question_answer 1)

Solve the following equations: (a) \[2y+\frac{5}{2}=\frac{37}{2}\]                            (b) \[5t+28=10\]                               (c) \[\frac{a}{5}+3=2\]                   (d) \[\frac{q}{4}+7=5\] (e) \[\frac{5}{2}x=-10\]                 (f) \[\frac{5}{2}x=\frac{25}{4}\]                 (g) \[7m+\frac{19}{2}=13\]          (h) \[6z+10=-2\] (i) \[\frac{2b}{3}-5=3\].                

Answer:

                (a) The given equation is \[2y+\frac{5}{2}=\frac{37}{2}\] Transposing \[\frac{5}{2}\] from L.H.S. to R.H.S., \[2y=\frac{37}{2}-\frac{5}{2}\]   \[\Rightarrow \]  \[2y=\frac{37-5}{2}\] \[\Rightarrow \]               \[2y=\frac{32}{2}\] \[\Rightarrow \]\[2y=16\] Divide both sides by 2, \[\frac{2y}{2}=\frac{16}{2}\] \[\Rightarrow \]\[y=8\] It is the required solution. Check. Put \[y=8\] in the given equation.                 \[\text{L}\text{.H}\text{.S = }2y+\frac{5}{2}=2\times 8+\frac{5}{2}=16+\frac{5}{2}\] \[=\frac{32+5}{2}=\frac{37}{2}=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (b) The given equation is \[5t+28=10\] Transposing 28 from L.H.S. to R.H.S.,          \[\Rightarrow \]               \[5t=10-28\]\[\Rightarrow \]\[5t=-18\] Divide both sides by 5, \[\frac{5t}{5}=-\frac{18}{5}\]\[\Rightarrow \]\[t=-\frac{18}{5}\] It is the required solution. Check. Put \[t=-\frac{18}{5}\] in the given equation.                 \[\text{L}\text{.H}\text{.S = 5t + 28 = 5 }\times \left( -\frac{18}{5} \right)+28\]                 \[=-18+28=10\] = R.H.S. as required. Hence, the solution is correct. (c) The given equation is \[\frac{a}{5}+3=2\] Transposing 3 from L.H.S. to R.H.S., \[\frac{a}{5}=2-3\]           \[\Rightarrow \]\[\frac{a}{5}=-1\] Multiply both sides by 5, \[\frac{a}{5}\times 5=(-1)\times 5\]        \[\Rightarrow \]\[a=-5\] It is the required solution. Check. Put \[\text{a}=-\text{ 5}\] in the given equation. \[\text{L}\text{.H}\text{.S =}\frac{a}{5}+3=-\frac{5}{5}+3\] \[=-1+3=2=\text{R}\text{.H}\text{.S }\] as required. Hence, the solution is correct. (d) The given equation is \[\frac{q}{4}+7=5\] Transposing 7 from L.H.S. to R.H.S., \[\frac{q}{4}=5-7\]          \[\Rightarrow \] \[\frac{q}{4}=-2\] Multiply both sides by 4, \[\frac{q}{4}\times 4=(-2)\times 4\] \[\Rightarrow \] \[q=-8\] It is the required solution. Check. Put \[\text{q}=-\text{ 8}\] in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{q}{4}+7=\frac{-8}{4}+7=-2+7=5=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (e) The given equation is \[\frac{5}{2}x=-10\] Multiply both sides by 2, \[\left( \frac{5}{2}x \right)\times 2=(-10)\times 2\] \[\Rightarrow \] \[5x=-20\] Divide both sides by 5 \[\frac{5x}{5}=\frac{-20}{5}\]\[\Rightarrow \] \[x=-4\] It is the required solution. Check. Put \[x=-\text{ }4\] in the given equation \[\text{L}\text{.H}\text{.S}=\frac{5}{2}x=\frac{5}{2}\times (-4)=5\times (-2)=-10\] = R.H.S. as required. Hence, the solution is correct. (f) The given equation is \[\frac{5}{2}x=\frac{25}{4}\] Multiply both sides by 2, \[\left( \frac{5}{2}x \right)\times 2=\frac{25}{4}\times 2\]\[\Rightarrow \]\[5x=\frac{25}{2}\] Divide both sides by 5, \[\frac{5x}{5}=\frac{25}{2\times 5}\]\[\Rightarrow \]\[x=\frac{5}{2}\] It is the required solution. Check. Put \[x=\frac{5}{2}\] in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{5}{2}x=\frac{5}{2}\times \frac{5}{2}=\frac{25}{4}=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (g) The given equation is \[7m+\frac{19}{2}=13\] Transposing \[\frac{19}{2}\]from L.H.S. to R.H.S., \[7m=13-\frac{19}{2}=\frac{26-19}{2}=\frac{7}{2}\] Divide both sides by 7 \[\frac{7m}{7}=\frac{7}{2\times 7}\]\[\Rightarrow \]\[m=\frac{1}{2}\] It is the required solution. Check. Put \[m=\frac{1}{2}\] in the given equation. \[\text{L}\text{.H}\text{.S}=7m+\frac{19}{2}=7\times \left( \frac{1}{2} \right)+\frac{19}{2}=\frac{7}{2}+\frac{19}{2}\] \[=\frac{7+19}{2}=\frac{26}{2}=13=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (h) The given equation is \[\text{6z}+10=-\text{2}\] Transposing 10 from L.H.S. to R.H.S., \[\text{6z}=-\text{2}-\text{1}0\]              \[\Rightarrow \] \[\text{6z}=-\text{12}\] Dividing both sides by 6, \[\frac{6z}{6}=-\frac{12}{6}\]\[\Rightarrow \]\[z=-2\] It is the required solution. Check. Put \[\text{z}=-\text{ 2}\] in the given equation. \[\text{L}.\text{H}.\text{S}.=\text{6z}+\text{1}0=\text{6}\times (-\text{ 2})+\text{1}0.\] \[=-\text{12}+\text{1}0=-\text{2}\] \[=\text{ R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct. (i) The given equation is \[\frac{3l}{2}=\frac{2}{3}\] Multiply both sides by 2, \[\frac{3l}{2}\times 2=\frac{2}{3}\times 2\]          \[\Rightarrow \]\[3l=\frac{4}{3}\] Divide both sides by 3, \[\frac{3l}{3}=\frac{4}{3\times 3}\]          \[\Rightarrow \]\[l=\frac{4}{9}\] It is the required solution. Check. Put \[l=\frac{4}{9}\] in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{3l}{2}=\frac{3}{2}\times \left( \frac{4}{9} \right)=\frac{2}{3}=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (j) The given equation is \[\frac{2b}{3}-5=3\] Transposing \[-5\] from L.H.S. to R.H.S., \[\frac{2b}{3}=3+5\]\[\Rightarrow \] \[\frac{2b}{3}=8\] Multiply both sides by 3, \[\left( \frac{2b}{3} \right)\times 3=8\times 3\] \[\Rightarrow \] \[2b=24\] Divide both sides by 2, \[\frac{2b}{2}=\frac{24}{2}\]\[\Rightarrow \]\[b=12\] It is the required solution. Check. Put b = 12 in the given equation. L.H.S. \[=\frac{2b}{3}-5=\frac{2}{3}\times 12-5\] \[=\text{2}\times \text{4}-\text{5}=\text{8}-\text{5}\] \[=\text{3}=\text{R}.\text{H}.\text{S}.\] as required. Hence, the solution is correct