How many solutions does the system of linear equations represented in the graph have?

Coordinate plane with one line that passes through the points negative 2 comma negative 3 and 0 comma negative 2 and another line that passes through the points 0 comma 3 and 1 comma 1.

One solution at (−1, 2)
One solution at (2, −1)
No solution
Infinitely many solutions
Question

1 answer

To determine how many solutions the system of linear equations has based on the given graph information, we need to analyze the two lines described.

  1. First Line: The first line passes through the points \((-2, -3)\) and \((0, -2)\). From these points, we can find the slope (\(m\)) and y-intercept (\(b\)) of the line.

    • The slope (\(m\)) can be calculated as: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-3)}{0 - (-2)} = \frac{1}{2}. \]
    • To find the equation of the line, we can use the slope-intercept form \(y = mx + b\). Using point \((0, -2)\) as it gives the y-intercept directly: \[ y = \frac{1}{2}x - 2. \]
  2. Second Line: The second line passes through the points \((0, 3)\) and \((1, 1)\).

    • Its slope is: \[ m = \frac{1 - 3}{1 - 0} = -2. \]
    • Starting from point \((0, 3)\), we can write the equation of the line as: \[ y = -2x + 3. \]

Now, we have the equations of both lines:

  1. First line: \(y = \frac{1}{2}x - 2\)
  2. Second line: \(y = -2x + 3\)

Now we need to find out whether these lines intersect, and if they do, where.

To find the intersection, we will set the two equations equal to each other: \[ \frac{1}{2}x - 2 = -2x + 3. \]

Multiplying everything by 2 to eliminate the fraction gives: \[ x - 4 = -4x + 6. \]

Combining like terms: \[ x + 4x = 6 + 4 \ 5x = 10 \ x = 2. \]

Now, substitute \(x = 2\) back into one of the line equations to find \(y\): Using the first line: \[ y = \frac{1}{2}(2) - 2 = 1 - 2 = -1. \]

Thus, the intersection point is \((2, -1)\).

Since the lines intersect at exactly one point \((2, -1)\), there is one solution to the system of equations.

Answer: One solution at \((2, -1)\).