise, when we express x and y in terms of a and b, we are essentially writing algebraic expressions for these variables.
An important skill when working with algebraic expressions is combining like terms and using distributive properties to simplify the expressions. In step 2 and step 4, as outlined in the solution, we use these techniques to simplify the algebraic expressions for x and y in terms of a and b. Simplifying algebraic expressions is an essential step in solving equations because it can significantly reduce the complexity of the calculations required to find a solution.
An important skill when working with algebraic expressions is combining like terms and using distributive properties to simplify the expressions. In step 2 and step 4, as outlined in the solution, we use these techniques to simplify the algebraic expressions for x and y in terms of a and b. Simplifying algebraic expressions is an essential step in solving equations because it can significantly reduce the complexity of the calculations required to find a solution.
Other exercises in this chapter
Problem 82
Perform the indicated operation. \(-3-(-9)\) (Section \(1.7,\) Table 1.7 )
View solution Problem 83
Perform the indicated operation. \(-3(-9)\) (Section \(1.7,\) Table 1.7 )
View solution Problem 84
Will help you prepare for the material covered in the next section. In each exercise, solve the given equation. $$4 x-3(-x-1)=24$$
View solution Problem 85
The point of intersection of the graphs of the equations \(A x-3 y=16\) and \(3 x+B y=7\) is \((5,-2) .\) Find \(A\) and \(B\)
View solution