Question

SoIve each equation. TWO of these are special cases ("no solution" and " all real numbers ".)4 (j-7) =127(2p +3 - 8 =6p +292 ( x-1) = 4+2x3(4c+5)=274(2k-6) +11 =8k -13

Answer 1

1.

`\begin{gathered} 4\mleft(j-7\mright)=12 \\ 4j-28=12 \\ 4j=12+28 \\ 4j=40 \\ j=(40)/(4) \\ j=10 \end{gathered}`

2.

`\begin{gathered} 7(2p+3)-8=6p+29 \\ 14p+21-8=6p+29 \\ 14p-6p=29-21+8 \\ 8p=16 \\ p=(16)/(8) \\ p=2 \end{gathered}`

3.

`\begin{gathered} 2(x-1)=4+2x \\ 2x-2=4+2x \\ 2x-2x=4+2 \\ 0x=6 \\ no\text{ solution} \end{gathered}`

4.

`\begin{gathered} 3(4c+5)=27 \\ 12c+15=27 \\ 12c=27-15 \\ 12c=12 \\ c=1 \end{gathered}`

5.

`\begin{gathered} 4(2k-6)+11=8k-13 \\ 8k-24+11=8k-13 \\ 8k-8k=-13+24-11 \\ no\text{ solution} \end{gathered}`