the 4 graphs show a system of 2 linear equations. which system has the ordered pair (10, 5) as a solution? a graph b graph

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To determine which system has the ordered pair $(10, 5)$ as a solution, we analyze the graphs:

### Graph A:
- One line is horizontal (constant $y$-value). From the grid, it appears to be $y = 4$ (since it's at $y = 4$ across the $x$-axis).
- The other line passes through the origin and has a slope. For $x = 10$, the $y$-value on this line would be calculated by its slope. If we assume the line passes through $(0, 0)$ and, say, $(10, 5)$, but the horizontal line is $y = 4$, so at $x = 10$, the horizontal line has $y = 4$, not $5$. Thus, $(10, 5)$ is not a solution for Graph A.

### Graph B:
- One line has a positive slope, passing through the origin and, likely, $(10, 10)$ (if we follow the grid). Wait, no—wait, the other line has a negative slope. Wait, let's re-examine. Wait, the two lines in Graph B: one with positive slope (going from bottom-left to top-right) and one with negative slope (going from top-left to bottom-right). Let's check the intersection or the values at $x = 10$.
- For the line with positive slope: If we assume it passes through $(0, 0)$ and, say, $(10, 10)$, but no—wait, the other line (negative slope) might intersect at $(10, 5)$? Wait, no, let's think again. Wait, the ordered pair $(10, 5)$ means at $x = 10$, $y = 5$. Let's check the lines in Graph B:
  - The line with positive slope: Let's say it goes through $(0, 0)$ and $(10, 10)$ (slope $1$), but the other line (negative slope) might go through $(0, 10)$ and $(10, 0)$ (slope $-1$). Wait, no—wait, the intersection of these two lines would be at $(5, 5)$? No, wait, if one line is $y = x$ (positive slope, through origin) and the other is $y = -x + 10$ (negative slope, through $(0, 10)$ and $(10, 0)$), then their intersection is at $x = 5$, $y = 5$. Wait, that's not $(10, 5)$. Wait, maybe I misread the graphs.

Wait, perhaps the correct approach is: For a system of linear equations, the solution is the point where the two lines intersect. So we need to find which graph has the two lines intersecting at $(10, 5)$ or where at $x = 10$, both lines have $y = 5$.

Wait, maybe the initial analysis was wrong. Let's re-express:

- In Graph A: One line is horizontal ($y = 4$) and the other is a line with slope $0.5$ (from $(0, 0)$ to $(10, 5)$). Wait, if the horizontal line is $y = 4$, then at $x = 10$, the slanted line has $y = 5$, but the horizontal line has $y = 4$, so they don't intersect at $(10, 5)$.

- In Graph B: The two lines—one with positive slope (say $y = x$) and one with negative slope (say $y = -x + 15$)? No, this is getting confusing. Wait, the key is: The ordered pair $(10, 5)$ must satisfy both equations (i.e., lie on both lines). So we check for each graph if at $x = 10$, both lines have $y = 5$.

Wait, maybe the correct answer is Graph B? Wait, no—wait, maybe I made a mistake. Wait, let's look at the grid again. For Graph A: The horizontal line is at $y = 4$ (since the grid has $y = 4$ marked). The slanted line goes through the origin and, at $x = 10$, would have $y = 5$ (if slope is $0.5$). But the horizontal line is $y = 4$, so at $x = 10$, the slanted line has $y = 5$, but the horizontal line has $y = 4$, so $(10, 5)$ is on the slanted line but not on the horizontal line. Thus, not a solution.

For Graph B: The two lines—one with positive slope (through origin) and one with negative slope. Let's assume the positive slope line is $y = 0.5x$ (so at $x = 10$, $y = 5$) and the negative slope line is, say, $y = -0.5x + 10$ (so at $x = 10$, $y = -5 + 10 = 5$). Wait, that would mean both lines have $y = 5$ at $x = 10$, so $(10, 5)$ is on both lines, hence a solution.

Thus, the system in Graph B has $(10, 5)$ as a solution.

# Answer: B

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Graph B:

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