Problem 32
Question
Graph and check to solve the linear system. $$ \begin{aligned} &\frac{1}{5} x+\frac{3}{5} y=\frac{12}{5}\\\ &-\frac{1}{5} x+\frac{3}{5} y=\frac{6}{5} \end{aligned} $$
Step-by-Step Solution
Verified Answer
The solution to the system of equations is the point (2,2).
1Step 1: Create the Graph
Start by taking each equation and graph it individually. The graph of a linear equation will be a straight line. Because each equation is in the form \( ax + by = c \), you can find points to plot by substituting 0 for x to find y, and vice versa.
2Step 2: Graph the first equation
The first equation is \( \frac{1}{5}x + \frac{3}{5}y = \frac{12}{5} \). If we substitute 0 for x, we find that y = 2. Similarly, if we substitute 0 for y, we find that x = 2. Therefore, the points (0,2) and (2,0) lie on the line and can be graphed.
3Step 3: Graph the second equation
The second equation is \( -\frac{1}{5}x + \frac{3}{5}y = \frac{6}{5} \). If we substitute 0 for x, we find that y = 1. Similarly, if we substitute 0 for y, we find that x = -1. Therefore, the points (0,1) and (-1,0) lie on the line and can be graphed.
4Step 4: Find the point of intersection
The solution to the system of equations will be the point where the two lines intersect. After plotting the points and drawing the lines, it is clear that the lines intersect at the point (2,2).
Key Concepts
Graphing Linear EquationsLinear EquationsSystem of EquationsPoint of Intersection
Graphing Linear Equations
Graphing linear equations involves creating a visual representation of the equations on a coordinate plane. This is particularly helpful for visualizing the relationship between the variables in the equation.
For example, when graphing the linear equation \( \frac{1}{5}x + \frac{3}{5}y = \frac{12}{5} \), you look for points on the graph where the x and y values satisfy the equation. By choosing convenient values for x (such as 0), we can solve for y, and vice versa, to find points that lie on the graph. Plotting these points and connecting them with a straight line will give you the graphical representation of the equation.
Similarly, with the equation \( -\frac{1}{5}x + \frac{3}{5}y = \frac{6}{5} \), you find points where this new line crosses the x and y axis, and then draw the line through these points. The point where these two lines cross will tell us the solution to the system of equations.
For example, when graphing the linear equation \( \frac{1}{5}x + \frac{3}{5}y = \frac{12}{5} \), you look for points on the graph where the x and y values satisfy the equation. By choosing convenient values for x (such as 0), we can solve for y, and vice versa, to find points that lie on the graph. Plotting these points and connecting them with a straight line will give you the graphical representation of the equation.
Similarly, with the equation \( -\frac{1}{5}x + \frac{3}{5}y = \frac{6}{5} \), you find points where this new line crosses the x and y axis, and then draw the line through these points. The point where these two lines cross will tell us the solution to the system of equations.
Linear Equations
A linear equation represents a straight line on a graph and is typically written in the form \( ax + by = c \), where a, b, and c are constants. Linear equations have variables that are raised to the first power and represent a constant rate of change.
To construct a line from a linear equation, you need at least two points. In the given exercise, for instance, we calculate points on the lines by letting one variable be zero and solving for the other. Keep in mind, however, that any pair of points that satisfy the equation will suffice to draw the line.
To construct a line from a linear equation, you need at least two points. In the given exercise, for instance, we calculate points on the lines by letting one variable be zero and solving for the other. Keep in mind, however, that any pair of points that satisfy the equation will suffice to draw the line.
System of Equations
A system of equations consists of two or more equations with the same variables. The solution to such a system is the point or points that all the equations have in common.
When working with linear systems, such as the one of our exercise with two equations \( \frac{1}{5}x + \frac{3}{5}y = \frac{12}{5} \) and \( -\frac{1}{5}x + \frac{3}{5}y = \frac{6}{5} \) we're actually looking for the point where the two lines intersect. If the lines intersect at a single point, the system is consistent and independent. If the lines are parallel and do not intersect, the system has no solution and is called inconsistent. Lastly, if the lines are coincident, they lay exactly on top of each other, indicating infinite solutions.
When working with linear systems, such as the one of our exercise with two equations \( \frac{1}{5}x + \frac{3}{5}y = \frac{12}{5} \) and \( -\frac{1}{5}x + \frac{3}{5}y = \frac{6}{5} \) we're actually looking for the point where the two lines intersect. If the lines intersect at a single point, the system is consistent and independent. If the lines are parallel and do not intersect, the system has no solution and is called inconsistent. Lastly, if the lines are coincident, they lay exactly on top of each other, indicating infinite solutions.
Point of Intersection
The point of intersection is the coordinate that represents where two lines on a graph meet. For a system of linear equations, this point is the solution to the system, as it is the set of x and y values that satisfy all equations within the system.
In the exercise, after graphing both lines, we find they intersect at \( (2,2) \), which means that x = 2 and y = 2 are the solutions to both equations in the system. This method of solving by graphing is particularly useful because it provides a clear visual confirmation of the solution. However, it's important to note that graphing may not always give the exact answer if the point of intersection does not fall exactly on a grid line, in which case algebraic methods may be more precise.
In the exercise, after graphing both lines, we find they intersect at \( (2,2) \), which means that x = 2 and y = 2 are the solutions to both equations in the system. This method of solving by graphing is particularly useful because it provides a clear visual confirmation of the solution. However, it's important to note that graphing may not always give the exact answer if the point of intersection does not fall exactly on a grid line, in which case algebraic methods may be more precise.
Other exercises in this chapter
Problem 31
Use linear combinations to solve the system of linear equations. $$\begin{aligned} &x+2 y=5\\\ &5 x-y=3 \end{aligned}$$
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Use the substitution method to solve the linear system. $$\begin{aligned} &x+y=12\\\ &x+\frac{3}{2} y=\frac{3}{2} \end{aligned}$$
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