Sign In
To solve the equation ax^2+bx+c=0, use the Quadratic Formula.
(- 11, - 91) and (9, - 71)
We want to solve the given system of equations using the Substitution Method. - x^2-x+19=y & (I) x=y+80 & (II) The y-variable is isolated in Equation (I). This allows us to substitute its value - x^2 - x +19 for y in Equation (II).
(II): y= - x^2-x+19
(II): LHS-x=RHS-x
(II): Rearrange equation
(II): Add terms
Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. - x^2-2x+99=0 ⇔ - 1x^2+( - 2)x+ 99=0
Substitute values
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
| x=2± 20/- 2 | |
|---|---|
| x_1=2+20/- 2 | x_2=2-20/- 2 |
| x_1=22/- 2 | x_2=- 18/- 2 |
| x_1=- 11 | x_2=9 |
Now, consider Equation (I). - x^2-x+19=y We can substitute x=- 11 and x=9 into the above equation to find the values for y. Let's start with x=- 11.
x= - 11
Calculate power
- (- a)=a
Add terms
Rearrange equation
We found that y=- 91 when x=- 11. One solution of the system, which is a point of intersection of the two parabolas, is (- 11,- 91). To find the other solution, we will substitute 9 for x in Equation (I) again.
We found that y=- 71 when x=9. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (9, - 71).
(I), (II): x= - 11, y= - 91
Since both equations produced true statements, the solution (- 11,- 91) is correct. Let's now check ( 9,- 71).
(I), (II): x= 9, y= - 71
Since again both equations produce true statements, the solution (9,- 71) is also correct.