Answer:
(a) The given equation is \[3l=42\] Divide both sides by 3, \[\frac{3l}{3}=\frac{42}{3}\] \[\Rightarrow \] \[l=14\] It is the required solution. Cheek. Put \[\text{t}=\text{14}\] in the given equation. L.H.S. \[~=3l=3\times 14=42=\] R.H.S. as required Hence, the solution correct. (b) The given equation is \[\frac{b}{2}=6\] Multiply both sides by 2, \[2\times \left( \frac{b}{2} \right)=2\times 6\] \[\Rightarrow \]\[b=12\] It is the required solution. Check. Put b = 12 in the given equation. L.H.S \[=\frac{b}{2}=\frac{12}{2}=6=\]R.H.S. as required Hence, the solution is correct. (c) The given equation is \[\frac{p}{7}=4\] Multiply both sides by 7, \[7\times \left( \frac{p}{7} \right)=7\times 4\] \[\Rightarrow \]\[p=28\] It is the required solution. Check. Put \[p=28\] in the given equation. \[\text{L}\text{.H}\text{.S =}\frac{p}{2}=\frac{28}{7}=4=\text{R}\text{.H}\text{.S }\]as required. Hence, the solution is correct. (d) The given equation is \[4x=25\] Divide both sides by 4, \[\frac{4x}{4}=\frac{25}{4}\] \[\Rightarrow \]\[x=\frac{25}{4}\] It is the required solution. Check. Put \[x=\frac{25}{4}\]in the given equation. \[\text{L}\text{.H}\text{.S}\text{.}=4x=4\times \left( \frac{25}{4} \right)=25=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (e) The given equation is \[8y=36\] Divide both sides by 8, \[\frac{8y}{8}=\frac{36}{8}\] \[\Rightarrow \] \[y=\frac{36}{8}\] \[\Rightarrow \] \[y=\frac{36\div 4}{8\div 2}\] \[\Rightarrow \] \[y=\frac{9}{2}\] It is the required solution. Check. Put \[y=\frac{9}{2}\] in the given equation. \[\text{L}.\text{H}.\text{S}.=8y=8\times \left( \frac{9}{2} \right)=4\times 9=36=\text{R}.\text{H}.\text{S}\] as required. Hence, the solution is correct. (f) The given equation is \[\frac{z}{3}=\frac{5}{4}\] Multiply both sides by 3, \[3\times \left( \frac{z}{3} \right)=3\times \left( \frac{5}{4} \right)\]\[\Rightarrow \] \[z=\frac{15}{4}\] It is the required solution. Check. Put \[z=\frac{15}{4}\]in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{z}{3}=\frac{1}{3}\times \frac{15}{4}=\frac{5}{4}=\text{R}\text{.H}\text{.S}\] as required. Hence, the solution is correct. (g) The given equation is \[\frac{a}{5}=\frac{7}{15}\] Multiply both sides by 5, \[\frac{a}{5}\times 5=\frac{7}{15}\times 5\] \[\Rightarrow \] \[a=\frac{7}{3}\] It is the required solution. Check. Put \[a=\frac{7}{3}\] in the given equation. \[\text{L}\text{.H}\text{.S}=\frac{a}{5}=\frac{1}{5}\times \left( \frac{7}{3} \right)=\frac{7}{15}=\text{R}\text{.H}\text{.S}\] as required Hence, the solution is correct. (h) The given equation is \[20t=-10\] Divide both sides by 20, \[\frac{20t}{20}=-\frac{10}{20}\] \[\Rightarrow \] \[t=-\frac{1}{2}\] It is the required solution. Check. Put\[t=-\frac{1}{2}\] in the given equation. L.H.S. \[=20t=20\times \left( -\frac{1}{2} \right)=-10=\]R.H.S. as required. Hence, the solution is correct.
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