question_answer 1)

Solve the following equations: (a) \[\text{4}=\text{5(p}-\text{2)}\]       (b) \[-4=5(p-2)\]               (c) \[\text{16}=\text{4}+\text{3(t}+\text{2)}\]         (d) \[\text{4}+\text{5(p}-\text{l)}=\text{34}\]          (e) \[0=16+4(m-6).\]                

Answer:

                (a) The given equation is \[\text{4}=\text{5(p}-\text{2)}\] \[\Rightarrow \]               \[\text{5(p}-\text{2)}=\text{4}\]     \[\left| \begin{align}   & \text{An equation remains the same},\text{ } \\  & \text{when the expressionson the left } \\  & \text{and on the right are interchanged}. \\ \end{align} \right.\] Divide both sides by 5, \[\frac{5(p-2)}{5}=\frac{4}{5}\]  \[\Rightarrow \]\[P-2=\frac{4}{5}\] Transposing \[-\text{ 2}\] from L.H.S. to R.H.S., \[P=\frac{4}{5}+2\] \[\Rightarrow \] \[P=\frac{4+10}{5}\] \[\Rightarrow \]               \[p=\frac{14}{5}\] It is the required solution. Check. Put \[p=\frac{14}{5}\] in the given equation. \[\text{R}\text{.H}\text{.S = 5(p-2) = 5 }\left( \frac{14}{5}-2 \right)\] \[=5\left( \frac{14-10}{5} \right)=5\left( \frac{4}{5} \right)=4\] \[=\text{L}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (b) The given equation is \[-4=5(p-2)\] \[\Rightarrow \]               \[5(p-2)=-4\] \[\left| \begin{align}   & \text{An equation remains the same when the} \\  & \text{expressions on the left and on the right} \\  & \text{are interchanged}. \\ \end{align} \right.\] Divide both sides by 5, \[\frac{5(p-2)}{5}=\frac{-4}{5}\]                \[\Rightarrow \] \[p-2=\frac{-4}{5}\] Transposing \[-\text{ 2}\] from L.H.S. to R.H.S., \[p=\frac{-4}{5}+2\]\[\Rightarrow \]\[p=\frac{-4+10}{5}\] \[\Rightarrow \]               \[p=\frac{6}{5}\] It is the required solution. Check. Put \[p=\frac{6}{5}\] in the given equation. \[R.H.S\text{ }=\text{ }5(p-2)\] \[=5\left( \frac{6}{5}-2 \right)=5\left( \frac{6-10}{5} \right)\] \[=5\left( -\frac{4}{5} \right)=-4\] \[=\text{L}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (c) The given equation is \[\text{16}=\text{4}+\text{3(t}+\text{2)}\] Transposing it from L.H.S. to R.H.S., \[\text{16}-\text{4}=\text{3(t}+\text{2)}\] \[\Rightarrow \]               \[12=3(t+2)\] Divide both sides by 3, \[\frac{12}{3}=\frac{3(t+2)}{3}\]\[\Rightarrow \]\[4=t+2\] \[\Rightarrow \]               \[t=2\] Transposing 2 from L.H.S. to R.H.S., \[4-2=t\]\[\Rightarrow \]  \[2=t\] \[\Rightarrow \]               \[t=2\] It is the required solution. Check. Put \[\text{t}=\text{2}\] in the given equation. R.H.S.    \[~=\text{4}+\text{3(2}+\text{2)}\] \[=\text{4}+\text{3(4)}\] \[=\text{4}+\text{12}\] \[=16\] = L.H.S. as required. Hence, the solution is correct. (d) The given equation is \[\text{4}+\text{5(p}-\text{l)}=\text{34}\] Transposing it from L.H.S. to R.H.S.,                                              \[5(p-1)=34-4=30\] Divide both sides by 5, \[\frac{5(p-1)}{5}=\frac{30}{5}\]                \[\Rightarrow \]               \[p-1=6\] Transposing \[-\text{ 1}\] from L.H.S. to R.H.S., \[\text{p}=\text{6}+\text{l}=\text{7}\] It is the required solution. Check. Put \[\text{p}=\text{7}\] in the given equation. \[\text{R}.\text{H}.\text{S}.=\text{4}+\text{5(7}-\text{1)}\] \[=\text{4}+\text{5(6)}\] \[=\text{4}+\text{3}0\] \[=\text{34}\] \[=\text{ L}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (e) The given equation is \[0=\text{16}+\text{4(m}-\text{6)}\] \[\Rightarrow \]               \[16+4(m-6)=0\] \[\left| \begin{align}   & \text{An equation remains the same when the expres } \\  & \text{sions on the left and on the right are interchanged} \\ \end{align} \right.\] Transposing 16 form L.H.S. to R.H.S., \[4(m-6)=-16\] Divide both sides by 4, \[\frac{4(m-6)}{4}=-\frac{16}{4}\] \[\Rightarrow \] \[m-6=-4\] Transposing - 6 from L.H.S. to R.H.S., \[m=-4+6\]  \[\Rightarrow \] \[m=2\] It is the required solution. Check. Put m = 2 in the given equation. \[\text{R}.\text{H}.\text{S}.=\text{16}+\text{4(m}-\text{6)}=\text{16}+\text{4(2}-\text{6)}\] \[=\text{16}+\text{4(}-\text{4)}=\text{16}-\text{16}=0\] \[=\text{L}.\text{H}.\text{S}.\] as required. Hence, the solution is correct.