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Sagot :
Let's solve the given system of simultaneous equations step-by-step:
[tex]\[ \begin{aligned} (1) \quad y^2 + xy + 3 &= 0 \\ (2) \quad x &= 6y + 5 \end{aligned} \][/tex]
### Part (a): Show that [tex]\(7 y^2 + 5 y + 3 = 0\)[/tex]
1. Substitute [tex]\(x\)[/tex] from equation (2) into equation (1):
Given [tex]\( x = 6y + 5 \)[/tex], we substitute [tex]\( x \)[/tex] in equation (1):
[tex]\[ y^2 + (6y + 5)y + 3 = 0 \][/tex]
2. Expand and simplify the equation:
[tex]\[ y^2 + 6y^2 + 5y + 3 = 0 \][/tex]
Combine like terms:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
This confirms that the equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] is indeed correct.
### Part (b): Work out how many solutions these simultaneous equations have
1. Solve the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] to find the values of [tex]\( y \)[/tex]:
The quadratic formula is given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 7 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 7 \cdot 3}}{2 \cdot 7} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 - 84}}{14} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{-59}}{14} \][/tex]
Since [tex]\(\sqrt{-59} = \sqrt{59}i\)[/tex] (where [tex]\(i\)[/tex] is the imaginary unit), we have:
[tex]\[ y = \frac{-5 \pm \sqrt{59}i}{14} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{-5 - \sqrt{59}i}{14}, \quad y_2 = \frac{-5 + \sqrt{59}i}{14} \][/tex]
2. Calculate the corresponding values of [tex]\( x \)[/tex] for each [tex]\( y \)[/tex] using equation (2):
Substituting [tex]\( y_1 \)[/tex] into equation (2):
[tex]\[ x_1 = 6y_1 + 5 = 6 \left(\frac{-5 - \sqrt{59}i}{14}\right) + 5 \][/tex]
[tex]\[ x_1 = \frac{-30 - 6\sqrt{59}i}{14} + 5 = \frac{-30 - 6\sqrt{59}i}{14} + \frac{70}{14} \][/tex]
[tex]\[ x_1 = \frac{-30 - 6\sqrt{59}i + 70}{14} = \frac{40 - 6\sqrt{59}i}{14} = \frac{20}{7} - \frac{3\sqrt{59}i}{7} \][/tex]
Similarly, for [tex]\( y_2 \)[/tex]:
[tex]\[ x_2 = 6y_2 + 5 = 6 \left(\frac{-5 + \sqrt{59}i}{14}\right) + 5 \][/tex]
[tex]\[ x_2 = \frac{-30 + 6\sqrt{59}i}{14} + 5 = \frac{-30 + 6\sqrt{59}i}{14} + \frac{70}{14} \][/tex]
[tex]\[ x_2 = \frac{-30 + 6\sqrt{59}i + 70}{14} = \frac{40 + 6\sqrt{59}i}{14} = \frac{20}{7} + \frac{3\sqrt{59}i}{7} \][/tex]
In summary, the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] has 2 solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{-5 - \sqrt{59}i}{14}, \quad y_2 = \frac{-5 + \sqrt{59}i}{14} \][/tex]
Correspondingly, the [tex]\( x \)[/tex] values are:
[tex]\[ x_1 = \frac{20}{7} - \frac{3\sqrt{59}i}{7}, \quad x_2 = \frac{20}{7} + \frac{3\sqrt{59}i}{7} \][/tex]
Thus, there are 2 solutions to the system of simultaneous equations.
[tex]\[ \begin{aligned} (1) \quad y^2 + xy + 3 &= 0 \\ (2) \quad x &= 6y + 5 \end{aligned} \][/tex]
### Part (a): Show that [tex]\(7 y^2 + 5 y + 3 = 0\)[/tex]
1. Substitute [tex]\(x\)[/tex] from equation (2) into equation (1):
Given [tex]\( x = 6y + 5 \)[/tex], we substitute [tex]\( x \)[/tex] in equation (1):
[tex]\[ y^2 + (6y + 5)y + 3 = 0 \][/tex]
2. Expand and simplify the equation:
[tex]\[ y^2 + 6y^2 + 5y + 3 = 0 \][/tex]
Combine like terms:
[tex]\[ 7y^2 + 5y + 3 = 0 \][/tex]
This confirms that the equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] is indeed correct.
### Part (b): Work out how many solutions these simultaneous equations have
1. Solve the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] to find the values of [tex]\( y \)[/tex]:
The quadratic formula is given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 7 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = 3 \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 7 \cdot 3}}{2 \cdot 7} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 - 84}}{14} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{-59}}{14} \][/tex]
Since [tex]\(\sqrt{-59} = \sqrt{59}i\)[/tex] (where [tex]\(i\)[/tex] is the imaginary unit), we have:
[tex]\[ y = \frac{-5 \pm \sqrt{59}i}{14} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{-5 - \sqrt{59}i}{14}, \quad y_2 = \frac{-5 + \sqrt{59}i}{14} \][/tex]
2. Calculate the corresponding values of [tex]\( x \)[/tex] for each [tex]\( y \)[/tex] using equation (2):
Substituting [tex]\( y_1 \)[/tex] into equation (2):
[tex]\[ x_1 = 6y_1 + 5 = 6 \left(\frac{-5 - \sqrt{59}i}{14}\right) + 5 \][/tex]
[tex]\[ x_1 = \frac{-30 - 6\sqrt{59}i}{14} + 5 = \frac{-30 - 6\sqrt{59}i}{14} + \frac{70}{14} \][/tex]
[tex]\[ x_1 = \frac{-30 - 6\sqrt{59}i + 70}{14} = \frac{40 - 6\sqrt{59}i}{14} = \frac{20}{7} - \frac{3\sqrt{59}i}{7} \][/tex]
Similarly, for [tex]\( y_2 \)[/tex]:
[tex]\[ x_2 = 6y_2 + 5 = 6 \left(\frac{-5 + \sqrt{59}i}{14}\right) + 5 \][/tex]
[tex]\[ x_2 = \frac{-30 + 6\sqrt{59}i}{14} + 5 = \frac{-30 + 6\sqrt{59}i}{14} + \frac{70}{14} \][/tex]
[tex]\[ x_2 = \frac{-30 + 6\sqrt{59}i + 70}{14} = \frac{40 + 6\sqrt{59}i}{14} = \frac{20}{7} + \frac{3\sqrt{59}i}{7} \][/tex]
In summary, the quadratic equation [tex]\( 7y^2 + 5y + 3 = 0 \)[/tex] has 2 solutions for [tex]\( y \)[/tex]:
[tex]\[ y_1 = \frac{-5 - \sqrt{59}i}{14}, \quad y_2 = \frac{-5 + \sqrt{59}i}{14} \][/tex]
Correspondingly, the [tex]\( x \)[/tex] values are:
[tex]\[ x_1 = \frac{20}{7} - \frac{3\sqrt{59}i}{7}, \quad x_2 = \frac{20}{7} + \frac{3\sqrt{59}i}{7} \][/tex]
Thus, there are 2 solutions to the system of simultaneous equations.
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