Problem 39
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. $$\left\\{\begin{array}{l} 4(3 x-y)=0 \\ 3(x+3)=10 y \end{array}\right.$$
Step-by-Step Solution
Verified Answer
The solution to these equations would be the values of x and y found in Step 4 and 5 respectively. These are the coordinates of the point where the two lines represented by the system of equations intersect and is the solution to the system. The solution would need to be expressed in set notation as per the requirements of the exercise.
1Step 1: Simplify the equations
First simplify the set of equations to a simpler form. This can be done by expanding the equations. After expansion, the system of equations becomes: \(\begin{array}{l} 12x - 4y = 0 \\ 3x + 9 = 10y \end{array}\)
2Step 2: Rearrange terms to have the system in standard form
The equations can be rearranged to align the like terms together and further simplify each equation in the system to obtain: \(\begin{array}{l} 12x - 4y = 0 \\ 3x - 10y = -9 \end{array}\) Notice that one equation is in the form $Ax - By = 0$. This can be used later in the following steps.
3Step 3: Add the two equations
As stated in the exercise, we should use the addition method, so we add these equations together. This gives: \(15x - 14y = -9\). Now, isolate y to find its value: \(y = \frac{15x + 9}{14}\)
4Step 4: Substitution
Now, substitute the value of y from the third step into one of the original equations. Choose the first equation to substitute y into: \(12x - 4(\frac{15x + 9}{14}) = 0\). Then, solve for x.
5Step 5: Find the value of y
After finding the value of x from step 4, substitute it back into the equation for y to find its value.
Key Concepts
Linear SystemsAlgebraic EquationsSet NotationSubstitution Method
Linear Systems
Linear systems consist of two or more linear equations involving the same set of variables. They represent lines on a Cartesian plane, and solving the system means finding the point or points where these lines intersect. This process involves determining the variable values that satisfy all equations in the system simultaneously. Most linear systems can be categorized as having one unique solution, no solution, or infinitely many solutions.
When solving a linear system by the addition method, the goal is to eliminate one variable by adding or subtracting the equations. This is possible when the coefficients of one variable are opposites or can be made opposites through multiplication. After one variable is eliminated, the remaining equation can be solved to find the value of the remaining variable.
When solving a linear system by the addition method, the goal is to eliminate one variable by adding or subtracting the equations. This is possible when the coefficients of one variable are opposites or can be made opposites through multiplication. After one variable is eliminated, the remaining equation can be solved to find the value of the remaining variable.
Algebraic Equations
Algebraic equations are mathematical expressions that equate two algebraic expressions using the equal sign. They include variables (unknowns), constants, and coefficients. To solve an algebraic equation, we must isolate the variable on one side of the equation to determine its value. Equations in a linear system are called linear because they graph as straight lines, and their variables are only raised to the first power.
In the context of solving systems, algebraic equations are manipulated so that they are in a standard form, often as Ax + By = C, to facilitate the use of the addition (also known as the elimination) method. Each operation performed to reach this form must maintain the equality by treating both sides of the equation equivalently.
In the context of solving systems, algebraic equations are manipulated so that they are in a standard form, often as Ax + By = C, to facilitate the use of the addition (also known as the elimination) method. Each operation performed to reach this form must maintain the equality by treating both sides of the equation equivalently.
Set Notation
Set notation is a method used to define collections of objects, numbers, or solutions. In algebra, we often use it to express the solution set of a system of equations, where the solution set is the collection of all possible solutions that satisfy the system. If a system has a unique solution, the set will contain a single ordered pair (x, y). If a system has infinitely many solutions, the set notation will typically describe all the ordered pairs that make up a line or a plane. If there is no solution—meaning the lines never intersect—the solution set is empty, denoted by \( \emptyset \).
Using set notation to express a solution is a concise way to communicate the result of solving a system. For example, the solution set {\( (x, y) \)} denotes a single solution, while {\( (x, \frac{15x+9}{14}) \)} could represent the solution set for a dependent system in terms of x.
Using set notation to express a solution is a concise way to communicate the result of solving a system. For example, the solution set {\( (x, y) \)} denotes a single solution, while {\( (x, \frac{15x+9}{14}) \)} could represent the solution set for a dependent system in terms of x.
Substitution Method
The substitution method is an algebraic technique used to solve systems of equations where one equation is solved for one variable in terms of the others, and this expression is then substituted into the other equation(s). This process reduces the system to one with fewer equations and fewer variables, making it simpler to solve.
In our original exercise, we could use the substitution method after using addition to eliminate one variable. However, it's often used as an alternative to the addition method, particularly when equations are already solved for a variable. It's most efficient when the coefficients of the variable being substituted are 1 or -1, reducing the complexity of the substitution.
In our original exercise, we could use the substitution method after using addition to eliminate one variable. However, it's often used as an alternative to the addition method, particularly when equations are already solved for a variable. It's most efficient when the coefficients of the variable being substituted are 1 or -1, reducing the complexity of the substitution.
Other exercises in this chapter
Problem 38
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