Determine Which Of The Ordered Pairs 0 1 2 0 0 1 8 5ar
Determine Which Of The Ordered Pairs0 1 2 0 0 1 8 5ar
Determine which of the given ordered pairs satisfy the equations provided. Specifically, identify the solutions for the equation \( x + 2y = 2 \) among the pairs (0, 1), (2, 0), (0, –1), and (–8, 5). Then, complete the missing components of given ordered pairs so that they become solutions to the equation \( 2x + y = 10 \). Additionally, analyze various points and slopes related to linear equations and graphs to deepen understanding of coordinate geometry concepts.
Paper For Above instruction
Introduction
Understanding solutions to linear equations and the calculation of slopes between points are essential skills in algebra and coordinate geometry. This paper explores methods to verify if given ordered pairs are solutions to specific equations, how to complete ordered pairs so they satisfy particular equations, and how to calculate the slopes of lines passing through given points. These concepts form the foundation for analyzing linear functions, graphing lines, and understanding their behaviors.
Identifying Solutions to Linear Equations
To determine whether an ordered pair satisfies a particular linear equation, substitute the coordinates into the equation and verify if the statement holds true. For the equation \( x + 2y = 2 \), test each pair:
- For (0, 1): \( 0 + 2(1) = 0 + 2 = 2 \). Since the result equals 2, (0, 1) is a solution.
- For (2, 0): \( 2 + 2(0) = 2 + 0 = 2 \). Solution confirmed.
- For (0, –1): \( 0 + 2(-1) = 0 - 2 = -2 \neq 2 \). Not a solution.
- For (–8, 5): \( -8 + 2(5) = -8 + 10 = 2 \). Solution confirmed.
Hence, the pairs (0, 1), (2, 0), and (–8, 5) satisfy the equation \( x + 2y = 2 \), while (0, –1) does not.
Completing Ordered Pairs for Solutions
To complete the pairs for the equation \( 2x + y= 10 \):
- For the first pair: \( (5, \_ ) \):
Substitute \( x=5 \):
\( 2(5) + y = 10 \Rightarrow 10 + y= 10 \Rightarrow y= 0 \).
So, the complete ordered pair is (5, 0).
- Second pair: \( (\_ , 10) \):
Substitute \( y=10 \):
\( 2x + 10 = 10 \Rightarrow 2x= 0 \Rightarrow x= 0 \).
The pair is (0, 10).
- Third pair: \( (\_ , -2) \):
\( 2x + (-2)= 10 \Rightarrow 2x= 12 \Rightarrow x= 6 \)
Complete pair: (6, –2).
- Fourth pair: (7, \_):
Substitute \( x=7 \):
\( 2(7) + y= 10 \Rightarrow 14 + y= 10 \Rightarrow y= -4 \).
Complete pair: (7, –4).
These completions verify the pairs as solutions.
Graph Coordinates and Slope Calculations
Various points are considered for their placement on the coordinate plane, such as the points (10, 7) and (8, –10), as well as points (-3, –2) and (-3, 0). With the points (10, 7) and (8, –10), the slope \( m \) is computed as:
\[
m= \frac{y_2 - y_1}{x_2 - x_1} = \frac{-10 - 7}{8 - 10} = \frac{-17}{-2} = 8.5
\]
Similarly, for the points (-3, –2) and (-3, 0):
\[
x_1= x_2= -3 \Rightarrow \text{The slope is undefined as } x_1= x_2,
\]
which indicates a vertical line.
For (-6, 3) and (5, 3):
\[
m= \frac{3 - 3}{5 - (-6)}= \frac{0}{11} = 0,
\]
indicating a horizontal line.
Calculating the slope between (3, 2) and (8, 11):
\[
m= \frac{11 - 2}{8 - 3} = \frac{9}{5} = 1.8
\]
Between (3, 7) and (–2, 11):
\[
m= \frac{11 - 7}{-2 - 3} = \frac{4}{-5} = -0.8
\]
Between (3, –2) and (–1, –6):
\[
m= \frac{-6 - (-2)}{-1 - 3} = \frac{-4}{-4} = 1
\]
The calculation of slopes helps visualize and analyze the orientation and steepness of lines graphed on the coordinate plane.
Graphing Linear Equations
Graphing the equation \( 3x + 2 y= 6 \) involves plotting points that satisfy the equation, such as those with integer coordinates. As an example, plotting points where \( x= 0 \):
\[
3(0)+ 2 y= 6 \Rightarrow 2 y= 6 \Rightarrow y= 3,
\]
gives (0, 3). Similarly, for \( y= 0 \):
\[
3 x + 0= 6 \Rightarrow x= 2,
\]
which provides (2, 0). Plotting these points and connecting them creates the graph of the line.
Verifying Solutions
To verify if specific points are solutions to given equations:
- For \( y= 3x - 5 \), check (0, 5):
\[
5= 3(0) - 5 \Rightarrow 5= -5,
\]
which is false. Therefore, (0, 5) is not a solution.
- For \( y= -2x + 7 \), check (–2, 3):
\[
3= -2(-2) + 7 \Rightarrow 3= 4 + 7= 11,
\]
not true; hence, not a solution.
- For \( -6x + 5 y= 0 \), check (1, 0):
\[
-6(1) + 5(0)= 0 \Rightarrow -6= 0,
\]
which is false.
- For \( 5x - 3 y= 0 \), check \( (12/5, -1) \):
\[
5 \times \frac{12}{5} - 3 \times (-1)= 12 + 3= 15 \neq 0,
\]
so not a solution.
- For \( y= -2x + 7 \), check (1, 5):
\[
5= -2(1)+ 7= -2+ 7= 5,
\]
which holds true, confirming the point as a solution.
- For \( y= x - 5\), check (–1, –8):
\[
-8= -1 - 5= -6,
\]
which is false; so, (–1, –8) is not a solution.
Conclusion
The process of identifying solutions among ordered pairs, completing pairs to satisfy linear equations, calculating slopes between points, and verifying solutions against equations are fundamental skills. These methods facilitate understanding of the behavior of linear functions, graph interpretation, and the geometric relationships encoded in algebraic expressions.
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