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Sagot :
To determine whether each equation has "no solution," "one solution," or "infinitely many solutions," we need to simplify each equation and analyze the resulting expressions for solvability. Here is a detailed, step-by-step solution for each equation:
### 1. Equation: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
First, simplify the right side:
[tex]\[ 1.4v - 3.1v = -1.7v \][/tex]
So the equation becomes:
[tex]\[ -1.7v + 2.8 = -1.7v + 2.8 \][/tex]
Next, subtract [tex]\(-1.7v + 2.8\)[/tex] from both sides:
[tex]\[ 0 = 0 \][/tex]
This equation is an identity, meaning it is true for all values of [tex]\(v\)[/tex]. Hence, it has infinitely many solutions.
### 2. Equation: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
Combine like terms on the left side:
[tex]\[ 4a + 2a - 3 = 6a - 3 \][/tex]
So the equation is:
[tex]\[ 6a - 3 = 7a - 2 \][/tex]
Subtract [tex]\(6a\)[/tex] from both sides:
[tex]\[ -3 = a - 2 \][/tex]
Add 2 to both sides to solve for [tex]\(a\)[/tex]:
[tex]\[ a = -1 \][/tex]
Since we found a unique value for [tex]\(a\)[/tex], this equation has one solution.
### 3. Equation: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
First, combine like terms on both sides by adding [tex]\(\frac{1}{5}f\)[/tex] to both sides:
[tex]\[ \frac{2}{5}f - \frac{2}{3} = \frac{2}{3} \][/tex]
Next, add [tex]\(\frac{2}{3}\)[/tex] to both sides:
[tex]\[ \frac{2}{5}f = \frac{4}{3} \][/tex]
Multiply both sides by [tex]\(\frac{5}{2}\)[/tex] to solve for [tex]\(f\)[/tex]:
[tex]\[ f = \frac{4}{3} \cdot \frac{5}{2} = \frac{20}{6} = \frac{10}{3} \][/tex]
Since we found a unique value for [tex]\(f\)[/tex], this equation has one solution.
### 4. Equation: [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]
First, distribute on the right side:
[tex]\[ 2y - 3 = 5 + 2y - 2 \][/tex]
Combine like terms:
[tex]\[ 2y - 3 = 2y + 3 \][/tex]
Subtract [tex]\(2y\)[/tex] from both sides:
[tex]\[ -3 = 3 \][/tex]
This is a contradiction; thus, the equation has no solution.
### 5. Equation: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
First, distribute on both sides:
[tex]\[ -3n - 12 + n = -2n - 12 \][/tex]
Combine like terms:
[tex]\[ -2n - 12 = -2n - 12 \][/tex]
This is an identity, so this equation is always true for any [tex]\(n\)[/tex]; thus, it has infinitely many solutions.
### Summary:
- [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]: infinitely many solutions
- [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]: one solution
- [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]: one solution
- [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]: no solution
- [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]: no solution
Therefore, the final classification of each equation is as follows:
| Equation | Solution Type |
|-----------------------------------------------|---------------------------|
| [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex] | infinitely many solutions |
| [tex]\(4a - 3 + 2a = 7a - 2\)[/tex] | one solution |
| [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex] | one solution |
| [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex] | no solution |
| [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex] | no solution |
### 1. Equation: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
First, simplify the right side:
[tex]\[ 1.4v - 3.1v = -1.7v \][/tex]
So the equation becomes:
[tex]\[ -1.7v + 2.8 = -1.7v + 2.8 \][/tex]
Next, subtract [tex]\(-1.7v + 2.8\)[/tex] from both sides:
[tex]\[ 0 = 0 \][/tex]
This equation is an identity, meaning it is true for all values of [tex]\(v\)[/tex]. Hence, it has infinitely many solutions.
### 2. Equation: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
Combine like terms on the left side:
[tex]\[ 4a + 2a - 3 = 6a - 3 \][/tex]
So the equation is:
[tex]\[ 6a - 3 = 7a - 2 \][/tex]
Subtract [tex]\(6a\)[/tex] from both sides:
[tex]\[ -3 = a - 2 \][/tex]
Add 2 to both sides to solve for [tex]\(a\)[/tex]:
[tex]\[ a = -1 \][/tex]
Since we found a unique value for [tex]\(a\)[/tex], this equation has one solution.
### 3. Equation: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
First, combine like terms on both sides by adding [tex]\(\frac{1}{5}f\)[/tex] to both sides:
[tex]\[ \frac{2}{5}f - \frac{2}{3} = \frac{2}{3} \][/tex]
Next, add [tex]\(\frac{2}{3}\)[/tex] to both sides:
[tex]\[ \frac{2}{5}f = \frac{4}{3} \][/tex]
Multiply both sides by [tex]\(\frac{5}{2}\)[/tex] to solve for [tex]\(f\)[/tex]:
[tex]\[ f = \frac{4}{3} \cdot \frac{5}{2} = \frac{20}{6} = \frac{10}{3} \][/tex]
Since we found a unique value for [tex]\(f\)[/tex], this equation has one solution.
### 4. Equation: [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]
First, distribute on the right side:
[tex]\[ 2y - 3 = 5 + 2y - 2 \][/tex]
Combine like terms:
[tex]\[ 2y - 3 = 2y + 3 \][/tex]
Subtract [tex]\(2y\)[/tex] from both sides:
[tex]\[ -3 = 3 \][/tex]
This is a contradiction; thus, the equation has no solution.
### 5. Equation: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
First, distribute on both sides:
[tex]\[ -3n - 12 + n = -2n - 12 \][/tex]
Combine like terms:
[tex]\[ -2n - 12 = -2n - 12 \][/tex]
This is an identity, so this equation is always true for any [tex]\(n\)[/tex]; thus, it has infinitely many solutions.
### Summary:
- [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]: infinitely many solutions
- [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]: one solution
- [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]: one solution
- [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]: no solution
- [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]: no solution
Therefore, the final classification of each equation is as follows:
| Equation | Solution Type |
|-----------------------------------------------|---------------------------|
| [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex] | infinitely many solutions |
| [tex]\(4a - 3 + 2a = 7a - 2\)[/tex] | one solution |
| [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex] | one solution |
| [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex] | no solution |
| [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex] | no solution |
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