Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our 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].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.