Answer :
Sure! Let's solve each part of the problem step-by-step.
### Problem 10.1.1
Equation: [tex]\( x(x - 1) = 20 \)[/tex]
1. Expand the left side to form a quadratic equation:
[tex]\[
x^2 - x = 20
\][/tex]
2. Rearrange the equation:
[tex]\[
x^2 - x - 20 = 0
\][/tex]
3. Now, we need to factor the quadratic equation. We are looking for two numbers whose product is [tex]\(-20\)[/tex] and sum is [tex]\(-1\)[/tex]. These numbers are [tex]\(-5\)[/tex] and [tex]\(4\)[/tex].
4. So, we can factor the equation as:
[tex]\[
(x - 5)(x + 4) = 0
\][/tex]
5. To solve for [tex]\(x\)[/tex], set each factor equal to zero:
- [tex]\(x - 5 = 0 \Rightarrow x = 5\)[/tex]
- [tex]\(x + 4 = 0 \Rightarrow x = -4\)[/tex]
Thus, the solutions for [tex]\(x(x - 1) = 20\)[/tex] are [tex]\(x = 5\)[/tex] and [tex]\(x = -4\)[/tex].
### Problem 10.1.2
Equation: [tex]\(\frac{3x - 2}{2} = x + 1\)[/tex]
1. Eliminate the fraction by multiplying all terms by 2:
[tex]\[
3x - 2 = 2(x + 1)
\][/tex]
2. Expand the right side:
[tex]\[
3x - 2 = 2x + 2
\][/tex]
3. Move all terms involving [tex]\(x\)[/tex] to one side and constant terms to the other:
[tex]\[
3x - 2x = 2 + 2
\][/tex]
4. Simplify both sides:
[tex]\[
x = 4
\][/tex]
Thus, the solution for [tex]\(\frac{3x - 2}{2} = x + 1\)[/tex] is [tex]\(x = 4\)[/tex].
Overall, the solutions for the given problems are:
- For [tex]\(x(x-1) = 20\)[/tex], [tex]\(x = 5\)[/tex] and [tex]\(x = -4\)[/tex].
- For [tex]\(\frac{3x-2}{2} = x+1\)[/tex], [tex]\(x = 4\)[/tex].
### Problem 10.1.1
Equation: [tex]\( x(x - 1) = 20 \)[/tex]
1. Expand the left side to form a quadratic equation:
[tex]\[
x^2 - x = 20
\][/tex]
2. Rearrange the equation:
[tex]\[
x^2 - x - 20 = 0
\][/tex]
3. Now, we need to factor the quadratic equation. We are looking for two numbers whose product is [tex]\(-20\)[/tex] and sum is [tex]\(-1\)[/tex]. These numbers are [tex]\(-5\)[/tex] and [tex]\(4\)[/tex].
4. So, we can factor the equation as:
[tex]\[
(x - 5)(x + 4) = 0
\][/tex]
5. To solve for [tex]\(x\)[/tex], set each factor equal to zero:
- [tex]\(x - 5 = 0 \Rightarrow x = 5\)[/tex]
- [tex]\(x + 4 = 0 \Rightarrow x = -4\)[/tex]
Thus, the solutions for [tex]\(x(x - 1) = 20\)[/tex] are [tex]\(x = 5\)[/tex] and [tex]\(x = -4\)[/tex].
### Problem 10.1.2
Equation: [tex]\(\frac{3x - 2}{2} = x + 1\)[/tex]
1. Eliminate the fraction by multiplying all terms by 2:
[tex]\[
3x - 2 = 2(x + 1)
\][/tex]
2. Expand the right side:
[tex]\[
3x - 2 = 2x + 2
\][/tex]
3. Move all terms involving [tex]\(x\)[/tex] to one side and constant terms to the other:
[tex]\[
3x - 2x = 2 + 2
\][/tex]
4. Simplify both sides:
[tex]\[
x = 4
\][/tex]
Thus, the solution for [tex]\(\frac{3x - 2}{2} = x + 1\)[/tex] is [tex]\(x = 4\)[/tex].
Overall, the solutions for the given problems are:
- For [tex]\(x(x-1) = 20\)[/tex], [tex]\(x = 5\)[/tex] and [tex]\(x = -4\)[/tex].
- For [tex]\(\frac{3x-2}{2} = x+1\)[/tex], [tex]\(x = 4\)[/tex].
