write a pair of equations that have exactly two solutions. (at least one equation is not linear) any links to help?
For the nonlinear equation, it can be something like:
(x + 2) * (x - 5)
and then FOIL it out like this:
x^2 - 3x - 10
You can substitute any two different integers for 2 and -5 there, and it will be a nonlinear function with two solutions.
For a linear equation, it could be as simple as the square root of x. Let's say we have the square root of 4. This has two solutions, since 2^2 = 4, and -2^2 = 4.
I'd send you a link, but I can't in this forum -- but Khan Academy Algebra has great resources on this.
a straight line can intersect a parabola in two places
... the intersection points are the simultaneous solutions
linear form ... y = m x + b ... m and/or b could be zero
quadratic form ... y = a x^2 + b x + c ... b and/or c could be zero
fill in some values and check with a graphing program
to tyger ... the √ function is not linear
but it makes no difference in this case
To create a pair of equations with exactly two solutions, we can combine a linear equation and a quadratic equation. Here's an example:
1. Equation 1 (Linear): y = 2x + 3
2. Equation 2 (Quadratic): y = x^2 - 4x + 3
In this example, Equation 1 is a linear equation and Equation 2 is a quadratic equation. By solving these equations simultaneously, we can find the values of x and y that satisfy both equations. Since Equation 1 is linear, it will always have exactly one solution. Equation 2, being quadratic, can have either zero, one, or two solutions.
To find the solutions, we need to solve these equations simultaneously by substituting one equation into the other. By substituting Equation 1 into Equation 2, we get:
2x + 3 = x^2 - 4x + 3
By rearranging the terms and simplifying, we obtain a quadratic equation:
x^2 - 6x = 0
Factoring out x, we get:
x(x - 6) = 0
From this equation, we find that the solutions for x are x = 0 and x = 6.
To find the corresponding y-values for each x-value, we substitute these solutions back into Equation 1:
When x = 0, y = 2(0) + 3 = 3. So, one solution is (0, 3).
When x = 6, y = 2(6) + 3 = 15. So, another solution is (6, 15).
Therefore, the pair of equations y = 2x + 3 and y = x^2 - 4x + 3 have exactly two solutions: (0, 3) and (6, 15).
For additional help or further practice, you can visit these resources:
- Khan Academy: Quadratic Equations - https://www.khanacademy.org/math/algebra/quadratics
- Purplemath: Quadratic Equations - https://www.purplemath.com/modules/quadform.htm