question_answer 1)

Solve the following equations: (a) \[\text{1}0\text{p}=\text{1}00\]                        (b) \[\text{1}0\text{p}+\text{1}0=\text{1}00\]   (c) \[\frac{p}{4}=5\]                          (d) \[\frac{-p}{3}=5\] (e) \[\frac{3p}{4}=6\]                     (f) \[\text{3s}=-\text{9}\]                            (g) \[\text{3s}+\text{12}=0\]                          (h) \[\text{3s}=0\] (i) \[\text{2q}=\text{6}\]                              (j) \[\text{2q}-\text{6}=0\]                          (k) \[\text{2q}+\text{6}=0\]                           (I) \[\text{2q}+\text{6}=\text{12}\].                

Answer:

                (a) The given equation is \[10p=100\] Divide both sides by 10, \[\frac{10p}{10}=\frac{100}{10}\]\[\Rightarrow \]\[p=10\] It is the required solution. Check. Put p = 10 in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{l0p}=\text{1}0\times \text{1}0=\text{1}00=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (b) The given equation is \[\text{l0p}+\text{1}0=\text{1}00\] Subtract 10 from both sides, \[\text{l0p}+\text{1}0-\text{1}0=\text{1}00-\text{1}0\] \[\Rightarrow \] \[10p=90\] Divide both sides by 10, \[\frac{10p}{10}=\frac{90}{10}\]                \[\Rightarrow \] \[p=9\] It is the required solution. Check. Put \[p=9\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{l0p}+\text{1}0=\text{1}0\times \text{9}+\text{1}0\] \[=\text{9}0+\text{1}0=\text{1}00=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (c) The given equation is \[\frac{p}{4}=5\] Multiply both sides by 4, \[\frac{p}{4}\times 4=5\times 4\]             \[\Rightarrow \]\[p=20\] It is the required solution. Check. Put \[p=20\] in the given equation. L.H.S. \[=\frac{p}{4}=\frac{20}{4}=5=\]R.H.S. as required. Hence, the solution is correct. (d) The given equation is \[-\frac{p}{3}=5\] Multiplying both sides by \[(-\text{ 3})\], \[\left( \frac{-p}{3} \right)\,\times (-3)=5\times (-3)\]\[\Rightarrow \]\[p=-15\] It is the required solution. Check. Put \[\text{p}=-\text{15}\] in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{-p}{3}=\frac{-(-15)}{3}=\frac{15}{3}=5=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (e) The given equation is \[\frac{3p}{4}=6\] Multiply both sides by 4, \[\frac{3p}{4}\times 4=6\times 4\]           \[\Rightarrow \] \[3p=24\] Divide both sides by 3, \[\frac{3p}{3}=\frac{24}{3}\] \[\Rightarrow \] \[p=8\] It is the required solution. Check. Put p = 8 in the given equation. \[\text{L}\text{.H}\text{.S =}\frac{3p}{4}=\frac{3}{4}\times 8=3\times 2=6=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (f) The given equation is \[3s=-9\] Divide both sides by 3, \[\frac{3s}{3}=-\frac{9}{3}\] \[\Rightarrow \] \[s=-3\] It is the required solution. Check. Put \[\text{s}=-\text{3}\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{3}\times (-\text{3})=-\text{9}=\text{ R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (g) The given equation is \[\text{3s}+\text{12}=0\] Subtract ?12 from both sides, \[\text{3s}+\text{12}-\text{12}=0-\text{12}\] \[\Rightarrow \]\[\text{3s}=-\text{ 12}\] Divide both sides by 3, \[\frac{3s}{3}=-\frac{12}{3}\]\[\Rightarrow \] \[s=-4\] It is the required solution. Check. Put \[\text{s}=-\text{ 4}\] in the given equation. L.H.S. \[~=\text{3s}+\text{12}=\text{3}\times (-\text{4})+\text{12}\] \[=-\text{12}+\text{12}=0=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (h) The given equation is \[\text{3s}=0\] Divide both sides by 3, \[\frac{3s}{3}=\frac{0}{3}\]          \[\Rightarrow \]\[s=0\] It is the required solution. Check. Put \[\text{s}=0\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{3s}=\text{3}\times 0=0=\text{ R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (i) The given equation is \[2q=6\] Divide both sides by 2, \[\frac{2q}{2}=\frac{6}{2}\]         \[\Rightarrow \]\[q=3\] It is the required solution. Check. Put \[q=3\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{2q}=\text{2}\times \text{3}=\text{6}=\text{R}.\text{H}.\text{S}.\] Hence, the solution is correct. (i) The given equation is \[2q-6=0\] Add 6 to both sides, \[2q-6+6=0+6\] \[\Rightarrow \] \[2q=6\] Divide both sides by 2, \[\frac{2q}{2}=\frac{6}{2}\]         \[\Rightarrow \] \[q=3\] It is the required solution. Check. Put \[q=3\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{2q}-\text{6}=\text{2}\times \text{3}-\text{6}=\text{6}-\text{6}=0=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (k) The given equation is \[\text{2q}+\text{6}=0\] Subtract 6 from both sides, \[2q+6-6=0-6\]\[\Rightarrow \]\[2q=-6\] Divide both sides by 2, \[\frac{2q}{2}=-\frac{6}{2}\]\[\Rightarrow \]\[q=-3\] It is the required solution. Check. Put \[\text{q}=-\text{ 3}\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{2q}+\text{6}=\text{2}\times (-\text{3})+\text{6}=-\text{6}+\text{6}=0=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (l) The given equation is \[\text{2q}+\text{6}=\text{12}\] Subtract 6 from both sides, \[2q+6-6=12-6\] \[\Rightarrow \]\[2q=6\] Divide both sides by 2, \[\frac{2q}{2}=\frac{6}{2}\]         \[\Rightarrow \]\[q=3\] It is the required solution. Check. Put \[q=3\] in the given equation. L.H.S. \[=\text{2q}+\text{6}=\text{2}\times \text{3}+\text{6}=\text{6}+\text{6}=\text{12}=\] R.H.S. as required. Hence, the solution is correct.