Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
