Let's solve the given system of equations step-by-step:
Given system of equations:
1) [tex]\(x + y = 8\)[/tex]
2) [tex]\(2x^2 - y = -5\)[/tex]
Step 1: Express [tex]\(y\)[/tex] from the first equation.
From [tex]\(x + y = 8\)[/tex]:
[tex]\[ y = 8 - x \][/tex]
Step 2: Substitute [tex]\(y\)[/tex] into the second equation.
Substitute [tex]\(y = 8 - x\)[/tex] into [tex]\(2x^2 - y = -5\)[/tex]:
[tex]\[ 2x^2 - (8 - x) = -5 \][/tex]
Step 3: Simplify the equation.
[tex]\[ 2x^2 - 8 + x = -5 \][/tex]
[tex]\[ 2x^2 + x - 8 = -5 \][/tex]
[tex]\[ 2x^2 + x - 3 = 0 \][/tex]
Step 4: Solve the quadratic equation.
The quadratic equation is [tex]\(2x^2 + x - 3 = 0\)[/tex]. Let's solve this using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -3\)[/tex].
Discriminant, [tex]\(\Delta = b^2 - 4ac\)[/tex]:
[tex]\[ \Delta = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \][/tex]
So, the roots are:
[tex]\[ x = \frac{-1 \pm \sqrt{25}}{2 \cdot 2} = \frac{-1 \pm 5}{4} \][/tex]
Thus, we have two solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-1 + 5}{4} = 1 \][/tex]
[tex]\[ x = \frac{-1 - 5}{4} = -\frac{3}{2} \][/tex]
Step 5: Find corresponding values of [tex]\(y\)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 8 - x = 8 - 1 = 7 \][/tex]
For [tex]\( x = -\frac{3}{2} \)[/tex]:
[tex]\[ y = 8 - \left(-\frac{3}{2}\right) = 8 + \frac{3}{2} = \frac{16}{2} + \frac{3}{2} = \frac{19}{2} \][/tex]
Step 6: Write the solution pairs.
The solutions to the system of equations are:
[tex]\[ \left(1, 7\right) \][/tex]
and
[tex]\[ \left(-\frac{3}{2}, \frac{19}{2}\right) \][/tex]
So the solutions of the system are:
[tex]\[ \boxed{\left(1, 7\right) \text{ and } \left(-\frac{3}{2}, \frac{19}{2}\right)} \][/tex]
