Answer :
Sure, let's classify each equation step by step.
### Equation 1: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
First, simplify the right-hand side:
[tex]\[ 1.4v - 3.1v + 2.8 = -1.7v + 2.8 \][/tex]
So the equation becomes:
[tex]\[ -1.7v + 2.8 = -1.7v + 2.8 \][/tex]
This equation is always true regardless of the value of [tex]\(v\)[/tex]. Therefore, it has infinitely many solutions.
### Equation 2: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
Combine like terms:
[tex]\[ 6a - 3 = 7a - 2 \][/tex]
To isolate the variable [tex]\(a\)[/tex], subtract [tex]\(6a\)[/tex] from both sides:
[tex]\[ -3 = a - 2 \][/tex]
Add 2 to both sides:
[tex]\[ -1 = a \][/tex]
So, [tex]\(a\)[/tex] equals [tex]\(-1\)[/tex]. This means the equation has one solution.
### Equation 3: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
First, collect like terms:
[tex]\[ \frac{1}{5}f + \frac{1}{5}f = \frac{2}{3} + \frac{2}{3} \][/tex]
[tex]\[ \frac{2}{5}f = \frac{4}{3} \][/tex]
Multiply both sides by [tex]\(5/2\)[/tex]:
[tex]\[ f = \frac{4}{3} \times \frac{5}{2} \][/tex]
[tex]\[ f = \frac{20}{6} \][/tex]
[tex]\[ f = \frac{10}{3} \][/tex]
Hence, [tex]\(f\)[/tex] equals [tex]\(\frac{10}{3}\)[/tex]. This means the equation has one solution.
### Equation 4: [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]
Simplify the right-hand side:
[tex]\[ 5 + 2(y - 1) = 5 + 2y - 2 = 2y + 3 \][/tex]
So the equation becomes:
[tex]\[ 2y - 3 = 2y + 3 \][/tex]
Subtract [tex]\(2y\)[/tex] from both sides:
[tex]\[ -3 = 3 \][/tex]
This is a contradiction. Therefore, the equation has no solution.
### Equation 5: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
First, distribute the terms:
[tex]\[ -3n - 12 + n = -2n - 12 \][/tex]
Combine like terms:
[tex]\[ -2n - 12 = -2n - 12 \][/tex]
This equation is always true regardless of the value of [tex]\(n\)[/tex]. Therefore, it has infinitely many solutions.
### Summary Table
| Equation [tex]\( \)[/tex] | 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] | Infinitely Many Solutions |
I hope this helps you understand how to classify the solutions of each equation!
### Equation 1: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
First, simplify the right-hand side:
[tex]\[ 1.4v - 3.1v + 2.8 = -1.7v + 2.8 \][/tex]
So the equation becomes:
[tex]\[ -1.7v + 2.8 = -1.7v + 2.8 \][/tex]
This equation is always true regardless of the value of [tex]\(v\)[/tex]. Therefore, it has infinitely many solutions.
### Equation 2: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
Combine like terms:
[tex]\[ 6a - 3 = 7a - 2 \][/tex]
To isolate the variable [tex]\(a\)[/tex], subtract [tex]\(6a\)[/tex] from both sides:
[tex]\[ -3 = a - 2 \][/tex]
Add 2 to both sides:
[tex]\[ -1 = a \][/tex]
So, [tex]\(a\)[/tex] equals [tex]\(-1\)[/tex]. This means the equation has one solution.
### Equation 3: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
First, collect like terms:
[tex]\[ \frac{1}{5}f + \frac{1}{5}f = \frac{2}{3} + \frac{2}{3} \][/tex]
[tex]\[ \frac{2}{5}f = \frac{4}{3} \][/tex]
Multiply both sides by [tex]\(5/2\)[/tex]:
[tex]\[ f = \frac{4}{3} \times \frac{5}{2} \][/tex]
[tex]\[ f = \frac{20}{6} \][/tex]
[tex]\[ f = \frac{10}{3} \][/tex]
Hence, [tex]\(f\)[/tex] equals [tex]\(\frac{10}{3}\)[/tex]. This means the equation has one solution.
### Equation 4: [tex]\(2y - 3 = 5 + 2(y - 1)\)[/tex]
Simplify the right-hand side:
[tex]\[ 5 + 2(y - 1) = 5 + 2y - 2 = 2y + 3 \][/tex]
So the equation becomes:
[tex]\[ 2y - 3 = 2y + 3 \][/tex]
Subtract [tex]\(2y\)[/tex] from both sides:
[tex]\[ -3 = 3 \][/tex]
This is a contradiction. Therefore, the equation has no solution.
### Equation 5: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
First, distribute the terms:
[tex]\[ -3n - 12 + n = -2n - 12 \][/tex]
Combine like terms:
[tex]\[ -2n - 12 = -2n - 12 \][/tex]
This equation is always true regardless of the value of [tex]\(n\)[/tex]. Therefore, it has infinitely many solutions.
### Summary Table
| Equation [tex]\( \)[/tex] | 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] | Infinitely Many Solutions |
I hope this helps you understand how to classify the solutions of each equation!
