Question 1

Check off all of the equations that would give one solution

Question 2

Solution:

Given the equations:

$\begin{gathered} 5x+10=5x-15 \\ collect\text{ like terms,} \\ 5x-5x=-15-10 \\ 0=-25 \\ thus,\text{ there's no solution.} \end{gathered}$
$\begin{gathered} 3x+12=4x-21 \\ collect\text{ like terms,} \\ 3x-4x=-21-12 \\ \Rightarrow-x=-33 \\ divide\text{ both sides by -1,} \\ -(x)/(-1)=-(33)/(-1) \\ \Rightarrow x=33 \\ thus,\text{ there's a solution} \end{gathered}$
$\begin{gathered} 3(x+4)=3x+12 \\ open\text{ parentheses,} \\ 3x+12=3x+12 \\ thus,\text{ there's infinitely many solutions.} \end{gathered}$
$\begin{gathered} 5x+10=6x-8 \\ collect\text{ like terms,} \\ 5x-6x=-8-10 \\ \Rightarrow-x=-18 \\ divide\text{ both sides by -1,} \\ -(x)/(-1)=-(18)/(-1) \\ \Rightarrow x=18 \\ thus,\text{ there's a solution.} \end{gathered}$
$\begin{gathered} 2(3x-4)=6x-8 \\ open\text{ parentheses,} \\ 6x-8=6x-8 \\ thus,\text{ there's infinitely many solutions.} \end{gathered}$
$\begin{gathered} 6x-8=2(3x-3) \\ open\text{ parentheses,} \\ 6x-8=6x-6 \\ collect\text{ like terms,} \\ 6x-6x=-6+8 \\ 0=2 \\ thus,\text{ there's no solution.} \end{gathered}$ Hence, the equations that would give one solution are:

Check off all of the equations that would give one solution-example-2

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