Problem 7 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether each ordered pair is a solution of the system of equations. $$\left\\{\begin{array}{l} 4 x-y=1 \\ 6 x+y=-6 \end{array}\right.$$ (a) (0,-3) (b) (-1,-5) (c) \(\left(-\frac{3}{2}, 3\right)\) (d) \(\left(-\frac{1}{2},-3\right)\)

Step-by-Step Solution

Answer

None of the given ordered pairs are solutions to the system of equations.

Step 1: Substitute ordered pair (0,-3) into the equations

In the first equation, substitute 0 for \(x\) and -3 for \(y\). This gives \(4(0) - (-3) = 1\), which simplifies to 3 = 1, a false statement. Therefore, (0,-3) is not a solution to the system of equations.

Step 2: Substitute ordered pair (-1,-5) into the equations

In the first equation, substitute -1 for \(x\) and -5 for \(y\). This gives \(4(-1) - (-5) = 1\), which simplifies to 1 = 1, a true statement. Repeat for the second equation: \(6(-1) + (-5) = -6\), which simplifies to -11 = -6, a false statement. Therefore, (-1,-5) is not a solution to the system of equations.

Step 3: Substitute ordered pair \(-\frac{3}{2}, 3\) into the equations

In the first equation, substitute \(-\frac{3}{2}, 3\) for \(x, y\) respectively. This gives \(4(-\frac{3}{2}) - 3 = 1\), which simplifies to -6 = 1, a false statement. Therefore, \(-\frac{3}{2}, 3\) is not a solution to the system of equations.

Step 4: Substitute ordered pair \(-\frac{1}{2},-3\) into the equations

In the first equation, substitute \(-\frac{1}{2}\) for \(x\) and -3 for \(y\). This gives \(4(-\frac{1}{2}) - (-3) = 1\), which simplifies to 2 = 1, a false statement. Therefore, \(-\frac{1}{2},-3\) is not a solution to the system of equations.

Key Concepts

Ordered Pairs and Systems of Equations

An ordered pair is a fundamental concept in mathematics where two numbers or variables are paired together to indicate a point on a two-dimensional coordinate plane. In the context of systems of equations, each ordered pair \(x, y\) represents a potential solution to the set of equations provided. The ordered pair conveys both a specific spot on the graph and potential values for the variables in the equations. Ensuring these values satisfy all the equations in the system determines if the ordered pair is, indeed, a solution. When working through problems involving systems of equations, pay close attention to the ordered pairs under consideration, as these are crucial in checking the validity of potential solutions.

Introduction to Linear Equations

Linear equations are mathematical expressions that represent straight lines when graphed on a coordinate plane. These equations usually take the form \(Ax + By = C\), where \(A, B,\) and \(C\) are constants, and \(x, y\) are variables. In the context of systems of equations, linear equations are combined in either two or more, where each equation corresponds to a line on the coordinate plane.Understanding how these lines interact—with the possibility of intersecting, being parallel, or coinciding—is essential for determining the solutions to the system.Some key characteristics of linear equations include:

• A consistent rate of change, depicted as the slope of the line
• The interaction of lines which can indicate no solution, one solution, or infinite solutions

Using Algebraic Substitution in Solving Equations

Algebraic substitution is an important technique used in solving systems of equations. This method involves replacing one variable with an expression derived from another equation within the same system. By breaking down the equations initially, we can simplify the complex relationships between the variables, making it easier to solve for one variable at a time. Here's a simple approach to applying substitution:

• Solve one of the equations for one variable in terms of the other.
• Substitute this expression into the other equation to find the value of one variable.
• Once you have a value for one variable, substitute it back into one of the original equations to find the value of the remaining variable.

This method is particularly helpful when dealing with systems of linear equations, as it reduces complex expressions to more manageable calculations. By successfully applying substitution, you can efficiently identify if an ordered pair is a genuine solution to the system.