Answer :
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].
