To solve the simultaneous equations:
[tex]\[
\begin{aligned}
&1. \quad y = x^2 - 9x + 21 \\
&2. \quad y = 2x - 3
\end{aligned}
\][/tex]
we will substitute the expression for \( y \) from the second equation into the first equation.
Given:
[tex]\[ y = 2x - 3 \][/tex]
Substitute \( y \) in the first equation:
[tex]\[ 2x - 3 = x^2 - 9x + 21 \][/tex]
Rearrange this to form a standard quadratic equation:
[tex]\[ x^2 - 9x - 2x + 21 + 3 = 0 \][/tex]
[tex]\[ x^2 - 11x + 24 = 0 \][/tex]
We need to solve this quadratic equation for \( x \). To do this, we factorize the quadratic equation:
[tex]\[ x^2 - 11x + 24 = 0 \][/tex]
The factors of 24 that add up to -11 are -3 and -8. Hence, we can write:
[tex]\[ (x - 3)(x - 8) = 0 \][/tex]
Setting each factor equal to zero gives the solutions for \( x \):
[tex]\[ x - 3 = 0 \][/tex]
[tex]\[ x = 3 \][/tex]
[tex]\[ x - 8 = 0 \][/tex]
[tex]\[ x = 8 \][/tex]
Now, we substitute these solutions back into the second original equation \( y = 2x - 3 \) to find the corresponding \( y \)-values.
1. When \( x = 3 \):
[tex]\[ y = 2(3) - 3 = 6 - 3 = 3 \][/tex]
2. When \( x = 8 \):
[tex]\[ y = 2(8) - 3 = 16 - 3 = 13 \][/tex]
Therefore, the solutions to the system of equations are:
[tex]\[ (x, y) = (3, 3) \][/tex]
[tex]\[ (x, y) = (8, 13) \][/tex]
These can be written as the solution set:
[tex]\[ \{(3, 3), (8, 13)\} \][/tex]
Hence, the simultaneous equations [tex]\( y = x^2 - 9x + 21 \)[/tex] and [tex]\( y = 2x - 3 \)[/tex] are solved by the points [tex]\((3, 3)\)[/tex] and [tex]\((8, 13)\)[/tex].
