Question

what system of equations is shown on the graph below
cumulative exam complete
1 2 3 4 5 6 7 8
-5 0 5
-5
2x+y=3 and 3x-4y=8 .
2x-y=3 and 3x+4y=8
2x+y=3 and 3x-4y=2
2x-y=3 and 3x-4y=-2

Explanation:

Step1: Analyze the slope of lines

First, we can rewrite each equation in slope - intercept form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept) to analyze the slopes of the lines.

For the first option:

• For \(2x + y=3\), rewrite it as \(y=- 2x + 3\), slope \(m_1=-2\) (negative slope, since the coefficient of \(x\) is negative).
• For \(3x-4y = 8\), rewrite it as \(4y=3x - 8\), then \(y=\frac{3}{4}x-2\), slope \(m_2 = \frac{3}{4}\) (positive slope).

For the second option:

• For \(2x - y=3\), rewrite it as \(y = 2x-3\), slope \(m_1 = 2\) (positive slope).
• For \(3x + 4y=8\), rewrite it as \(4y=-3x + 8\), then \(y=-\frac{3}{4}x + 2\), slope \(m_2=-\frac{3}{4}\) (negative slope).

For the third option:

• For \(2x + y=3\), \(y=-2x + 3\), slope \(m_1=-2\) (negative slope).
• For \(3x-4y = 2\), \(4y=3x - 2\), \(y=\frac{3}{4}x-\frac{1}{2}\), slope \(m_2=\frac{3}{4}\) (positive slope).

For the fourth option:

• For \(2x - y=3\), \(y = 2x-3\), slope \(m_1 = 2\) (positive slope).
• For \(3x-4y=-2\), \(4y=3x + 2\), \(y=\frac{3}{4}x+\frac{1}{2}\), slope \(m_2=\frac{3}{4}\) (positive slope).

From the graph, we can see that one line has a negative slope and one line has a positive slope. So we can eliminate options 2 and 4 because in option 2, one slope is positive (\(2\)) and one is negative (\(-\frac{3}{4}\)), but let's check the y - intercepts as well. In option 4, both slopes are positive.

Step2: Analyze the y - intercepts

For the first option:

• For \(y=-2x + 3\), the y - intercept \(b_1 = 3\) (positive y - intercept, since when \(x = 0\), \(y=3\)).
• For \(y=\frac{3}{4}x-2\), the y - intercept \(b_2=-2\) (negative y - intercept, since when \(x = 0\), \(y=-2\)).

For the third option:

• For \(y=-2x + 3\), y - intercept \(b_1 = 3\).
• For \(y=\frac{3}{4}x-\frac{1}{2}\), y - intercept \(b_2=-\frac{1}{2}\).

Now, looking at the graph, the blue line (with positive slope) has a y - intercept that is negative (close to \(- 2\)) and the red line (with negative slope) has a y - intercept that is positive (close to \(3\)).

Let's also check the intersection point. Let's solve the system for option 1:

From \(2x + y=3\), we have \(y = 3 - 2x\). Substitute into \(3x-4y=8\):

\(3x-4(3 - 2x)=8\)

\(3x-12 + 8x=8\)

\(11x=20\)

\(x=\frac{20}{11}\approx1.82\)

\(y=3-2\times\frac{20}{11}=3-\frac{40}{11}=\frac{33 - 40}{11}=-\frac{7}{11}\approx - 0.64\)

For option 3:

From \(2x + y=3\), \(y = 3 - 2x\). Substitute into \(3x-4y=2\):

\(3x-4(3 - 2x)=2\)

\(3x-12 + 8x=2\)

\(11x=14\)

\(x=\frac{14}{11}\approx1.27\)

\(y=3-2\times\frac{14}{11}=3-\frac{28}{11}=\frac{33 - 28}{11}=\frac{5}{11}\approx0.45\)

Looking at the graph, the intersection point seems to be around \(x\approx2\) and \(y\approx - 1\) (approximate from the grid). The solution of option 1 is closer to this approximation. Also, the slopes and y - intercepts of option 1 match the visual cues from the graph (one line with negative slope, y - intercept positive; one line with positive slope, y - intercept negative).

Answer:

A. \(2x + y = 3\) and \(3x-4y = 8\)