What is the solution of the linear-quadratic system of equations?{y=x^2+5x−3y−x=2
To find the solution of the linear-quadratic system of equations, we'll solve the two equations simultaneously.
1) Rearrange the first equation:
y = x^2 + 5x - 3
2) Substitute the rearranged first equation into the second equation:
x^2 + 5x - 3 - x = 2
3) Simplify the equation:
x^2 + 4x - 5 = 0
4) Factor the equation:
(x + 5)(x - 1) = 0
5) Set each factor equal to zero and solve for x:
x + 5 = 0 or x - 1 = 0
x = -5 or x = 1
6) Substitute the values of x back into the first equation to find the corresponding y values:
For x = -5:
y = (-5)^2 + 5(-5) - 3
y = 25 - 25 - 3
y = -3
For x = 1:
y = 1^2 + 5(1) - 3
y = 1 + 5 - 3
y = 3
Therefore, the solution to the linear-quadratic system of equations is:
x = -5, y = -3
x = 1, y = 3
To find the solution of the linear-quadratic system of equations, we need to solve the equations simultaneously. Let's use the method of substitution.
Given equations:
1) y = x^2 + 5x - 3
2) -y - x = 2
Let's solve equation 2) for y:
-y = x + 2
y = -x - 2
Now we substitute y in equation 1) with -x - 2:
-x - 2 = x^2 + 5x - 3
Rearranging the equation:
x^2 + 6x - 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 6, and c = -1. Plugging in these values:
x = (-6 ± sqrt(6^2 - 4(1)(-1))) / (2(1))
x = (-6 ± sqrt(36 + 4)) / 2
x = (-6 ± sqrt(40)) / 2
x = (-6 ± 2sqrt(10)) / 2
Simplifying:
x = -3 ± sqrt(10)
So the possible solutions for x are x = -3 + sqrt(10) and x = -3 - sqrt(10).
To find the corresponding y values, we substitute these x values into equation 1):
For x = -3 + sqrt(10):
y = (-3 + sqrt(10))^2 + 5(-3 + sqrt(10)) - 3
Simplifying,
y = 18 - 6sqrt(10)
For x = -3 - sqrt(10):
y = (-3 - sqrt(10))^2 + 5(-3 - sqrt(10)) - 3
Simplifying,
y = 18 + 6sqrt(10)
Therefore, the solutions for the linear-quadratic system of equations are:
(-3 + sqrt(10), 18 - 6sqrt(10))
(-3 - sqrt(10), 18 + 6sqrt(10))
To find the solution of the linear-quadratic system of equations, let's start by writing the given equations in a standard form.
Equation 1: y = x^2 + 5x - 3
Equation 2: -y - x = 2
We can rearrange Equation 2 to solve for y:
-y = x + 2
y = -x - 2
Now we can substitute this expression for y into Equation 1:
-x - 2 = x^2 + 5x - 3
Next, let's rearrange the equation to bring all terms to one side:
x^2 + 6x - 1 = 0
Now we have a quadratic equation. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Comparing the quadratic equation to the standard form (ax^2 + bx + c = 0), we have:
a = 1, b = 6, c = -1
Plugging these values into the quadratic formula:
x = (-(6) ± √((6)^2 - 4(1)(-1))) / (2(1))
x = (-6 ± √(36 + 4)) / 2
x = (-6 ± √40) / 2
x = (-6 ± 2√10) / 2
x = -3 ± √10
So, the solutions for x are x = -3 + √10 and x = -3 - √10.
Now that we have the values of x, we can substitute them back into either equation to find the corresponding y values. Let's use Equation 1:
For x = -3 + √10:
y = (-3 + √10)^2 + 5(-3 + √10) - 3
y = 10 + 6√10 + 10 - 15 + 5√10 - 3
y = 2 + 11√10
For x = -3 - √10:
y = (-3 - √10)^2 + 5(-3 - √10) - 3
y = 10 - 6√10 + 10 + 15 - 5√10 - 3
y = 12 - √10
So, the solutions to the given linear-quadratic system of equations are:
(-3 + √10, 2 + 11√10) and (-3 - √10, 12 - √10).