Algebra

To solve the system of equations, we can use the addition method. First, add the two equations together to get (3x = 9). Dividing both sides by 3, we get (x = 3). Substituting this value of (x) back into the first equation, we get (3 - y = 2), which simplifies to (y = 1). Therefore, the solution to the system of equations is ((x, y) = (3, 1)).

Vieta's formula is a well-known formula in algebra that states that the product of the two roots of a quadratic equation is equal to the constant term.

Rewrite the equation as (2^5 = x + 3), then solve for (x) to get (x = 31).

To solve the equation (x^2 - 5x + 6 = 0), we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Plugging in the values of (a = 1, b = -5, c = 6), we get (x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = 2, 3).

To solve the equation, we can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Plugging in the values $a = 1$, $b = -4$, and $c = 3$, we get $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2}$. Therefore, the solutions are $x = 1$ and $x = 3$.

To solve the equation, we can use the rational root theorem or Descartes' rule of signs. Using the rational root theorem, we find that $x = 1$ is a root. Dividing the polynomial by $x - 1$, we get $2x^3 - 3x^2 + x - 2$. Factoring this cubic, we get $(2x - 1)(x^2 - x + 2)$. Solving the quadratic $x^2 - x + 2 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(2)}}{2(1)} = \frac{1 \pm \sqrt{-7}}{2}$. Therefore, the roots of the polynomial are $x = 1, 2, -1, -2$.

To solve the equation, we can use the rational root theorem or Descartes' rule of signs. Using the rational root theorem, we find that $x = 1$ and $x = -1$ are roots. Dividing the polynomial by $x - 1$, we get $x^5 - x^4 - 2x^3 + x^2 + 3x - 7$. Dividing this polynomial by $x + 1$, we get $x^4 - 2x^3 - 3x^2 + 4x + 7$. Factoring this quartic, we get $(x^2 - x + 7)(x^2 - x - 1)$. Solving the quadratic $x^2 - x + 7 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(7)}}{2(1)} = \frac{1 \pm \sqrt{-27}}{2}$. Solving the quadratic $x^2 - x - 1 = 0$, we get $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$. Therefore, the roots of the polynomial are $x = 1, -1, 2, -2, i, -i$.

Bhaskara II's formula for solving a quadratic equation is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a), (b), and (c) are the coefficients of the quadratic equation.

Bhaskara II's formula for solving quadratic equations is $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$.

To solve the quadratic equation x^2 + 4x + 3 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 4, and c = 3. Plugging these values into the formula, we get x = (-4 ± √(4^2 - 4(1)(3))) / 2(1). Simplifying this expression, we get x = (-4 ± √(16 - 12)) / 2. Further simplifying, we get x = (-4 ± √4) / 2. Finally, we get x = (-4 ± 2) / 2. This gives us two solutions: x = -1 and x = -3.

To solve the equation 2x - 7 = 1, we need to isolate the variable x on one side of the equation. We can do this by adding 7 to both sides of the equation, which gives us 2x = 8. Then, we can divide both sides of the equation by 2, which gives us x = 4.

To solve the quadratic equation x^2 - 5x + 6 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -5, and c = 6. Plugging these values into the formula, we get x = (-(-5) ± √((-5)^2 - 4(1)(6))) / 2(1). Simplifying this expression, we get x = (5 ± √(25 - 24)) / 2. Further simplifying, we get x = (5 ± √1) / 2. Finally, we get x = (5 ± 1) / 2. This gives us two solutions: x = 2 and x = 3.

To solve the quadratic equation x^2 - 4x + 4 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -4, and c = 4. Plugging these values into the formula, we get x = (-(-4) ± √((-4)^2 - 4(1)(4))) / 2(1). Simplifying this expression, we get x = (4 ± √(16 - 16)) / 2. Further simplifying, we get x = (4 ± √0) / 2. Finally, we get x = (4 ± 0) / 2. This gives us two solutions: x = 2 and x = 2.

To solve the equation 6x + 1 = 13, we need to isolate the variable x on one side of the equation. We can do this by subtracting 1 from both sides of the equation, which gives us 6x = 12. Then, we can divide both sides of the equation by 6, which gives us x = 2.

To solve the quadratic equation x^2 + 6x + 9 = 0, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = 6, and c = 9. Plugging these values into the formula, we get x = (-6 ± √(6^2 - 4(1)(9))) / 2(1). Simplifying this expression, we get x = (-6 ± √(36 - 36)) / 2. Further simplifying, we get x = (-6 ± √0) / 2. Finally, we get x = (-6 ± 0) / 2. This gives us two solutions: x = -3 and x = -3.