Question
Identify the pair of linear equations that have the same solution set.(1 point)
Responses
4=k2+5
and 4j−143+5=3
4 equals Start Fraction k over 2 End Fraction plus 5 and Start Fraction 4 j minus 14 over 3 End Fraction plus 5 equals 3
2(3g+5)−2g=2(4−g)
and −36h6=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
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
4k−5=2k
and 8j−2=3j
Responses
4=k2+5
and 4j−143+5=3
4 equals Start Fraction k over 2 End Fraction plus 5 and Start Fraction 4 j minus 14 over 3 End Fraction plus 5 equals 3
2(3g+5)−2g=2(4−g)
and −36h6=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
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
4k−5=2k
and 8j−2=3j
Answers
GPT-4o mini
To find the pair of linear equations that have the same solution set, we need to determine which pairs can be manipulated to yield equivalent equations. Let’s evaluate each pair systematically:
1. **First Pair:**
- **Equation 1:** \( 4 = k^2 + 5 \) ➔ Rearranging gives \( k^2 = -1 \) (no real solutions, since \( k^2 \) cannot be negative).
- **Equation 2:** \( 4j - \frac{14}{3} + 5 = 3 \) ➔ Rearranging gives \( 4j - \frac{14}{3} + 5 - 3 = 0 \), which simplifies to a specific linear equation in terms of \( j \).
- **Conclusion:** No solution set in common because one has no real solutions.
2. **Second Pair:**
- **Equation 1:** \( 2(3g + 5) - 2g = 2(4 - g) \) ➔ This simplifies to a linear equation.
- **Equation 2:** \( -\frac{36h}{6} = 2 \) ➔ This simplifies to \( -6h = 2 \) or \( h = -\frac{1}{3} \) (finding a specific value for \( h \)).
- **Conclusion:** Different solution types (one is variable linear, the other has a specific value).
3. **Third Pair:**
- **Equation 1:** \( 3(m + 1) = 10 \) ➔ This simplifies to a linear equation in terms of \( m \).
- **Equation 2:** \( 6n + 3 = 6 - n \) ➔ This will also simplify to a linear equation in terms of \( n \).
- Let’s solve both:
- \( 3(m + 1) = 10 \) ⇒ \( m + 1 = \frac{10}{3} \) ⇒ \( m = \frac{10}{3} - 1 = \frac{7}{3} \).
- \( 6n + 3 = 6 - n \) ⇒ \( 7n = 3 \) ⇒ \( n = \frac{3}{7} \).
- **Conclusion:** Different solutions.
4. **Fourth Pair:**
- **Equation 1:** \( 4k - 5 = 2k \) ➔ Rearranging gives \( 2k = 5 \) ⇒ \( k = \frac{5}{2} \) (specific value).
- **Equation 2:** \( 8j - 2 = 3j \) ➔ Rearranging gives \( 5j = 2 \) ⇒ \( j = \frac{2}{5} \) (another specific value).
- **Conclusion:** Different solutions.
After evaluating all pairs, the rigorous comparison indicates they all lead to unique solutions, thus none have the same solution set.
However, apparently pair #3 had similar linear forms but resulted in different singular values.
Thus the answer seems to lead toward the **Third Pair** as the closest, since with equal manipulation of linear equations, outputs differ in constraints rather than forms.
In a typical scenario, without constraints applied, none satisfy the same solution sets.
- Final assessment: **None of the pairs are technically correct** by given equational forms for sameness in solution. Choose by closest alignment of clarity & algebra structure.
1. **First Pair:**
- **Equation 1:** \( 4 = k^2 + 5 \) ➔ Rearranging gives \( k^2 = -1 \) (no real solutions, since \( k^2 \) cannot be negative).
- **Equation 2:** \( 4j - \frac{14}{3} + 5 = 3 \) ➔ Rearranging gives \( 4j - \frac{14}{3} + 5 - 3 = 0 \), which simplifies to a specific linear equation in terms of \( j \).
- **Conclusion:** No solution set in common because one has no real solutions.
2. **Second Pair:**
- **Equation 1:** \( 2(3g + 5) - 2g = 2(4 - g) \) ➔ This simplifies to a linear equation.
- **Equation 2:** \( -\frac{36h}{6} = 2 \) ➔ This simplifies to \( -6h = 2 \) or \( h = -\frac{1}{3} \) (finding a specific value for \( h \)).
- **Conclusion:** Different solution types (one is variable linear, the other has a specific value).
3. **Third Pair:**
- **Equation 1:** \( 3(m + 1) = 10 \) ➔ This simplifies to a linear equation in terms of \( m \).
- **Equation 2:** \( 6n + 3 = 6 - n \) ➔ This will also simplify to a linear equation in terms of \( n \).
- Let’s solve both:
- \( 3(m + 1) = 10 \) ⇒ \( m + 1 = \frac{10}{3} \) ⇒ \( m = \frac{10}{3} - 1 = \frac{7}{3} \).
- \( 6n + 3 = 6 - n \) ⇒ \( 7n = 3 \) ⇒ \( n = \frac{3}{7} \).
- **Conclusion:** Different solutions.
4. **Fourth Pair:**
- **Equation 1:** \( 4k - 5 = 2k \) ➔ Rearranging gives \( 2k = 5 \) ⇒ \( k = \frac{5}{2} \) (specific value).
- **Equation 2:** \( 8j - 2 = 3j \) ➔ Rearranging gives \( 5j = 2 \) ⇒ \( j = \frac{2}{5} \) (another specific value).
- **Conclusion:** Different solutions.
After evaluating all pairs, the rigorous comparison indicates they all lead to unique solutions, thus none have the same solution set.
However, apparently pair #3 had similar linear forms but resulted in different singular values.
Thus the answer seems to lead toward the **Third Pair** as the closest, since with equal manipulation of linear equations, outputs differ in constraints rather than forms.
In a typical scenario, without constraints applied, none satisfy the same solution sets.
- Final assessment: **None of the pairs are technically correct** by given equational forms for sameness in solution. Choose by closest alignment of clarity & algebra structure.