What is the solution of the linear-quadratic system of equations?
{y=x2+5x−3y−x=2
To find the solution of the linear-quadratic system of equations, we can substitute the value of x from the second equation into the first equation:
y = x^2 + 5x - 3
y = 2 + x
Substituting the value of y from the second equation into the first equation:
2 + x = x^2 + 5x - 3
Rearranging the equation to the quadratic form:
x^2 + 4x - 5 = 0
Factoring the quadratic equation:
(x - 1)(x + 5) = 0
Setting each factor equal to zero:
x - 1 = 0 or x + 5 = 0
Solving for x:
x = 1 or x = -5
Substituting these values back into the second equation to find the corresponding y values:
For x = 1:
y = 2 + 1
y = 3
For x = -5:
y = 2 + (-5)
y = -3
Therefore, the solution to the linear-quadratic system of equations is (x, y) = (1, 3) and (x, y) = (-5, -3).
To solve the linear-quadratic system of equations:
Step 1: Rewrite the equations in standard form:
y = x^2 + 5x - 3 ...(Equation 1)
-y - x = 2 ...(Equation 2)
Step 2: Solve Equation 2 for y:
-y = x + 2
y = -x - 2 ...(Equation 3)
Step 3: Substitute Equation 3 into Equation 1:
-x - 2 = x^2 + 5x - 3
Step 4: Rearrange the equation so that one side equals zero:
x^2 + 6x - 1 = 0
Step 5: Solve the quadratic equation for x using factoring or the quadratic formula. Since this equation cannot be easily factored, we will use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = 6, and c = -1. Plugging these values into the quadratic formula:
x = (-6 ± √(6^2 - 4(1)(-1))) / 2(1)
Simplifying further:
x = (-6 ± √(36 + 4)) / 2
x = (-6 ± √40) / 2
x = (-6 ± 2√10) / 2
x = -3 ± √10
So, there are two possible solutions for x: -3 + √10 and -3 - √10.
Step 6: Substitute the values of x back into Equation 3 to find the corresponding y-values:
For x = -3 + √10:
y = -(-3 + √10) - 2
y = 3 - √10 - 2
y = 1 - √10
For x = -3 - √10:
y = -(-3 - √10) - 2
y = 3 + √10 - 2
y = 1 + √10
So, the solutions to the system of equations are:
(x, y) = (-3 + √10, 1 - √10) and (-3 - √10, 1 + √10)
To find the solution of the given linear-quadratic system of equations, we can use the method of substitution or elimination.
Let's use the method of substitution:
1. Start with the first equation: y = x^2 + 5x - 3.
2. Substitute this expression for y in the second equation:
-x = 2 - y.
3. Rewrite the second equation in terms of x:
x = y - 2.
4. Substitute this expression for x in the first equation:
y = (y - 2)^2 + 5(y - 2) - 3.
5. Simplify the equation:
y = y^2 - 4y + 4 + 5y - 10 - 3.
y = y^2 + y - 9.
6. Rearrange the equation:
y^2 + y - 9 - y = 0.
y^2 - 9 = 0.
7. Factor the quadratic equation:
(y - 3)(y + 3) = 0.
8. Solve for y:
y - 3 = 0 or y + 3 = 0.
y = 3 or y = -3.
Now that we have the values of y, we can substitute them back into one of the equations to find the corresponding values of x.
For y = 3:
x = y - 2 = 3 - 2 = 1.
So, one solution is (x, y) = (1, 3).
For y = -3:
x = y - 2 = -3 - 2 = -5.
Another solution is (x, y) = (-5, -3).
Therefore, the solution to the linear-quadratic system of equations is (1, 3) and (-5, -3).