Question

What is the solution set of this system of equations?y= x^2+ 2x + 2y = x +1The solution set of a system of equations is the set of all values that satisfy all theequations.

Answer 1

Hello there. To solve this question, we'll have to remember some properties about system of equations.

Given the following system of equations:

`\begin{cases}y=x^2+2x+2 \\ y=x+1\end{cases}`

In the R² plane, we have that the solutions to this system are the points of intersection of the parabola y = x² + 2x + 2 and the line y = x + 1.

In fact, there must be two points of intersection and, therefore, two solutions for this system of equation.

Let's plug the first equation into the second equation (called the method of substitution):

`x^2+2x+2=x+1`

Subtract x + 1 on both sides of the equation

`\begin{gathered} x^2+2x+2-(x+1)=x+1-(x+1) \\ x^2+2x+2-x-1=0 \\ x^2+x+1=0 \end{gathered}`

Now, this is a quadratic equation of the form:

`ax^2+bx+c=0,a\\e0`

Its solutions are given by the quadratic formula:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Plugging the coefficients a = b = c = 1, we get:

`x=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot1}}{2\cdot1}`

Square the number and multiply the values. Add everything inside the radical.

`x=\frac{-1\pm\sqrt[]{1-4}_{}}{2}=\frac{-1\pm\sqrt[]{-3}_{}}{2}`

In this case, we got a negative number inside the square root. This is not defined in the real numbers, therefore we don't have solutions for this system of equations.

Graphically, and this is why I omitted it from the beginning, is that in fact the parabola and the line won't have any intersection points, as you can see in the following image:

If we were to solve this system of equations in the set of the complex numbers, then we would have solutions.

First, remember:

`\sqrt[]{-1}=i`

Such that:

`x=\frac{-1\pm\sqrt[]{-3}}{2}=\frac{-1\pm\sqrt[]{3}\cdot\sqrt[]{-1}}{2}=\frac{-1\pm\sqrt[]{3}i}{2}`

And the values of y can be found by plugging it in the second equation (the easier one)

`y=\frac{-1\pm\sqrt[]{3}i}{2}+1=\frac{-1\pm\sqrt[]{3}i+2_{}}{2}=\frac{1\pm\sqrt[]{3}i}{2}`

And the ordered pairs (x, y) would have been:

`\begin{gathered} \mleft((-1+√(3)i)/(2),(1+√(3)i)/(2)\mright) \\ \mleft((-1-√(3)i)/(2),(1-√(3)i)/(2)\mright) \end{gathered}`

But as we're talking about real numbers, we say that this is an impossible system of equations, that has no solutions.