What is the solution of the linear-quadratic system of equations?
{y = x^2 + 5x - 3
{y - x = 2
So confused about how to do this any help is appreciated.
just use substitution. you know that
y = x+2, so now plug that into the other equation to get
x+2 = x^2+5x-3
x^2+4x-5 = 0
(x-1)(x+5) = 0
x = -5 or 1
So, y = -2 or 3
To find the solution of the linear-quadratic system of equations, you need to solve the equations simultaneously. Let's start by using the second equation to find the value of y in terms of x.
Given:
y - x = 2
Rearrange the equation to solve for y:
y = x + 2
Now, substitute this value of y in the first equation:
x + 2 = x^2 + 5x - 3
Rearrange the equation to put it in standard quadratic form:
x^2 + 4x - 5 = 0
Now, we can solve this quadratic equation to find the values of x. There are a few methods you can use to solve a quadratic equation, such as factoring, completing the square, or using the quadratic formula.
In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = 4, and c = -5. Plugging in these values, we get:
x = (-4 ± √(4^2 - 4(1)(-5))) / (2(1))
x = (-4 ± √(16 + 20)) / 2
x = (-4 ± √36) / 2
x = (-4 ± 6) / 2
This gives us two possible values for x:
x = (-4 + 6) / 2 = 1
x = (-4 - 6) / 2 = -5
Now that we have the values of x, we can substitute them back into the second equation to find the corresponding values of y.
When x = 1, y = x + 2 = 1 + 2 = 3. So one solution is (x,y) = (1,3).
When x = -5, y = x + 2 = -5 + 2 = -3. So the second solution is (x,y) = (-5,-3).
Therefore, the solution of the linear-quadratic system of equations is (x,y) = (1,3) and (x,y) = (-5,-3).