PLEASE HELP ME ASAP!!!!!!!!!!!! :) I really need to remember on how to do this equation.
Solve the system algebraically.
y=25x^2+x
9x+y=3
I would use substitution
from the 2nd: y = 3 - 9x
sub into the 1st
3-9x = 25x^2 + x
25x^2 + 10x - 3 = 0
(5x-1)(5x+3) = 0
x = 1/5 or x = -3/5
if x = 1/5 , y = 3-9(1/5) = 6/5
if x = -3/5 , y = 3 - 9(-3/5) = 42/5
Thank you so much I really appreciate it.
Sure, I'd be happy to help you solve the system of equations algebraically.
The given equations are:
1) y = 25x^2 + x
2) 9x + y = 3
To solve the system, we need to find the values of x and y that satisfy both equations.
One way to solve this system is by substitution. Let's solve for y in equation 1) and substitute it into equation 2).
From equation 1), we have:
y = 25x^2 + x
Now, substitute y with 25x^2 + x in equation 2):
9x + (25x^2 + x) = 3
Simplifying the equation:
9x + 25x^2 + x = 3
25x^2 + 10x + 3 = 0
Now, we have a quadratic equation. To solve quadratic equations, we can either factor, complete the square, or use the quadratic formula.
If we attempt to factor the quadratic equation, we find that it cannot be factored easily. Therefore, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation 25x^2 + 10x + 3 = 0, we have:
a = 25, b = 10, c = 3
Substituting these values into the quadratic formula, we get:
x = (-10 ± √(10^2 - 4 * 25 * 3)) / (2 * 25)
x = (-10 ± √(100 - 300)) / 50
x = (-10 ± √(-200)) / 50
Since we have a negative value under the square root (√(-200)), the solutions to this equation are imaginary. This means that there are no real solutions for x that satisfy both equations.
Therefore, the given system of equations has no solution algebraically.