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Let's analyze each equation step by step to determine whether it has no solution, one solution, or infinitely many solutions:
1. Equation 1: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
- Simplify the right side: [tex]\(1.4v - 3.1v + 2.8 = -1.7v + 2.8\)[/tex]
- The equation becomes: [tex]\(-1.7v + 2.8 = -1.7v + 2.8\)[/tex]
- This is an identity, meaning it is true for all values of [tex]\(v\)[/tex].
- Conclusion: Infinitely Many Solutions
2. Equation 2: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
- Combine like terms on the left: [tex]\(6a - 3 = 7a - 2\)[/tex]
- Subtract [tex]\(6a\)[/tex] from both sides: [tex]\(-3 = a - 2\)[/tex]
- Add 2 to both sides: [tex]\(-1 = a\)[/tex]
- This equation has a unique solution, [tex]\(a = -1\)[/tex].
- Conclusion: One Solution
3. Equation 3: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
- Multiply both sides by 15 (the LCM of 5 and 3) to clear the fractions: [tex]\(3f - 10 = -3f + 10\)[/tex]
- Combine like terms: [tex]\(3f + 3f = 10 + 10\)[/tex]
- Simplify: [tex]\(6f = 20\)[/tex]
- Divide by 6: [tex]\(f = \frac{20}{6} = \frac{10}{3}\)[/tex]
- This equation has a unique solution, [tex]\(f = \frac{10}{3}\)[/tex].
- Conclusion: One Solution
4. Equation 4: [tex]\(2y - 3 = 5 + 2(y - 3)\)[/tex]
- Distribute on the right side: [tex]\(2y - 3 = 5 + 2y - 6\)[/tex]
- Simplify the right side: [tex]\(2y - 3 = 2y - 1\)[/tex]
- Subtract [tex]\(2y\)[/tex] from both sides: [tex]\(-3 = -1\)[/tex]
- This is a contradiction, indicating that there are no values of [tex]\(y\)[/tex] that satisfy the equation.
- Conclusion: No Solution
5. Equation 5: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
- Distribute both sides: [tex]\(-3n - 12 + n = -2n - 12\)[/tex]
- Combine like terms: [tex]\(-2n - 12 = -2n - 12\)[/tex]
- This is an identity, meaning it is true for all values of [tex]\(n\)[/tex].
- Conclusion: Infinitely Many Solutions
Now, we can fill in the table:
[tex]\[ \begin{tabular}{|l|l|} \hline No Solution & \(2y - 3 = 5 + 2(y - 3)\) \\ \hline One Solution & \(4a - 3 + 2a = 7a - 2\), \(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\) \\ \hline Infinitely Many Solutions & \(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\), \(-3(n + 4) + n = -2(n + 6)\) \\ \hline \end{tabular} \][/tex]
1. Equation 1: [tex]\(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\)[/tex]
- Simplify the right side: [tex]\(1.4v - 3.1v + 2.8 = -1.7v + 2.8\)[/tex]
- The equation becomes: [tex]\(-1.7v + 2.8 = -1.7v + 2.8\)[/tex]
- This is an identity, meaning it is true for all values of [tex]\(v\)[/tex].
- Conclusion: Infinitely Many Solutions
2. Equation 2: [tex]\(4a - 3 + 2a = 7a - 2\)[/tex]
- Combine like terms on the left: [tex]\(6a - 3 = 7a - 2\)[/tex]
- Subtract [tex]\(6a\)[/tex] from both sides: [tex]\(-3 = a - 2\)[/tex]
- Add 2 to both sides: [tex]\(-1 = a\)[/tex]
- This equation has a unique solution, [tex]\(a = -1\)[/tex].
- Conclusion: One Solution
3. Equation 3: [tex]\(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\)[/tex]
- Multiply both sides by 15 (the LCM of 5 and 3) to clear the fractions: [tex]\(3f - 10 = -3f + 10\)[/tex]
- Combine like terms: [tex]\(3f + 3f = 10 + 10\)[/tex]
- Simplify: [tex]\(6f = 20\)[/tex]
- Divide by 6: [tex]\(f = \frac{20}{6} = \frac{10}{3}\)[/tex]
- This equation has a unique solution, [tex]\(f = \frac{10}{3}\)[/tex].
- Conclusion: One Solution
4. Equation 4: [tex]\(2y - 3 = 5 + 2(y - 3)\)[/tex]
- Distribute on the right side: [tex]\(2y - 3 = 5 + 2y - 6\)[/tex]
- Simplify the right side: [tex]\(2y - 3 = 2y - 1\)[/tex]
- Subtract [tex]\(2y\)[/tex] from both sides: [tex]\(-3 = -1\)[/tex]
- This is a contradiction, indicating that there are no values of [tex]\(y\)[/tex] that satisfy the equation.
- Conclusion: No Solution
5. Equation 5: [tex]\(-3(n + 4) + n = -2(n + 6)\)[/tex]
- Distribute both sides: [tex]\(-3n - 12 + n = -2n - 12\)[/tex]
- Combine like terms: [tex]\(-2n - 12 = -2n - 12\)[/tex]
- This is an identity, meaning it is true for all values of [tex]\(n\)[/tex].
- Conclusion: Infinitely Many Solutions
Now, we can fill in the table:
[tex]\[ \begin{tabular}{|l|l|} \hline No Solution & \(2y - 3 = 5 + 2(y - 3)\) \\ \hline One Solution & \(4a - 3 + 2a = 7a - 2\), \(\frac{1}{5}f - \frac{2}{3} = -\frac{1}{5}f + \frac{2}{3}\) \\ \hline Infinitely Many Solutions & \(-1.7v + 2.8 = 1.4v - 3.1v + 2.8\), \(-3(n + 4) + n = -2(n + 6)\) \\ \hline \end{tabular} \][/tex]
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