Enter your answer and show all the steps that you use to solve this problem in the space provided.
What is the solution of the linear-quadratic system of equations?
{ y = x^2 + 5x -3
{ y - x = 2
To solve the linear-quadratic system of equations, we need to find the values of x and y that satisfy both equations.
First, let's solve the second equation for y:
y - x = 2
y = x + 2
Now we substitute this expression for y into the first equation:
x^2 + 5x - 3 = x + 2
Simplifying the equation:
x^2 + 4x - 5 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 1, b = 4, c = -5
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 -> x = 1
x = (-4 - 6) / 2 -> x = -5
Now substitute these values back into the second equation to find the corresponding y-values:
For x = 1:
y = 1 + 2 -> y = 3
So one solution is (1, 3).
For x = -5:
y = -5 + 2 -> y = -3
So another solution is (-5, -3).
Therefore, the solution of the linear-quadratic system of equations is: (1, 3) and (-5, -3).
To find the solution of the linear-quadratic system of equations, we can solve it by substitution or elimination method. Let's first solve it by substitution:
Step 1: Start with the given system of equations:
{ y = x^2 + 5x - 3
{ y - x = 2
Step 2: Solve the second equation for "y":
y = x + 2
Step 3: Substitute the value of "y" from Step 2 into the first equation:
x + 2 = x^2 + 5x - 3
Step 4: Rearrange the equation to standard quadratic form:
x^2 + 4x - 5 = 0
Step 5: Factorize the quadratic equation:
(x + 5)(x - 1) = 0
Step 6: Set each factor equal to zero and solve for "x":
x + 5 = 0 or x - 1 = 0
For x + 5 = 0: x = -5
For x - 1 = 0: x = 1
Step 7: Substitute the values of "x" into one of the original equations to find the corresponding values of "y":
For x = -5: y = (-5)^2 + 5(-5) - 3 = 25 - 25 - 3 = -3
For x = 1: y = (1)^2 + 5(1) - 3 = 1 + 5 - 3 = 3
Step 8: The solution to the system of equations is:
(x, y) = (-5, -3) and (1, 3)
Therefore, the solution to the linear-quadratic system of equations is (-5, -3) and (1, 3).
To find the solution of the linear-quadratic system of equations, we can use the method of substitution or elimination. In this case, let's use the substitution method.
Step 1: Solve one equation for one variable in terms of the other.
We have the second equation: y - x = 2.
Solving for y, we get: y = x + 2.
Step 2: Substitute the expression for y in the other equation.
The first equation is: y = x^2 + 5x - 3.
Substituting y with x + 2, we get: x + 2 = x^2 + 5x - 3.
Step 3: Simplify and rearrange the equation.
Rearranging the equation, we have: x^2 + 4x - 5 = 0.
Step 4: Solve the resulting quadratic equation.
To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, the equation can be factored as: (x - 1)(x + 5) = 0.
Setting each factor equal to zero, we get two possible solutions:
x - 1 = 0 or x + 5 = 0.
Solving these equations, we find:
x = 1 or x = -5.
Step 5: Find the corresponding values of y.
Substituting the values of x back into either of the original equations, we can find the corresponding values of y.
For x = 1:
Using y = x + 2, we get y = 1 + 2 = 3.
So one solution is (x, y) = (1, 3).
For x = -5:
Using y = x + 2, we get y = -5 + 2 = -3.
So another solution is (x, y) = (-5, -3).
Therefore, the solution to the linear-quadratic system of equations is:
(x, y) = (1, 3) and (x, y) = (-5, -3).