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Solve the system graphically. $$\left\\{\begin{array}{r}x-y=0 \\ 5 x-2 y=6\end{array}\right.$$

Short Answer

Expert verified
The solution of the system of equations is the point where the two lines intersect, which can be determined by graphing the equations on the same graph.

Step by step solution

01

Re-write Equations in Slope-Intercept Form

First, let's re-write each equation in the slope-intercept form (y = mx + b). Starting with the first equation: \[x-y=0 \rightarrow y=x\] And for the second equation: \[5x - 2y = 6 \rightarrow y = \frac{5}{2}x - 3\]
02

Graph the Equations

Now, we can graph both functions. The first line has a slope of 1 and passes through the origin. The second line has a slope of \(\frac{5}{2}\) and crosses the y-axis at -3.
03

Identify The Point of Intersection

After plotting both lines on the same graph, find the point where both lines intersect. This point is the solution to the system of equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
Understanding the slope-intercept form is essential for graphing linear equations and solving systems of equations. It is usually expressed as \( y = mx + b \), where \( m \) represents the slope of the line, and \( b \) indicates the y-intercept, the point where the line crosses the y-axis.

For example, in the given system of equations, the first one can be written in slope-intercept form simply as \( y = x \). This implies a slope of 1 (each step right is one step up) and a y-intercept of 0, meaning it crosses the y-axis at the origin. The second equation, once reorganized to \( y = \frac{5}{2}x - 3 \), indicates a steeper slope of \( \frac{5}{2} \); for every two steps right, the line goes up five. Its y-intercept is -3, showing where it crosses the y-axis below the origin.

The slope-intercept form not only simplifies graphing but also aids in visual comparison of different lines to foresee their intersections and understand their relative steepness and direction.
Graphing Linear Equations
Graphing linear equations is the process of drawing a line on the Cartesian coordinate system based on the equation of a line. The slope-intercept form makes this process straightforward.

For the equations given in the example, the initial step involves plotting the y-intercept on the graph. Next, the slope determines the direction and steepness of the line. Following the slope from the intercept, additional points on the line can be plotted, which, when connected, reveal the complete line.

Drawing the First Line

The first equation, being in the form \( y = x \), suggests that for every one unit increase in x, y increases by the same amount. This line passes through the origin (0,0) and is drawn at a 45-degree angle to the axes.

Drawing the Second Line

For the second equation, start at the y-intercept at (0, -3), then move to the right two units along the x-axis and up five units in the y-direction. Doing this repeatedly allows you to place more points, resulting in the second line.
Point of Intersection
The point of intersection is where two or more lines on a graph cross each other. It represents a set of coordinates that is a solution to each equation in a system. Finding the point of intersection graphically involves drawing both lines and looking for the coordinates where they meet.

In the given system, after mapping out the two equations, the point where they intersect is the visual answer to the exercise. It's where both equations hold true simultaneously. The coordinates of this point give the values of x and y that solve the system. This method is particularly beneficial when solutions are integers or easily estimated from the graph.

In the example provided, the graphing would reveal the point where the lines representing each equation meet. Although not all solutions may be accurate through graphing, especially if they're non-integer values, it's a powerful tool for understanding the relationship between different linear equations in a system.

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Most popular questions from this chapter

The given linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the maximum value of the objective function and where it occurs. Objective function: \(z=-x+2 y\) Constraints: \(\begin{array}{rr}x & \geq 0 \\ y & \geq 0 \\ x & \leq 10 \\ x+y & \leq 7\end{array}\)

Sketch the region determined by the constraints. Then find the minimum anc maximum values of the objective function and where they occur, subject to the indicated constraints. Objective function: $$ z=x+2 y $$ Constraints: $$ \begin{aligned} x & \geq 0 \\ y & \geq 0 \\ x+2 y & \leq 40 \\ x+y & \leq 30 \\ 2 x+3 y & \leq 65 \end{aligned} $$

Graph the solution set of the system of inequalities. $$\left\\{\begin{array}{l}x<2 y-y^{2} \\ 0

Optimal Profit A manufacturer produces two models of bicycles. The times (in hours) required for assembling, painting, and packaging each model are shown in the table. $$ \begin{array}{|l|c|c|} \hline \text { Process } & \text { Model A } & \text { Model B } \\ \hline \text { Assembling } & 2 & 2.5 \\ \hline \text { Painting } & 4 & 1 \\ \hline \text { Packaging } & 1 & 0.75 \\ \hline \end{array} $$ The total times available for assembling, painting, and packaging are 4000 hours, 4800 hours, and 1500 hours, respectively. The profits per unit are \(\$ 50\) for model \(\mathrm{A}\) and \(\$ 75\) for model \(\mathrm{B}\). What is the optimal production level for each model? What is the optimal profit?

Maximize the objective function subject to the constraints \(3 x+y \leq 15,4 x+3 y \leq 30\) \(x \geq 0\), and \(y \geq 0\) $$z=3 x+y$$

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