IDNLearn.com: Your go-to resource for finding precise and accurate answers. Our experts provide timely, comprehensive responses to ensure you have the information you need.
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
