Choose The System Of Equations That Matches The Following Graph

The graph shows two lines intersecting at a single point, which means the system has one unique solution. To find the correct system, we need to identify the equations of both lines by examining their slopes and y-intercepts.

First, let's analyze the line that crosses the y-axis at 3 and has a negative slope. This line decreases as it moves from left to right, indicating a slope between 0 and -1. The equation for this line is in the form y = mx + b, where m is the slope and b is the y-intercept. Here, b = 3, so the equation becomes y = mx + 3. By observing the steepness, we can estimate the slope to be approximately -1/2, giving us y = -1/2x + 3.

Next, we look at the second line, which crosses the y-axis at -1 and has a positive slope. This line increases as it moves from left to right, indicating a slope greater than 1. Using the same form y = mx + b, we have b = -1, so the equation is y = mx - 1. The steepness suggests a slope of about 2, resulting in y = 2x - 1.

Now, let's verify these equations by checking if they match the graph. For the first line, y = -1/2x + 3, we can substitute x = 0 to get y = 3, which matches the y-intercept. For the second line, y = 2x - 1, substituting x = 0 gives y = -1, also matching the y-intercept.

To ensure these are the correct equations, we can solve the system algebraically. Setting the two equations equal to each other: -1/2x + 3 = 2x - 1 Adding 1/2x to both sides: 3 = 2.5x - 1 Adding 1 to both sides: 4 = 2.5x Dividing both sides by 2.5: x = 4/2.5 = 1.6

Substituting x = 1.6 into either equation to find y: y = -1/2(1.6) + 3 = -0.8 + 3 = 2.2 y = 2(1.6) - 1 = 3.2 - 1 = 2.2

Both equations give the same y-value, confirming that the lines intersect at the point (1.6, 2.2), which aligns with the graph's intersection point.

Therefore, the system of equations that matches the graph is: y = -1/2x + 3 y = 2x - 1

This system has one unique solution at the point (1.6, 2.2), as shown by the intersection of the two lines on the graph.

To further confirm the accuracy of the derived equations, we can test additional points that lie visibly on each line in the graph. For the descending line, selecting x = 2 yields y = ‑½·2 + 3 = 2, which matches the point (2, 2) observed on the graph. Choosing x = ‑2 gives y = ‑½·(‑2) + 3 = 1 + 3 = 4, corresponding to the point (‑2, 4) also evident on the line. These checks reinforce that the slope of ‑½ and intercept 3 are correct.

For the ascending line, substituting x = 1 produces y = 2·1 ‑ 1 = 1, aligning with the point (1, 1) shown. Likewise, x = ‑1 gives y = 2·(‑1) ‑ 1 = ‑2 ‑ 1 = ‑3, matching the point (‑1, ‑3). Such consistent agreement across multiple coordinates validates the slope estimate of 2 and intercept ‑1.

An alternative algebraic approach uses the elimination method. Multiplying the first equation by 2 to clear the fraction gives ‑x + 6 = 4x ‑ 2. Bringing all terms to one side yields ‑x ‑ 4x = ‑2 ‑ 6, or ‑5x = ‑8, leading again to x = 8/5 = 1.6. Substituting this x value into either original equation reproduces y = 11/5 = 2.2, confirming the intersection point as (8/5, 11/5).

The process demonstrates how visual inspection of a graph can be translated into precise algebraic representations, and how subsequent verification—whether by point testing, substitution, or elimination—ensures the solution’s reliability. In practical contexts, such as determining break‑even points in economics or computing equilibrium states in physics, the ability to extract and solve a system from a graphical model is indispensable.

In conclusion, the system of equations

[ \begin{cases} y = -\frac12 x + 3\[4pt] y = 2x - 1 \end{cases} ]

accurately captures the two lines depicted in the graph, intersecting uniquely at the point (\left(\frac{8}{5},\frac{11}{5}\right)) (approximately (1.6, 2.2)). This single solution underscores the consistency and independence of the linear equations, affirming that the graphical and algebraic perspectives are fully aligned.

s by 2.5: x = 4/2.5 = 1.6

Substituting x = 1.6 into either equation to find y: y = -1/2(1.6) + 3 = -0.8 + 3 = 2.2 y = 2(1.6) - 1 = 3.2 - 1 = 2.2

Both equations give the same y-value, confirming that the lines intersect at the point (1.6, 2.2), which aligns with the graph's intersection point.

Therefore, the system of equations that matches the graph is: y = -1/2x + 3 y = 2x - 1

This system has one unique solution at the point (1.6, 2.2), as shown by the intersection of the two lines on the graph.

To further confirm the accuracy of the derived equations, we can test additional points that lie visibly on each line in the graph. For the descending line, selecting x = 2 yields y = -½·2 + 3 = 2, which matches the point (2, 2) observed on the graph. Choosing x = -2 gives y = -½·(-2) + 3 = 1 + 3 = 4, corresponding to the point (-2, 4) also evident on the line. These checks reinforce that the slope of -½ and intercept 3 are correct.

For the ascending line, substituting x = 1 produces y = 2·1 - 1 = 1, aligning with the point (1, 1) shown. Likewise, x = -1 gives y = 2·(-1) - 1 = -2 - 1 = -3, matching the point (-1, -3). Such consistent agreement across multiple coordinates validates the slope estimate of 2 and intercept -1.

An alternative algebraic approach uses the elimination method. Multiplying the first equation by 2 to clear the fraction gives -x + 6 = 4x - 2. Bringing all terms to one side yields -x - 4x = -2 - 6, or -5x = -8, leading again to x = 8/5 = 1.6. Substituting this x value into either original equation reproduces y = 11/5 = 2.2, confirming the intersection point as (8/5, 11/5).

The process demonstrates how visual inspection of a graph can be translated into precise algebraic representations, and how subsequent verification—whether by point testing, substitution, or elimination—ensures the solution's reliability. In practical contexts, such as determining break-even points in economics or computing equilibrium states in physics, the ability to extract and solve a system from a graphical model is indispensable.

In conclusion, the system of equations

[ \begin{cases} y = -\frac12 x + 3\[4pt] y = 2x - 1 \end{cases} ]

accurately captures the two lines depicted in the graph, intersecting uniquely at the point (\left(\frac{8}{5},\frac{11}{5}\right)) (approximately (1.6, 2.2)). This single solution underscores the consistency and independence of the linear equations, affirming that the graphical and algebraic perspectives are fully aligned.

This methodical approach—moving from visual estimation to algebraic formulation, then rigorously confirming through multiple independent checks—exemplifies a robust problem-solving framework applicable across disciplines. The convergence of graphical intuition, substitution, elimination, and point validation leaves no ambiguity about the system’s structure or its solution. Moreover, the uniqueness of the intersection point confirms that the two lines are neither parallel nor coincident, but rather independent and consistent, a fundamental property with implications for understanding system behavior in fields ranging from circuit design to resource allocation.

Ultimately, this exercise reinforces a critical mathematical principle: a graph is not merely a picture but a precise repository of quantitative relationships. By learning to decode that visual information into accurate equations and then subjecting the result to layered verification, one develops a transferable skill set. It cultivates attention to detail, logical sequencing, and the habit of cross-validation—all essential for tackling complex real-world problems where initial assumptions must be tested and confirmed. The seamless integration of graphical interpretation and algebraic rigor thus stands as a cornerstone of analytical competence.