Answer :
Certainly! Let's solve each of the equations step-by-step.
### Linear Equations
1. [tex]\( 20x + 80 = -8x + 276 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 20x + 8x = 276 - 80 \][/tex]
[tex]\[ 28x = 196 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 28:
[tex]\[ x = \frac{196}{28} = 7 \][/tex]
So, the solution is [tex]\( x = 7 \)[/tex].
2. [tex]\( 16x - 4 = -16x + 60 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 16x + 16x = 60 + 4 \][/tex]
[tex]\[ 32x = 64 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 32:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
So, the solution is [tex]\( x = 2 \)[/tex].
3. [tex]\( 28x - 64 = -20x + 80 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 28x + 20x = 80 + 64 \][/tex]
[tex]\[ 48x = 144 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 48:
[tex]\[ x = \frac{144}{48} = 3 \][/tex]
So, the solution is [tex]\( x = 3 \)[/tex].
### Quadratic Equations
1. [tex]\( x^2 - 9 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 9 to both sides:
[tex]\[ x^2 = 9 \][/tex]
2. Take the square root of both sides:
[tex]\[ x = \pm 3 \][/tex]
So, the solutions are [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. [tex]\( x^2 - 49 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 49 to both sides:
[tex]\[ x^2 = 49 \][/tex]
2. Take the square root of both sides:
[tex]\[ x = \pm 7 \][/tex]
So, the solutions are [tex]\( x = -7 \)[/tex] and [tex]\( x = 7 \)[/tex].
3. [tex]\( 3x^2 - 1200 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 1200 to both sides:
[tex]\[ 3x^2 = 1200 \][/tex]
2. Divide both sides by 3:
[tex]\[ x^2 = 400 \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm 20 \][/tex]
So, the solutions are [tex]\( x = -20 \)[/tex] and [tex]\( x = 20 \)[/tex].
### Summary of Solutions
- For the linear equations:
- [tex]\( 20x + 80 = -8x + 276 \)[/tex]: [tex]\( x = 7 \)[/tex]
- [tex]\( 16x - 4 = -16x + 60 \)[/tex]: [tex]\( x = 2 \)[/tex]
- [tex]\( 28x - 64 = -20x + 80 \)[/tex]: [tex]\( x = 3 \)[/tex]
- For the quadratic equations:
- [tex]\( x^2 - 9 = 0 \)[/tex]: [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex]
- [tex]\( x^2 - 49 = 0 \)[/tex]: [tex]\( x = -7 \)[/tex] and [tex]\( x = 7 \)[/tex]
- [tex]\( 3x^2 - 1200 = 0 \)[/tex]: [tex]\( x = -20 \)[/tex] and [tex]\( x = 20 \)[/tex]
### Linear Equations
1. [tex]\( 20x + 80 = -8x + 276 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 20x + 8x = 276 - 80 \][/tex]
[tex]\[ 28x = 196 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 28:
[tex]\[ x = \frac{196}{28} = 7 \][/tex]
So, the solution is [tex]\( x = 7 \)[/tex].
2. [tex]\( 16x - 4 = -16x + 60 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 16x + 16x = 60 + 4 \][/tex]
[tex]\[ 32x = 64 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 32:
[tex]\[ x = \frac{64}{32} = 2 \][/tex]
So, the solution is [tex]\( x = 2 \)[/tex].
3. [tex]\( 28x - 64 = -20x + 80 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Combine like terms by moving all terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 28x + 20x = 80 + 64 \][/tex]
[tex]\[ 48x = 144 \][/tex]
2. Solve for [tex]\( x \)[/tex] by dividing both sides by 48:
[tex]\[ x = \frac{144}{48} = 3 \][/tex]
So, the solution is [tex]\( x = 3 \)[/tex].
### Quadratic Equations
1. [tex]\( x^2 - 9 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 9 to both sides:
[tex]\[ x^2 = 9 \][/tex]
2. Take the square root of both sides:
[tex]\[ x = \pm 3 \][/tex]
So, the solutions are [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. [tex]\( x^2 - 49 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 49 to both sides:
[tex]\[ x^2 = 49 \][/tex]
2. Take the square root of both sides:
[tex]\[ x = \pm 7 \][/tex]
So, the solutions are [tex]\( x = -7 \)[/tex] and [tex]\( x = 7 \)[/tex].
3. [tex]\( 3x^2 - 1200 = 0 \)[/tex]
To solve for [tex]\( x \)[/tex]:
1. Add 1200 to both sides:
[tex]\[ 3x^2 = 1200 \][/tex]
2. Divide both sides by 3:
[tex]\[ x^2 = 400 \][/tex]
3. Take the square root of both sides:
[tex]\[ x = \pm 20 \][/tex]
So, the solutions are [tex]\( x = -20 \)[/tex] and [tex]\( x = 20 \)[/tex].
### Summary of Solutions
- For the linear equations:
- [tex]\( 20x + 80 = -8x + 276 \)[/tex]: [tex]\( x = 7 \)[/tex]
- [tex]\( 16x - 4 = -16x + 60 \)[/tex]: [tex]\( x = 2 \)[/tex]
- [tex]\( 28x - 64 = -20x + 80 \)[/tex]: [tex]\( x = 3 \)[/tex]
- For the quadratic equations:
- [tex]\( x^2 - 9 = 0 \)[/tex]: [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex]
- [tex]\( x^2 - 49 = 0 \)[/tex]: [tex]\( x = -7 \)[/tex] and [tex]\( x = 7 \)[/tex]
- [tex]\( 3x^2 - 1200 = 0 \)[/tex]: [tex]\( x = -20 \)[/tex] and [tex]\( x = 20 \)[/tex]
