Alright class, let's solve these systems of equations algebraically.
### Part (a)
We are given the following system of equations:
1. [tex]\( 5x^2 + 3y = -3 - x \)[/tex]
2. [tex]\( 2x^2 - x = -4 - 2y \)[/tex]
First, we rewrite these equations for clarity:
1. [tex]\( 5x^2 + 3y + x + 3 = 0 \)[/tex]
2. [tex]\( 2x^2 - x + 2y + 4 = 0 \)[/tex]
This gives us the system:
[tex]\[
\begin{cases}
5x^2 + 3y + x + 3 = 0 \\
2x^2 - x + 2y + 4 = 0
\end{cases}
\][/tex]
To solve this system, we solve the equations simultaneously for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. The solutions are:
[tex]\[ (x, y) = \left(-2, -7\right) \text{ and } \left(\frac{3}{4}, -\frac{35}{16}\right) \][/tex]
So, for part (a) we find that the solutions are:
1. [tex]\((-2, -7)\)[/tex]
2. [tex]\(\left(\frac{3}{4}, -\frac{35}{16}\right)\)[/tex]
### Part (b)
We are given the following system of equations:
1. [tex]\( y = 7x - 11 \)[/tex]
2. [tex]\( 5x^2 - 3x - y = 6 \)[/tex]
First, substitute [tex]\( y = 7x - 11 \)[/tex] from the first equation into the second equation:
[tex]\[ 5x^2 - 3x - (7x - 11) = 6 \][/tex]
Simplify the equation:
[tex]\[ 5x^2 - 3x - 7x + 11 = 6 \][/tex]
[tex]\[ 5x^2 - 10x + 11 = 6 \][/tex]
[tex]\[ 5x^2 - 10x + 5 = 0 \][/tex]
[tex]\[ x^2 - 2x + 1 = 0 \][/tex]
[tex]\[ (x - 1)^2 = 0 \][/tex]
So, [tex]\( x = 1 \)[/tex].
Now, substitute [tex]\( x = 1 \)[/tex] back into [tex]\( y = 7x - 11 \)[/tex]:
[tex]\[ y = 7(1) - 11 \][/tex]
[tex]\[ y = 7 - 11 \][/tex]
[tex]\[ y = -4 \][/tex]
Therefore, the solution for part (b) is [tex]\( (x, y) = (1, -4) \)[/tex].
In conclusion, the solutions to the systems of equations are as follows:
### Part (a)
1. [tex]\((-2, -7)\)[/tex]
2. [tex]\(\left(\frac{3}{4}, -\frac{35}{16}\right)\)[/tex]
### Part (b)
1. [tex]\((1, -4)\)[/tex]
