Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Alright, let's solve the system of equations step by step:
Given equations:
1) [tex]\(3y = 12 - 4x\)[/tex]
2) [tex]\((x+1)^2 + (y-2)^2 = 4\)[/tex]
### Step 1: Solve the first equation for [tex]\(y\)[/tex]
[tex]\[3y = 12 - 4x\][/tex]
To isolate [tex]\(y\)[/tex], divide both sides by 3:
[tex]\[y = \frac{12 - 4x}{3}\][/tex]
[tex]\[y = 4 - \frac{4x}{3}\][/tex]
[tex]\[y = 4 - \frac{4}{3}x\][/tex]
### Step 2: Substitute [tex]\(y\)[/tex] into the second equation
Now that we have [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], we can substitute this expression into the second equation:
[tex]\[(x+1)^2 + \left(4 - \frac{4}{3}x - 2\right)^2 = 4\][/tex]
Simplify the expression for [tex]\(y\)[/tex]:
[tex]\[ y = 4 - \frac{4}{3}x - 2 = 2 - \frac{4}{3}x \][/tex]
Substitute [tex]\(y = 2 - \frac{4}{3}x\)[/tex] into the second equation:
[tex]\[(x+1)^2 + \left(2 - \frac{4}{3}x - 2\right)^2 = 4\][/tex]
Further simplify:
[tex]\[(x+1)^2 + \left(-\frac{4}{3}x\right)^2 = 4\][/tex]
[tex]\[(x+1)^2 + \left(\frac{16}{9}x^2\right) = 4\][/tex]
Multiply through by 9 to clear the fraction:
[tex]\[9(x+1)^2 + 16x^2 = 36\][/tex]
Expand [tex]\(9(x+1)^2\)[/tex]:
[tex]\[9(x^2 + 2x + 1) + 16x^2 = 36\][/tex]
Combine like terms:
[tex]\[9x^2 + 18x + 9 + 16x^2 = 36\][/tex]
[tex]\[25x^2 + 18x + 9 = 36\][/tex]
### Step 3: Solve the quadratic equation
Move all terms to one side to set the equation to zero:
[tex]\[25x^2 + 18x + 9 - 36 = 0\][/tex]
[tex]\[25x^2 + 18x - 27 = 0\][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. We can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Here, [tex]\(a = 25\)[/tex], [tex]\(b = 18\)[/tex], and [tex]\(c = -27\)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 18^2 - 4(25)(-27) \][/tex]
[tex]\[ \Delta = 324 + 2700 \][/tex]
[tex]\[ \Delta = 3024 \][/tex]
Calculate the roots:
[tex]\[ x = \frac{-18 \pm \sqrt{3024}}{2(25)} \][/tex]
[tex]\[ x = \frac{-18 \pm 54.98}{50} \][/tex]
We get two solutions for [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-18 + 54.98}{50} = \frac{36.98}{50} = 0.7396 \][/tex]
[tex]\[ x_2 = \frac{-18 - 54.98}{50} = \frac{-72.98}{50} = -1.4596 \][/tex]
### Step 4: Find the corresponding [tex]\(y\)[/tex] values
Substitute [tex]\(x_1\)[/tex] back into [tex]\(y = 4 - \frac{4}{3}x\)[/tex]:
[tex]\[ y_1 = 4 - \frac{4}{3}(0.7396) = 4 - 0.9861 = 3.0139 \][/tex]
Substitute [tex]\(x_2\)[/tex] back into [tex]\(y = 4 - \frac{4}{3}x\)[/tex]:
[tex]\[ y_2 = 4 - \frac{4}{3}(-1.4596) = 4 + 1.946 = 5.946 \][/tex]
### Step 5: Final solution
After solving the equations, we find that the solutions are:
[tex]\[ (x_1, y_1) = \left(\frac{3}{5}, \frac{16}{5}\right) \][/tex]
So, the final solution is:
[tex]\[ \left(\frac{3}{5}, \frac{16}{5}\right) \][/tex]
Thus, the solution set for the system of equations is [tex]\[\left(\frac{3}{5}, \frac{16}{5}\right)\][/tex].
Given equations:
1) [tex]\(3y = 12 - 4x\)[/tex]
2) [tex]\((x+1)^2 + (y-2)^2 = 4\)[/tex]
### Step 1: Solve the first equation for [tex]\(y\)[/tex]
[tex]\[3y = 12 - 4x\][/tex]
To isolate [tex]\(y\)[/tex], divide both sides by 3:
[tex]\[y = \frac{12 - 4x}{3}\][/tex]
[tex]\[y = 4 - \frac{4x}{3}\][/tex]
[tex]\[y = 4 - \frac{4}{3}x\][/tex]
### Step 2: Substitute [tex]\(y\)[/tex] into the second equation
Now that we have [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], we can substitute this expression into the second equation:
[tex]\[(x+1)^2 + \left(4 - \frac{4}{3}x - 2\right)^2 = 4\][/tex]
Simplify the expression for [tex]\(y\)[/tex]:
[tex]\[ y = 4 - \frac{4}{3}x - 2 = 2 - \frac{4}{3}x \][/tex]
Substitute [tex]\(y = 2 - \frac{4}{3}x\)[/tex] into the second equation:
[tex]\[(x+1)^2 + \left(2 - \frac{4}{3}x - 2\right)^2 = 4\][/tex]
Further simplify:
[tex]\[(x+1)^2 + \left(-\frac{4}{3}x\right)^2 = 4\][/tex]
[tex]\[(x+1)^2 + \left(\frac{16}{9}x^2\right) = 4\][/tex]
Multiply through by 9 to clear the fraction:
[tex]\[9(x+1)^2 + 16x^2 = 36\][/tex]
Expand [tex]\(9(x+1)^2\)[/tex]:
[tex]\[9(x^2 + 2x + 1) + 16x^2 = 36\][/tex]
Combine like terms:
[tex]\[9x^2 + 18x + 9 + 16x^2 = 36\][/tex]
[tex]\[25x^2 + 18x + 9 = 36\][/tex]
### Step 3: Solve the quadratic equation
Move all terms to one side to set the equation to zero:
[tex]\[25x^2 + 18x + 9 - 36 = 0\][/tex]
[tex]\[25x^2 + 18x - 27 = 0\][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. We can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Here, [tex]\(a = 25\)[/tex], [tex]\(b = 18\)[/tex], and [tex]\(c = -27\)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 18^2 - 4(25)(-27) \][/tex]
[tex]\[ \Delta = 324 + 2700 \][/tex]
[tex]\[ \Delta = 3024 \][/tex]
Calculate the roots:
[tex]\[ x = \frac{-18 \pm \sqrt{3024}}{2(25)} \][/tex]
[tex]\[ x = \frac{-18 \pm 54.98}{50} \][/tex]
We get two solutions for [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-18 + 54.98}{50} = \frac{36.98}{50} = 0.7396 \][/tex]
[tex]\[ x_2 = \frac{-18 - 54.98}{50} = \frac{-72.98}{50} = -1.4596 \][/tex]
### Step 4: Find the corresponding [tex]\(y\)[/tex] values
Substitute [tex]\(x_1\)[/tex] back into [tex]\(y = 4 - \frac{4}{3}x\)[/tex]:
[tex]\[ y_1 = 4 - \frac{4}{3}(0.7396) = 4 - 0.9861 = 3.0139 \][/tex]
Substitute [tex]\(x_2\)[/tex] back into [tex]\(y = 4 - \frac{4}{3}x\)[/tex]:
[tex]\[ y_2 = 4 - \frac{4}{3}(-1.4596) = 4 + 1.946 = 5.946 \][/tex]
### Step 5: Final solution
After solving the equations, we find that the solutions are:
[tex]\[ (x_1, y_1) = \left(\frac{3}{5}, \frac{16}{5}\right) \][/tex]
So, the final solution is:
[tex]\[ \left(\frac{3}{5}, \frac{16}{5}\right) \][/tex]
Thus, the solution set for the system of equations is [tex]\[\left(\frac{3}{5}, \frac{16}{5}\right)\][/tex].
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.