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Answer:
The first 3 are correct on Edg
Step-by-step explanation:
You can use the method of substitution and the formula for finding the roots of the quadratic equations to find the solution to the given system of equations.
- Using substitution, the system of equations can be rewritten as 2x² – x – 3 = 0.
- There are two real number solutions.
How to find the solution to the given system of equation?
For that , we will try solving it first using the method of substitution in which we express one variable in other variable's form and then you can substitute this value in other equation to get linear equation in one variable.
If there comes a = a situation for any a, then there are infinite solutions.
If there comes wrong equality, say for example, 3=2, then there are no solutions, else there is one unique solution to the given system of equations.
How to find the solution to a standard quadratic equation?
Suppose the given quadratic equation is
[tex]ax^2 + bx + c = 0[/tex]
Then its solutions are given as
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
How to use discriminant to find the property of solutions of given quadratic equation?
The quantity D is called discriminant.
The solution contains the term [tex]D = \sqrt{b^2 - 4ac}[/tex] which will be real only if discriminant is 0 or positive. It came out to be < 0, thus, the solutions to the given quadratic equation became non real (imaginary solutions)
Using the above facts to find the solution of the given system of equations:
The given system of equations is
[tex]y = 2x^2 + 3\\y - x = 6[/tex]
Using the second equation to get y in terms of x
[tex]y - x = 6\\y = x + 6[/tex]
Using the above expression for y to substitute in place of y in the first equation we get
[tex]y= 2x^2 + 3\\x + 6 = 2x^2 + 3\\2x^2 -x - 3 = 0[/tex]
Comparing this equation with the standard form of the quadratic equation [tex]ax^2 + bx + c = 0[/tex] , we get the solution(roots) of the quadratic equation, and thus, of the given system of equations as
for a = 2, b = -1. c = - 3
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{1 \pm \sqrt{1 - 4 \times 2 \times -3}}{4} = \dfrac{1 \pm \sqrt{25}}{4} = \dfrac{1 \pm 5}{4}\\\\x = \dfrac{3}{2},\\ or\\x = -1[/tex]
Thus, the two solutions came out as real values (as the discriminant was non negative).
From the obtained values of x, we get value of y from the expression obtained for it as
For x = 3/2
[tex]y = 6 + x = 6 + 3/2 = 15/2[/tex]
For x = -1
[tex]y = 6 + x = 6-1 = 5[/tex]
Thus,
The solutions to the system of equations are:
[tex]x = 3/2, y = 15/2\\x = -1, y = 5[/tex]
From all these we have the options that apply as:
- Using substitution, the system of equations can be rewritten as 2x² – x – 3 = 0.
- There are two real number solutions.
The quadratic equation was not in standard form [tex]ax^2 + bx + c = 0[/tex] , but can be arranged to be in so.
Learn more here about solutions to the system of equations here:
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