Question
Identify the pair of linear equations that have the same solution set.(1 point)
Responses
3(m+1)=10
and 6n+3=6−n
3 left parenthesis m plus 1 right parenthesis equals 10 and 6 n plus 3 equals 6 minus n
2(3g+5)−2g=2(4−g)
and −36h/6=2
2 left parenthesis 3 g plus 5 right parenthesis minus 2 g equals 2 left parenthesis 4 minus g right parenthesis and Start Fraction negative 36 h over 6 End Fraction equals 2
4k−5=2k
and 8j−2=3j
4 k minus 5 equals 2 k and 8 j minus 2 equals 3 j
4=k2+5
and 4j−14/3+5=3
Responses
3(m+1)=10
and 6n+3=6−n
3 left parenthesis m plus 1 right parenthesis equals 10 and 6 n plus 3 equals 6 minus n
2(3g+5)−2g=2(4−g)
and −36h/6=2
2 left parenthesis 3 g plus 5 right parenthesis minus 2 g equals 2 left parenthesis 4 minus g right parenthesis and Start Fraction negative 36 h over 6 End Fraction equals 2
4k−5=2k
and 8j−2=3j
4 k minus 5 equals 2 k and 8 j minus 2 equals 3 j
4=k2+5
and 4j−14/3+5=3
Answers
GPT-4o mini
To identify the pair of linear equations that have the same solution set, we need to simplify each pair of equations to see if they yield the same solution.
1. **Equations:**
\(3(m + 1) = 10\)
\(6n + 3 = 6 - n\)
- For the first equation:
\[
3(m + 1) = 10
\]
\[
m + 1 = \frac{10}{3}
\]
\[
m = \frac{10}{3} - 1 = \frac{7}{3}
\]
- For the second equation:
\[
6n + 3 = 6 - n
\]
\[
7n = 3
\]
\[
n = \frac{3}{7}
\]
- **Results:** Different solutions.
2. **Equations:**
\(2(3g + 5) - 2g = 2(4 - g)\)
\(-\frac{36h}{6} = 2\)
- For the first equation:
\[
2(3g + 5) - 2g = 2(4 - g)
\]
\[
6g + 10 - 2g = 8 - 2g
\]
\[
4g + 10 = 8 - 2g
\]
\[
6g = -2
\]
\[
g = -\frac{1}{3}
\]
- For the second equation:
\[
-\frac{36h}{6} = 2
\]
\[
-6h = 2
\]
\[
h = -\frac{1}{3}
\]
- **Results:** Same solution \((-1/3)\).
3. **Equations:**
\(4k - 5 = 2k\)
\(8j - 2 = 3j\)
- For the first equation:
\[
4k - 5 = 2k
\]
\[
2k = 5
\]
\[
k = \frac{5}{2}
\]
- For the second equation:
\[
8j - 2 = 3j
\]
\[
5j = 2
\]
\[
j = \frac{2}{5}
\]
- **Results:** Different solutions.
4. **Equations:**
\(4 = k^2 + 5\)
\(4j - \frac{14}{3} + 5 = 3\)
- For the first equation:
\[
k^2 + 5 = 4
\]
\[
k^2 = -1
\]
(No real solution)
- For the second equation:
\[
4j - \frac{14}{3} + 5 = 3
\]
Find a common denominator:
\[
4j - \frac{14}{3} + \frac{15}{3} = 3
\]
\[
4j + \frac{1}{3} = 3
\]
\[
4j = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}
\]
\[
j = \frac{2}{3}
\]
- **Results:** No real solution for the first, valid solution for the second.
**Conclusion:** The pair of linear equations that have the same solution set is:
**\(2(3g + 5) - 2g = 2(4 - g)\) and \(-\frac{36h}{6} = 2\)** with the common solution \(g = h = -\frac{1}{3}\).
1. **Equations:**
\(3(m + 1) = 10\)
\(6n + 3 = 6 - n\)
- For the first equation:
\[
3(m + 1) = 10
\]
\[
m + 1 = \frac{10}{3}
\]
\[
m = \frac{10}{3} - 1 = \frac{7}{3}
\]
- For the second equation:
\[
6n + 3 = 6 - n
\]
\[
7n = 3
\]
\[
n = \frac{3}{7}
\]
- **Results:** Different solutions.
2. **Equations:**
\(2(3g + 5) - 2g = 2(4 - g)\)
\(-\frac{36h}{6} = 2\)
- For the first equation:
\[
2(3g + 5) - 2g = 2(4 - g)
\]
\[
6g + 10 - 2g = 8 - 2g
\]
\[
4g + 10 = 8 - 2g
\]
\[
6g = -2
\]
\[
g = -\frac{1}{3}
\]
- For the second equation:
\[
-\frac{36h}{6} = 2
\]
\[
-6h = 2
\]
\[
h = -\frac{1}{3}
\]
- **Results:** Same solution \((-1/3)\).
3. **Equations:**
\(4k - 5 = 2k\)
\(8j - 2 = 3j\)
- For the first equation:
\[
4k - 5 = 2k
\]
\[
2k = 5
\]
\[
k = \frac{5}{2}
\]
- For the second equation:
\[
8j - 2 = 3j
\]
\[
5j = 2
\]
\[
j = \frac{2}{5}
\]
- **Results:** Different solutions.
4. **Equations:**
\(4 = k^2 + 5\)
\(4j - \frac{14}{3} + 5 = 3\)
- For the first equation:
\[
k^2 + 5 = 4
\]
\[
k^2 = -1
\]
(No real solution)
- For the second equation:
\[
4j - \frac{14}{3} + 5 = 3
\]
Find a common denominator:
\[
4j - \frac{14}{3} + \frac{15}{3} = 3
\]
\[
4j + \frac{1}{3} = 3
\]
\[
4j = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}
\]
\[
j = \frac{2}{3}
\]
- **Results:** No real solution for the first, valid solution for the second.
**Conclusion:** The pair of linear equations that have the same solution set is:
**\(2(3g + 5) - 2g = 2(4 - g)\) and \(-\frac{36h}{6} = 2\)** with the common solution \(g = h = -\frac{1}{3}\).