Sure! Let's solve the given simultaneous equations step by step.
The given system of equations is:
1) [tex]\( y = x + 6 \)[/tex]
2) [tex]\( x^2 + y^2 = 26 \)[/tex]
Part (a): Show that [tex]\( x^2 + 6x + 5 = 0 \)[/tex]
To start, substitute [tex]\( y = x + 6 \)[/tex] from the first equation into the second equation.
[tex]\[
x^2 + (x + 6)^2 = 26
\][/tex]
Now, expand the squared term:
[tex]\[
x^2 + (x^2 + 12x + 36) = 26
\][/tex]
Combine like terms:
[tex]\[
x^2 + x^2 + 12x + 36 = 26
\][/tex]
Simplify the equation by combining the [tex]\( x^2 \)[/tex] terms and constant terms:
[tex]\[
2x^2 + 12x + 36 = 26
\][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[
2x^2 + 12x + 36 - 26 = 0
\][/tex]
Simplify the constants:
[tex]\[
2x^2 + 12x + 10 = 0
\][/tex]
Next, divide the entire equation by 2 for simplification:
[tex]\[
x^2 + 6x + 5 = 0
\][/tex]
This matches the required form, hence we have shown that:
[tex]\[
x^2 + 6x + 5 = 0
\][/tex]
Part (b): Use part (a) to solve the simultaneous equations
Now that we have [tex]\( x^2 + 6x + 5 = 0 \)[/tex], we can solve this quadratic equation.
The quadratic equation [tex]\( x^2 + 6x + 5 = 0 \)[/tex] can be factored as:
[tex]\[
(x + 1)(x + 5) = 0
\][/tex]
Setting each factor equal to zero gives the solutions for [tex]\( x \)[/tex]:
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
[tex]\[
x + 5 = 0 \implies x = -5
\][/tex]
Now, substitute these [tex]\( x \)[/tex] values back into the equation [tex]\( y = x + 6 \)[/tex] to find the corresponding [tex]\( y \)[/tex] values.
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = -1 + 6 = 5
\][/tex]
For [tex]\( x = -5 \)[/tex]:
[tex]\[
y = -5 + 6 = 1
\][/tex]
Therefore, the solutions to the simultaneous equations are:
1. [tex]\( (x, y) = (-1, 5) \)[/tex]
2. [tex]\( (x, y) = (-5, 1) \)[/tex]
So, the pair of solutions for the simultaneous equations are:
1. [tex]\( (-5, 1) \)[/tex]
2. [tex]\( (-1, 5) \)[/tex]
No decimals were produced in this solution, so we are done.
