To use the elimination method, we need to set up the two equations in the system so that the coefficients of either x or y are the same or opposite. We can rearrange the second equation to have the form y=mx+b:
y = -2x+5
Now, we have the following system:
y = x^2 - 4x + 10
y = -2x + 5
To use the elimination method, we need to eliminate one variable by adding or subtracting the two equations. In this case, we can subtract the second equation from the first equation:
(x^2 - 4x + 10) - (-2x + 5) = 0
Simplifying, we get:
x^2 - 4x + 10 + 2x - 5 = 0
x^2 - 2x + 5 = 0
At this point, we have a quadratic equation. We can try to solve it by factoring or using the quadratic formula:
x^2 - 2x + 5 = 0
(x - 1)(x - 5) = 0
Setting each factor equal to zero, we get:
x - 1 = 0 or x - 5 = 0
Solving for x, we find two solutions:
x = 1 or x = 5
To find the corresponding y-values, we substitute these x-values back into either of the original equations. Let's use the second equation:
y = -2x + 5
For x = 1:
y = -2(1) + 5
y = -2 + 5
y = 3
For x = 5:
y = -2(5) + 5
y = -10 + 5
y = -5
Therefore, the system has two solutions: (1, 3) and (5, -5).
Use the elimination method to determine whether the linear-quadratic system y=x2−4x+10, y=−2x+5 has zero, one, or two solutions.
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1 answer