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Alecsey [184]
1 year ago
3

3 Drag each equation to the correct location on the table. Determine the number of solutions to each equation. Then place each e

quation in the box that corresponds to its number of solutions. 35 = 2+ +1 2 – 1 = 45 + 3 31 – 2 35 + 1 2x + 3 = 35 – 1 1 2x + 1 = 21 No Solutions 1 Solution 2 Solutions Reset Next All rights reserved. i NE
Mathematics
1 answer:
Bad White [126]1 year ago
5 0
\begin{gathered} 3^x=2^x+1 \\ \text{Let x=1, then 3=2+1} \end{gathered}

Then, it has just 1 solution, and it should be placed in the second column.

\begin{gathered} 2^x-1=4^x+3. \\ \text{This has no solution} \end{gathered}\begin{gathered} 3x-2=3^x+1 \\ \text{This has no solution.} \end{gathered}

Next;

\begin{gathered} \frac{1}{2}x+3=3^x-1 \\ \text{This has no solution. It should be in the first column} \end{gathered}\begin{gathered} 2x+1=2^x \\ \text{Let x=0,} \\ 2(0)+1=2^0=1 \end{gathered}

This has one solution, and it should be placed in the second column.

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