Answer :
To solve the given simultaneous equations, let's go through the problem step-by-step.
### Given Equations:
1. [tex]\( y^2 + xy + 3 = 0 \)[/tex]
2. [tex]\( x = 6y + 5 \)[/tex]
### Part (a) Show that [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex]
#### Step 1: Substitute [tex]\( x \)[/tex] from the second equation into the first equation.
Given [tex]\( x = 6y + 5 \)[/tex], replace [tex]\( x \)[/tex] in the first equation:
[tex]\[ y^2 + (6y + 5)y + 3 = 0 \][/tex]
#### Step 2: Simplify the equation.
[tex]\[ y^2 + 6y^2 + 5y + 3 = 0 \][/tex]
Combine like terms:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
This demonstrates that indeed the substitution results in the equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex], as requested.
### Part (b) Determine the number of solutions to the simultaneous equations.
#### Step 1: Solve the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex].
To determine the solutions for [tex]\( y \)[/tex], we solve the quadratic equation:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
#### Step 2: Use the quadratic formula [tex]\( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex], where [tex]\( a = 7 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = 3 \)[/tex].
The discriminant [tex]\( \Delta \)[/tex] is calculated as follows:
[tex]\[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 7 \cdot 3 = 25 - 84 = -59 \][/tex]
Since the discriminant is negative ([tex]\( -59 \)[/tex]), the quadratic equation has two complex (imaginary) roots.
The solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{{-5 \pm \sqrt{-59}}}{14} = \frac{{-5 \pm \sqrt{59}i}}{14} \][/tex]
#### Step 3: Find the corresponding [tex]\( x \)[/tex] values for each solution of [tex]\( y \)[/tex].
Using the equation [tex]\( x = 6y + 5 \)[/tex]:
For [tex]\( y_1 = \frac{{-5 - \sqrt{59}i}}{14} \)[/tex]:
[tex]\[ x_1 = 6 \left( \frac{{-5 - \sqrt{59}i}}{14} \right) + 5 = \frac{{-30 - 6 \sqrt{59}i}}{14} + 5 = \frac{{-30 - 6 \sqrt{59}i + 70}}{14} = \frac{{40 - 6 \sqrt{59}i}}{14} = \frac{{20 - 3 \sqrt{59}i}}{7} \][/tex]
For [tex]\( y_2 = \frac{{-5 + \sqrt{59}i}}{14} \)[/tex]:
[tex]\[ x_2 = 6 \left( \frac{{-5 + \sqrt{59}i}}{14} \right) + 5 = \frac{{-30 + 6 \sqrt{59}i}}{14} + 5 = \frac{{-30 + 6 \sqrt{59}i + 70}}{14} = \frac{{40 + 6 \sqrt{59}i}}{14} = \frac{{20 + 3 \sqrt{59}i}}{7} \][/tex]
Thus, the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] has exactly two complex solutions for [tex]\( y \)[/tex], and correspondingly, there are two complex solutions for [tex]\( x \)[/tex].
### Conclusion
The simultaneous equations have two complex solutions for the pairs [tex]\((x, y)\)[/tex]:
1. [tex]\( y = \frac{{-5 - \sqrt{59}i}}{14} \)[/tex] and [tex]\( x = \frac{{20 - 3 \sqrt{59}i}}{7} \)[/tex]
2. [tex]\( y = \frac{{-5 + \sqrt{59}i}}{14} \)[/tex] and [tex]\( x = \frac{{20 + 3 \sqrt{59}i}}{7} \)[/tex]
### Given Equations:
1. [tex]\( y^2 + xy + 3 = 0 \)[/tex]
2. [tex]\( x = 6y + 5 \)[/tex]
### Part (a) Show that [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex]
#### Step 1: Substitute [tex]\( x \)[/tex] from the second equation into the first equation.
Given [tex]\( x = 6y + 5 \)[/tex], replace [tex]\( x \)[/tex] in the first equation:
[tex]\[ y^2 + (6y + 5)y + 3 = 0 \][/tex]
#### Step 2: Simplify the equation.
[tex]\[ y^2 + 6y^2 + 5y + 3 = 0 \][/tex]
Combine like terms:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
This demonstrates that indeed the substitution results in the equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex], as requested.
### Part (b) Determine the number of solutions to the simultaneous equations.
#### Step 1: Solve the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex].
To determine the solutions for [tex]\( y \)[/tex], we solve the quadratic equation:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
#### Step 2: Use the quadratic formula [tex]\( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex], where [tex]\( a = 7 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = 3 \)[/tex].
The discriminant [tex]\( \Delta \)[/tex] is calculated as follows:
[tex]\[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 7 \cdot 3 = 25 - 84 = -59 \][/tex]
Since the discriminant is negative ([tex]\( -59 \)[/tex]), the quadratic equation has two complex (imaginary) roots.
The solutions for [tex]\( y \)[/tex] are:
[tex]\[ y = \frac{{-5 \pm \sqrt{-59}}}{14} = \frac{{-5 \pm \sqrt{59}i}}{14} \][/tex]
#### Step 3: Find the corresponding [tex]\( x \)[/tex] values for each solution of [tex]\( y \)[/tex].
Using the equation [tex]\( x = 6y + 5 \)[/tex]:
For [tex]\( y_1 = \frac{{-5 - \sqrt{59}i}}{14} \)[/tex]:
[tex]\[ x_1 = 6 \left( \frac{{-5 - \sqrt{59}i}}{14} \right) + 5 = \frac{{-30 - 6 \sqrt{59}i}}{14} + 5 = \frac{{-30 - 6 \sqrt{59}i + 70}}{14} = \frac{{40 - 6 \sqrt{59}i}}{14} = \frac{{20 - 3 \sqrt{59}i}}{7} \][/tex]
For [tex]\( y_2 = \frac{{-5 + \sqrt{59}i}}{14} \)[/tex]:
[tex]\[ x_2 = 6 \left( \frac{{-5 + \sqrt{59}i}}{14} \right) + 5 = \frac{{-30 + 6 \sqrt{59}i}}{14} + 5 = \frac{{-30 + 6 \sqrt{59}i + 70}}{14} = \frac{{40 + 6 \sqrt{59}i}}{14} = \frac{{20 + 3 \sqrt{59}i}}{7} \][/tex]
Thus, the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] has exactly two complex solutions for [tex]\( y \)[/tex], and correspondingly, there are two complex solutions for [tex]\( x \)[/tex].
### Conclusion
The simultaneous equations have two complex solutions for the pairs [tex]\((x, y)\)[/tex]:
1. [tex]\( y = \frac{{-5 - \sqrt{59}i}}{14} \)[/tex] and [tex]\( x = \frac{{20 - 3 \sqrt{59}i}}{7} \)[/tex]
2. [tex]\( y = \frac{{-5 + \sqrt{59}i}}{14} \)[/tex] and [tex]\( x = \frac{{20 + 3 \sqrt{59}i}}{7} \)[/tex]
