Q:

30 points need help asap!!!!!!!!! someone in 8th grade k12 please help!!!!Assignment: Variables on Both Sides of an Equation Investigation Transform each of the following equations to determine whether each has one solution, infinitely many solutions or no solution. Use the result of your transformation to state the number of solutions.1. 2x + 4 = 3(x – 2) + 1 2. 4(x + 3) = 3x + 17 3. 3x + 4 + 2x = 5(x – 2) + 7 4. 2(1 + 5x) = 5(2x – 1)5. –18 + 15x = 3(4x – 6) + 3x 6. –10 + 5x = 5(x – 3) + 5 7. Describe the different types of results that will occur when performing transformations on equations that have one solution, no solutions or infinitely many solutions.8. Write an equation of your own that has one real solution. Transform the equation to demonstrate that it has one solution.9. Write an equation of your own that has infinitely many solutions. Transform the equation to demonstrate that it has infinitely many solutions.10. Write an equation of your own that has no solution. Transform the equation to demonstrate that it has no solution.

Accepted Solution

A:
2x + 4 = 3(x - 2) + 1
2x + 4 = 3x - 6 + 1
2x + 4 = 3x - 5
2x - 3x = -5 - 4
-x = -9
x = 9
one solution
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4(x + 3) = 3x + 17
4x + 12 = 3x + 17
4x - 3x = 17 - 12
x = 5
one solution
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3x + 4 + 2x = 5(x - 2) + 7
5x + 4 = 5x - 10 + 7
5x + 4 = 5x - 3
5x - 5x = -3 - 4
0 = -7
no solutions
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2(1 + 5x) = 5(2x - 1)
2 + 10x = 10x - 5
10x - 10x = -5 - 2
0 = -7
no solution
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-18 + 15x = 3(4x - 6) + 3x
-18 + 15x = 12x - 18 + 3x
-18 + 15x = 15x - 18
15x - 15x = -18 + 18
0 = 0
infinite solutions
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-10 + 5x = 5(x - 3) + 5
-10 + 5x = 5x - 15 + 5
-10 + 5x = 5x - 10
5x - 5x = -10 + 10
0 = 0
infinite solutions
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if the variables cancel out leaving u with an untrue statement, like 0 = 4 or 2 = 6, then u will have no solutions
if the variables cancel out leaving u with a true statement, like 2 = 2 or 4 = 4, then u will have infinite solutions
if u end up with a variable equaling a number, then u have 1 solution
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3x + 4 = 2x + 2
3x - 2x = 2 - 4
x = -2
one solution
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3(x + 2) + 1 = 3x + 7
3x + 6 + 1 = 3x + 7
3x + 7 = 3x + 7
3x - 3x = 7 - 7
0 = 0
infinite solutions
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2x + 3 = 2x + 4 + 1
2x + 3 = 2x + 5
2x - 2x = 5 - 3
0 = 2
no solutions