At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
