What is the solution of the linear-quadratic system of equations?
{y=x^2+5x−3
{y−x=2
To solve the system of equations, we can substitute the expression for y from the second equation into the first equation:
y - x = 2
x^2 + 5x - 3 - x = 2
x^2 + 4x - 5 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula. Factoring it, we have:
(x+5)(x-1) = 0
Setting each factor equal to zero, we get:
x+5 = 0 or x-1 = 0
x = -5 or x = 1
Substituting these values back into the second equation, we can solve for y:
For x = -5:
y - (-5) = 2
y + 5 = 2
y = -3
For x = 1:
y - 1 = 2
y = 3
Therefore, the solution to the system of equations is x = -5, y = -3 and x = 1, y = 3.
To find the solution of the linear-quadratic system of equations, we can use the method of substitution.
Step 1: Solve the second equation for y in terms of x.
y - x = 2
y = x + 2
Step 2: Substitute the value of y in the first equation with x + 2.
x + 2 = x^2 + 5x - 3
Step 3: Rearrange the quadratic equation to one side to solve it.
x^2 + 4x - 5 = 0
Step 4: Factor the quadratic equation.
(x + 5)(x - 1) = 0
Step 5: Set each factor equal to zero and solve for x.
x + 5 = 0 or x - 1 = 0
x = -5 or x = 1
Step 6: Substitute the values of x into one of the original equations 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
Hence, the solution to the linear-quadratic system of equations is (x, y) = (-5, -3) and (1, 3).
To find the solution to the linear-quadratic system of equations, we need to solve the equations simultaneously. Let's start by rewriting the second equation to make it easier to substitute into the first equation.
The second equation, y - x = 2, can be rearranged to express y in terms of x. Adding x to both sides, we have y = 2 + x.
Now, we can substitute this expression for y in the first equation:
x^2 + 5x - 3 = 2 + x
To solve this quadratic equation, let's rearrange it to set it equal to zero:
x^2 + 4x - 5 = 0
Next, we can factor this quadratic equation. The factors will give us the possible solutions for x:
(x + 5)(x - 1) = 0
Setting each factor equal to zero and solving for x, we get:
x + 5 = 0 or x - 1 = 0
Solving each equation, we find:
x = -5 or x = 1
Now that we have the values of x, we can substitute them back into either equation to find the corresponding values of y.
For x = -5:
y = 2 + (-5) = -3
So one solution to the system is (-5, -3).
For x = 1:
y = 2 + 1 = 3
So another solution to the system is (1, 3).
Therefore, the solution to the linear-quadratic system of equations is (-5, -3) and (1, 3).
What are the solutions of the quadratic equation?
2x^2 – 16x + 32 = 0
To find the solutions of the quadratic equation 2x^2 - 16x + 32 = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation.
In this case, the coefficients are a = 2, b = -16, and c = 32. Plugging these values into the quadratic formula, we have:
x = (-(-16) ± √((-16)^2 - 4(2)(32))) / (2(2))
x = (16 ± √(256 - 256)) / 4
x = (16 ± √0) / 4
x = (16 ± 0) / 4
x = 16/4
x = 4
Therefore, the solutions of the quadratic equation 2x^2 - 16x + 32 = 0 are x = 4.
Simplify the expression.
−5+i/2i
To simplify the expression (-5+i)/(2i), we can multiply the numerator and denominator by the conjugate of 2i, which is -2i. This will help eliminate the complex denominator.
(-5+i)/(2i) * (-2i)/(-2i)
Applying the distributive property, this becomes:
(10i-2i^2)/(-4i^2)
Now, let's simplify i^2, which is equal to -1:
(10i-2(-1))/(-4(-1))
Simplifying further:
(10i+2)/4
Dividing both terms by 2:
5i/2 + 1
The simplified expression is (5i/2) + 1.