Problem 25
Question
In Exercises \(1-44,\) solve each system by the addition method. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets. $$\left\\{\begin{array}{l} 2 x-y=3 \\ 4 x+4 y=-1 \end{array}\right.$$
Step-by-Step Solution
Verified Answer
The solution to the system of equations is \[{(x, y) | x = 0.92, y = -1.17}\]
1Step 1: Simplify the System of Equations
The first step in solving this problem is to simplify the system of equations. The aim is to manipulate the equations into a form that makes it easier to solve for the variables. The second equation can be simplified by dividing every term by 4. This yields the following system of equations: \[2x - y = 3 \ x + y = - \frac{1}{4}\]
2Step 2: Apply the Addition Method
The addition method involves adding the two equations together in order to combine them into a single equation. In this case, the y terms in both equations will cancel each other out when added together because one is negative and the other is positive. This yields: \[2x - y + x + y = 3 - \frac{1}{4}\] Simplifying the left side of the equation gives: \[3x = 2.75\]
3Step 3: Solve for x
To isolate x, divide both sides of the equation by 3. This gives: \[x = \frac{2.75}{3}\] to get \(x = 0.92\) (rounded to two decimal places).
4Step 4: Substitute x into One of the Original Equations to Solve for y
Now that x is known, it can be substituted into either of the original equations to solve for y. Using the second equation: \[4*(0.92) + 4y = -1\] Simplify and solve for y. so, \(y = -1 - 3.68 = -4.68 / 4 = -1.17\).
5Step 5: Express the Solution in Set Notation
Lastly, we have to express the solution in set notation. The solution to the system is the ordered pair \((x, y)\), or \((0.92, -1.17)\). In set notation, this can be expressed as: \[{(x, y) | x = 0.92, y = -1.17}\] where the '|' symbol reads as 'such that'.
Key Concepts
Addition Method in AlgebraSystem of Linear EquationsSet NotationSolving for Variables
Addition Method in Algebra
The addition method in algebra, also known as the elimination method, is a strategic process used to solve systems of linear equations. This technique involves adding or subtracting the equations from each other with the goal of eliminating one of the variables.
Consider a system of two equations. To implement the addition method, align the equations vertically and aim to make the coefficient of one of the variables the same in both equations, but with opposite signs. When the equations are then added together, that variable will cancel out, leaving a single equation with just one variable.
Consider a system of two equations. To implement the addition method, align the equations vertically and aim to make the coefficient of one of the variables the same in both equations, but with opposite signs. When the equations are then added together, that variable will cancel out, leaving a single equation with just one variable.
Example of Elimination
For instance, if you have two equations, \(3x + 2y = 5\) and \(2x - 2y = 4\), you would simply add them to get \(5x = 9\), thereby eliminating \(y\). Next, you'd solve for \(x\), and then backtrack to find \(y\) using one of the original equations. It's an elegant dance of numbers that streamlines the problem-solving process.System of Linear Equations
A system of linear equations consists of two or more linear equations that have the same set of variables and are considered together. Each equation in a system depicts a line, and the solution of the system corresponds to the point(s) where the lines intersect.
Systems can have one unique solution, no solution if the lines are parallel (and therefore never intersect), or infinitely many solutions if the lines overlap completely (indicating the equations are dependent).
Linear equations are often written in the form \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants. Graphically, systems can be solved by plotting each equation on a coordinate grid and identifying the intersection point(s). Algebraically, methods like substitution, graphing, or addition (elimination) method are used for solving.
Systems can have one unique solution, no solution if the lines are parallel (and therefore never intersect), or infinitely many solutions if the lines overlap completely (indicating the equations are dependent).
Linear equations are often written in the form \(Ax + By = C\), where \(A\), \(B\), and \(C\) are constants. Graphically, systems can be solved by plotting each equation on a coordinate grid and identifying the intersection point(s). Algebraically, methods like substitution, graphing, or addition (elimination) method are used for solving.
Set Notation
Set notation is a standardized method of describing collections of elements (numbers, variables, etc.) that belong to a group based on a specific property or rule. In the context of solving systems of equations, set notation is used to express solution sets succinctly and clearly.
To write the solution of a system, use curly braces \({\}\) to enclose the solutions, and the vertical bar \(|\) to mean 'such that'. For example, the set of solutions that contains only the number 3 is written as \({x|x=3}\). In the context of the exercise provided, the solution \({(x, y) | x = 0.92, y = -1.17}\) tells us there's a single unique solution to the system, where the value of \(x\) is 0.92 and the value of \(y\) is -1.17.
To write the solution of a system, use curly braces \({\}\) to enclose the solutions, and the vertical bar \(|\) to mean 'such that'. For example, the set of solutions that contains only the number 3 is written as \({x|x=3}\). In the context of the exercise provided, the solution \({(x, y) | x = 0.92, y = -1.17}\) tells us there's a single unique solution to the system, where the value of \(x\) is 0.92 and the value of \(y\) is -1.17.
Solving for Variables
Solving for variables means to find the value(s) of the variable(s) that make the equation or system of equations true. In systems of linear equations, the aim is typically to find the values for each variable that satisfy all equations simultaneously.
Once you have simplified the system (like using the addition method to eliminate a variable), you are left with an equation in one variable. Solve this by isolating the variable — that is, getting the variable by itself on one side of the equation — and calculating its value. Then, use this value to find the others, by substituting back into one of the original equations.
Nonetheless, it's always important to check the solutions in both original equations to ensure they indeed satisfy the system. If they don't, it's possible that you may have made a calculation error or that the system does not have a solution.
Once you have simplified the system (like using the addition method to eliminate a variable), you are left with an equation in one variable. Solve this by isolating the variable — that is, getting the variable by itself on one side of the equation — and calculating its value. Then, use this value to find the others, by substituting back into one of the original equations.
Nonetheless, it's always important to check the solutions in both original equations to ensure they indeed satisfy the system. If they don't, it's possible that you may have made a calculation error or that the system does not have a solution.
Other exercises in this chapter
Problem 24
Solve each system by the substitution method. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets. $
View solution Problem 24
Graph the solution set of each system of linear inequalities. $$\left\\{\begin{array}{l}x \geq-2 \\\y
View solution Problem 25
Solve each system by graphing. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets. $$\left\\{\begin
View solution Problem 25
Solve each system by the substinuion method. If there is no solution or an infinite number of solutions, so state. Use set notation to express solution sets.$ $
View solution