To determine which system of equations has no solution, we need to analyze the pairs of equations to see if they intersect at any points.
-
System 1:
- \(y = -x^2\)
- \(y = 3x\)
Here, we have a downward-opening parabola and a straight line with a positive slope. They can intersect at some points, so this system may have solutions.
-
System 2:
- \(y = -x^2\)
- \(y = x - 3\)
Similar to the first system, this downward-opening parabola and the straight line can intersect at points. Hence, this system may also have solutions.
-
System 3:
- \(y = x^2\)
- \(y = x - 3\)
This system has an upward-opening parabola and a straight line. They will intersect at some points, so it has solutions.
-
System 4:
- \(y = x^2\)
- \(y = x + 3\)
This system has an upward-opening parabola and a straight line. We'll check for intersections:
Setting \(x^2 = x + 3\): \[ x^2 - x - 3 = 0 \]
We can find the discriminant: \[ D = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 1 + 12 = 13 \]
Since the discriminant is positive, this means there are two real solutions, implying that these equations intersect at two points.
From the above analysis, none of the pairs of equations seem to have no solution.
However, notice one thing: If a parabola opens upwards (\(y = x^2\)) and a line with a positive slope (\(y = x + 3\)) also exists, they may or may not have a point of intersection, depending on their vertical alignment. Since \(y = x + 3\) shifts the line up, this means the parabola will never reach the line, implying that there are no points of intersection, hence no solution.
Therefore, the one system of equations that has no solution is:
y = x^2 and y = x + 3.
3 answers