Question

Identify the pair of linear equations that have the same solution set.(1 point) Responses 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−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 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 −36h6=2

Answer 1

To identify the pair of linear equations that have the same solution set, we need to analyze each pair of equations.

1. First pair: \(4k−5=2k\) and \(8j−2=3j\)

   • \(4k - 5 = 2k\) rearranges to \(2k = 5\) -> \(k = \frac{5}{2}\)
   • \(8j - 2 = 3j\) rearranges to \(5j = 2\) -> \(j = \frac{2}{5}\)

   Different solutions.

2. Second pair: \(4 = k^2 + 5\) and \(4j−14 + 5 = 3\)

   • \(4 = k^2 + 5\) rearranges to \(k^2 = -1\) (no real solutions)
   • \(4j - 14 + 5 = 3\) simplifies to \(4j - 9 = 3\) -> \(4j = 12\) -> \(j = 3\)

   Different solutions.

3. Third pair: \(3(m+1)=10\) and \(6n + 3 = 6 - n\)

   • \(3(m + 1) = 10\) simplifies to \(m + 1 = \frac{10}{3}\) -> \(m = \frac{7}{3}\)
   • \(6n + 3 = 6 - n\) rearranges to \(7n = 3\) -> \(n = \frac{3}{7}\)

   Different solutions.

4. Fourth pair: \(2(3g + 5) - 2g = 2(4 - g)\) and \(-36h = 6\)

   • First equation: \(6g + 10 - 2g = 8 - 2g\) simplifies to \(6g + 10 - 8 + 2g = 0\) -> \(6g + 2g + 2 = 0\) -> \(8g + 2 = 0\) -> \(g = -\frac{1}{4}\)
   • Second equation: \(-36h = 6\) gives \(h = -\frac{1}{6}\)

   Different solutions.

After reviewing all pairs, none of them have the same solution.

Thus, none of the pairs provided have the same solution set. If you meant to specify pairs differently or needed to see another aspect, please clarify!

Answer 2

which one is the aswer

Answer 3

Let's analyze the pairs of equations again and check for a correct solution set that matches.

1. First Pair:

   • \(4k−5=2k\)
     • Rearranging gives: \(2k = 5 \Rightarrow k = \frac{5}{2}\)
   • \(8j−2=3j\)
     • Rearranging gives: \(5j = 2 \Rightarrow j = \frac{2}{5}\)

   (Different solutions)

2. Second Pair:

   • \(4 = k^2 + 5\)
     • Rearranging gives: \(k^2 = -1\) (no real solutions)
   • \(4j−14 + 5 = 3\)
     • Rearranging gives: \(4j - 9 = 3 \Rightarrow 4j = 12 \Rightarrow j = 3\)

   (Different solutions)

3. Third Pair:

   • \(3(m+1)=10\)
     • Rearranging gives: \(m + 1 = \frac{10}{3} \Rightarrow m = \frac{7}{3}\)
   • \(6n + 3 = 6 - n\)
     • Rearranging gives: \(7n = 3 \Rightarrow n = \frac{3}{7}\)

   (Different solutions)

4. Fourth Pair:

   • \(2(3g + 5) - 2g = 2(4 - g)\)
     • Expanding gives: \(6g + 10 - 2g = 8 - 2g\)
     • Rearranging gives: \(8g + 2 = 0 \Rightarrow g = -\frac{1}{4}\)
   • \(-36h = 6\)
     • Rearranging gives: \(h = -\frac{1}{6}\)

   (Different solutions)

Now, let's check if any pairs are equivalent or express the same mathematical relationship.

After reviewing each one again, it seems I rushed in my initial judgment about "same solutions." The pairs do not yield matching values for different equations in terms of their variable assignments, but they do hold their individual solutions.

Given the original pairs provided, each does appear distinct in their outcomes yielding that none of them have an equivalent solution set.

If the goal was specifically to find pairs that are identical, I can confirm no correctly specified pairs match.

However, if you have additional specific equations or pairs in mind that might have been omitted from the problem set or might represent what needs answering, I'd be glad to look at those directly! Please reply or clarify as needed.