Answer :
Certainly! Let's go through the five equations one by one and solve them step-by-step.
### 1. Solve the equation: [tex]\(2x + 3y = 10\)[/tex]
First, we'll express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ 2x = 10 - 3y \][/tex]
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
So, the solution to this equation is:
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
This represents a family of solutions parameterized by [tex]\(y\)[/tex].
### 2. Solve the equation: [tex]\(3x^2 + 5x + 10 = 0\)[/tex]
This is a quadratic equation in standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with [tex]\(a = 3\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 10\)[/tex]. We use the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Let's calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ \Delta = 5^2 - 4 \cdot 3 \cdot 10 = 25 - 120 = -95 \][/tex]
Since the discriminant is negative, the solutions are complex:
[tex]\[ x = \frac{-5 \pm \sqrt{-95}}{6} = \frac{-5 \pm \sqrt{95}i}{6} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-5 - \sqrt{95}i}{6} \quad \text{and} \quad x = \frac{-5 + \sqrt{95}i}{6} \][/tex]
### 3. Solve the equation: [tex]\(x^2 + 5x = 10\)[/tex]
First, we rewrite this as a standard quadratic equation:
[tex]\[ x^2 + 5x - 10 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -10\)[/tex].
The discriminant ([tex]\(b^2 - 4ac\)[/tex]) is:
[tex]\[ \Delta = 5^2 - 4 \cdot 1 \cdot (-10) = 25 + 40 = 65 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-5 \pm \sqrt{65}}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-5 + \sqrt{65}}{2} \quad \text{and} \quad x = \frac{-5 - \sqrt{65}}{2} \][/tex]
### 4. Solve the equation: [tex]\(3x = 10\)[/tex]
Here, we simply isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{10}{3} \][/tex]
So, the solution is:
[tex]\[ x = \frac{10}{3} \][/tex]
### 5. Solve the equation: [tex]\(6x - 5 + 3x^2 = 0\)[/tex]
Rewriting it in standard quadratic form:
[tex]\[ 3x^2 + 6x - 5 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -5\)[/tex].
The discriminant ([tex]\(b^2 - 4ac\)[/tex]) is:
[tex]\[ \Delta = 6^2 - 4 \cdot 3 \cdot (-5) = 36 + 60 = 96 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-6 \pm \sqrt{96}}{6} = \frac{-6 \pm 4\sqrt{6}}{6} = \frac{-6 \pm 2\sqrt{6}}{3} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-6 + 2\sqrt{6}}{3} = -1 + \frac{2\sqrt{6}}{3} \][/tex]
[tex]\[ x = \frac{-6 - 2\sqrt{6}}{3} = -1 - \frac{2\sqrt{6}}{3} \][/tex]
### Summary of solutions
1. [tex]\(2x + 3y = 10 \)[/tex]
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
2. [tex]\(3x^2 + 5x + 10 = 0\)[/tex]
[tex]\[ x = \frac{-5 - \sqrt{95}i}{6} \text{ and } x = \frac{-5 + \sqrt{95}i}{6} \][/tex]
3. [tex]\(x^2 + 5x = 10\)[/tex]
[tex]\[ x = \frac{-5 + \sqrt{65}}{2} \text{ and } x = \frac{-5 - \sqrt{65}}{2} \][/tex]
4. [tex]\(3x = 10\)[/tex]
[tex]\[ x = \frac{10}{3} \][/tex]
5. [tex]\(6x - 5 + 3x^2 = 0\)[/tex]
[tex]\[ x = -1 + \frac{2\sqrt{6}}{3} \text{ and } x = -1 - \frac{2\sqrt{6}}{3} \][/tex]
### 1. Solve the equation: [tex]\(2x + 3y = 10\)[/tex]
First, we'll express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ 2x = 10 - 3y \][/tex]
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
So, the solution to this equation is:
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
This represents a family of solutions parameterized by [tex]\(y\)[/tex].
### 2. Solve the equation: [tex]\(3x^2 + 5x + 10 = 0\)[/tex]
This is a quadratic equation in standard form [tex]\(ax^2 + bx + c = 0\)[/tex] with [tex]\(a = 3\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 10\)[/tex]. We use the quadratic formula, [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
Let's calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ \Delta = 5^2 - 4 \cdot 3 \cdot 10 = 25 - 120 = -95 \][/tex]
Since the discriminant is negative, the solutions are complex:
[tex]\[ x = \frac{-5 \pm \sqrt{-95}}{6} = \frac{-5 \pm \sqrt{95}i}{6} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-5 - \sqrt{95}i}{6} \quad \text{and} \quad x = \frac{-5 + \sqrt{95}i}{6} \][/tex]
### 3. Solve the equation: [tex]\(x^2 + 5x = 10\)[/tex]
First, we rewrite this as a standard quadratic equation:
[tex]\[ x^2 + 5x - 10 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -10\)[/tex].
The discriminant ([tex]\(b^2 - 4ac\)[/tex]) is:
[tex]\[ \Delta = 5^2 - 4 \cdot 1 \cdot (-10) = 25 + 40 = 65 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-5 \pm \sqrt{65}}{2} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-5 + \sqrt{65}}{2} \quad \text{and} \quad x = \frac{-5 - \sqrt{65}}{2} \][/tex]
### 4. Solve the equation: [tex]\(3x = 10\)[/tex]
Here, we simply isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{10}{3} \][/tex]
So, the solution is:
[tex]\[ x = \frac{10}{3} \][/tex]
### 5. Solve the equation: [tex]\(6x - 5 + 3x^2 = 0\)[/tex]
Rewriting it in standard quadratic form:
[tex]\[ 3x^2 + 6x - 5 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 3\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -5\)[/tex].
The discriminant ([tex]\(b^2 - 4ac\)[/tex]) is:
[tex]\[ \Delta = 6^2 - 4 \cdot 3 \cdot (-5) = 36 + 60 = 96 \][/tex]
So, the solutions are:
[tex]\[ x = \frac{-6 \pm \sqrt{96}}{6} = \frac{-6 \pm 4\sqrt{6}}{6} = \frac{-6 \pm 2\sqrt{6}}{3} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{-6 + 2\sqrt{6}}{3} = -1 + \frac{2\sqrt{6}}{3} \][/tex]
[tex]\[ x = \frac{-6 - 2\sqrt{6}}{3} = -1 - \frac{2\sqrt{6}}{3} \][/tex]
### Summary of solutions
1. [tex]\(2x + 3y = 10 \)[/tex]
[tex]\[ x = \frac{10 - 3y}{2} \][/tex]
2. [tex]\(3x^2 + 5x + 10 = 0\)[/tex]
[tex]\[ x = \frac{-5 - \sqrt{95}i}{6} \text{ and } x = \frac{-5 + \sqrt{95}i}{6} \][/tex]
3. [tex]\(x^2 + 5x = 10\)[/tex]
[tex]\[ x = \frac{-5 + \sqrt{65}}{2} \text{ and } x = \frac{-5 - \sqrt{65}}{2} \][/tex]
4. [tex]\(3x = 10\)[/tex]
[tex]\[ x = \frac{10}{3} \][/tex]
5. [tex]\(6x - 5 + 3x^2 = 0\)[/tex]
[tex]\[ x = -1 + \frac{2\sqrt{6}}{3} \text{ and } x = -1 - \frac{2\sqrt{6}}{3} \][/tex]
