We want to solve the given system of equations using the substitution method.
y=x^2-3x-3 & (I) y=-2x^2-x+5 & (II)
The y-variable is isolated in Equation (II). This allows us to substitute its value, -2x^2-x+5, for y in Equation (I).
Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable.
3x^2-2x-8=0 ⇔ 3x^2+( - 2)x+( - 8)=0Now, recall the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
We can substitute a= 3, b= - 2, and c= - 8 into this formula to solve the quadratic equation.
We found that y=- 5 when x=2. One solution of the system, which is a point of intersection of two parabolas, is (2,- 5). To find the other solution, we will substitute - 43 for x in Equation (II).
We found that y= 259 when x=- 43. Therefore our second solution, which is the other point of intersection of the two parabolas, is ( - 43, 259).
Checking Our Answer
Checking the answer
We can check our answers by substituting the points into both equations. If they both produce a true statement, our solutions are correct. Let's start by checking (2,- 5). We will substitute 2 and - 5 for x and y, respectively, in Equation (I) and Equation (II).
y=x^2-3x-3 y=-2 x^2-x+5
(I), (II): x= 2, y= - 5
-5? =( 2)^2-3( 2)-3 -5? =-2 ( 2)^2- 2+5
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Simplify
(I), (II): Calculate power
-5? =4-3(2)-3 -5? =-2(4)-2+5
(I), (II): (- a)b = - ab
-5? =4-6-3 -5? =-8-2+5
(I), (II): Add and subtract terms
-5=-5 âś“ -5 = -5 âś“
Since both equations produce true statements, the solution (2,- 5) is correct. Let's now check ( - 43, 259 ).
