NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (EX 4.3) Exercise 4.3

Q.1

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)2\mathrm{y}+\frac{5}{2}=\frac{37}{2}\left(\mathrm{b}\right)5\mathrm{t}+28=10\left(\mathrm{c}\right)\frac{\mathrm{a}}{5}+3=2\\ \left(\mathrm{d}\right)\frac{\mathrm{q}}{4}+7=5\left(\mathrm{e}\right)\frac{5}{2}\mathrm{x}=10\left(\mathrm{f}\right)\frac{5}{2}\mathrm{x}=\frac{25}{4}\\ \left(\mathrm{g}\right)7\mathrm{m}+\frac{19}{2}=13\left(\mathrm{h}\right)6\mathrm{z}+10=-2\left(\mathrm{i}\right)\frac{3\mathrm{l}}{2}=\frac{2}{3}\\ \left(\mathrm{j}\right)\frac{2\mathrm{b}}{3}-5=3\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{2}y+\frac{5}{2}=\frac{37}{2}\\ 2y=\frac{37}{2}-\frac{5}{2}=\frac{37-5}{2}=\frac{32}{2}=16\\ 2y=16\\ y=\frac{16}{2}=8\\ Thus,\boxed{y=8}\\ \left(\text{b}\right)\text{5t}+\text{28}=\text{1}0\\ 5t=10-28\\ 5t=-18\\ \boxed{t=\frac{-18}{5}}\\ \left(\text{c}\right)\frac{\text{a}}{5}+3=2\\ \frac{a}{5}=2-3\\ \frac{a}{5}=-1\\ \boxed{a=-5}\\ \left(\text{d}\right)\frac{q}{4}+7=5\\ \frac{q}{4}=5-7\\ \frac{q}{4}=-2\\ \boxed{q=-8}\\ \left(\text{e}\right)\frac{5}{2}x=-10\\ 5x=-10\times 2\\ x=\frac{-10\times 2}{5}=\frac{-\cancel{5}\times 2\times 2}{\cancel{5}}=-4\\ Thus,\boxed{x=-4}\\ \left(\text{f}\right)\frac{5}{2}x=\frac{25}{4}\\ 5x=\frac{25\times 2}{4}\\ x=\frac{\cancel{5}\times 5\times \cancel{2}}{\cancel{2}\times 2\times \cancel{5}}\\ Thus,\boxed{x=\frac{5}{2}}\\ \left(\text{g}\right)7m+\frac{19}{2}=13\\ 7m=13-\frac{19}{2}=\frac{26-19}{2}=\frac{7}{2}\\ 7m=\frac{7}{2}\\ m=\frac{\cancel{7}}{2\times \cancel{7}}\\ Thus,\boxed{m=\frac{1}{2}}\\ \left(\text{h}\right)\text{6z}+\text{1}0=-\text{2}\\ 6z=-2-10\\ 6z=-12\\ z=\frac{-12}{6}=\frac{-2\times \cancel{6}}{\cancel{6}}0\\ Thus,\boxed{z=-2}\\ \left(\text{i}\right)\frac{3l}{2}=\frac{2}{3}\\ 3l=\frac{4}{3}\\ \boxed{l=\frac{4}{9}}\\ \left(\text{j}\right)\frac{2b}{3}-5=3\\ \frac{2b}{3}=8\\ 2b=24\\ b=\frac{24}{2}=\frac{\cancel{2}\times 12}{\cancel{2}}\\ Thus,\boxed{b=12}\end{array}

Q.2

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}:\\ \left(\mathrm{a}\right)2(\mathrm{x}+4)=12\left(\mathrm{b}\right)3(\mathrm{n}-5)=21\left(\mathrm{c}\right)3(\mathrm{n}-5)=21\\ \left(\mathrm{d}\right)3-2(2-\mathrm{y})=7\left(\mathrm{e}\right)-4(2-\mathrm{x})=9\left(\mathrm{f}\right)4(2-\mathrm{x})=9\\ \left(\mathrm{g}\right)4+5(\mathrm{p}-1)=34\left(\mathrm{h}\right)34-5(\mathrm{p}-1)=4\end{array}$

Ans.

\begin{array}{l}\text{(a})2(x+4)=12\\ \text{Divide both sides by 2 to get}\\ x+4=6\\ \text{Thus,}\boxed{x=2}\\ \left(\text{b}\right)3(n-5)=-21\\ \text{Divide both sides by 3 to get}\\ n-5=-\text{7}\\ \text{Thus,}\boxed{n=-2}\\ \left(\text{c}\right)3(n-5)=-\text{21}\\ \text{Divide both sides by 2 to get}\\ n-5=-7\\ Thus,\boxed{n=-12}\\ \left(\text{d}\right)\text{3}-\text{2}(\text{2}-y)=\text{7}\\ -2\left(2-y\right)=4\\ \text{Divide both sides by}-\text{2 to get}\\ 2-y=-2\\ \text{Multiply both sides by}-\text{1 to get}\\ y-2=2\\ \text{Thus,}\boxed{y=4}\\ \left(\text{e}\right)-4(2-x)=\text{9}\\ \text{Divide both sides by}-\text{4 to get}\\ \text{2}-x=\frac{-9}{4}\\ \text{Multiply both sides by}-\text{1 to get}\\ x-2=\frac{9}{4}\\ x=\frac{9}{4}+2=\frac{9+8}{4}\\ Thus,\boxed{x=\frac{17}{4}}\\ \left(\text{f}\right)4(2-x)=\text{9}\\ \text{Divide both sides by 4 to get}\\ \text{2}-x=\frac{9}{4}\\ \text{Multiply both sides by -1 to get}\\ \text{x}-2=\frac{-9}{4}\\ x=\frac{-9}{4}+2=\frac{-9+8}{4}\\ \text{Thus},\boxed{x=\frac{-1}{4}}\\ \left(\text{g}\right)4+5(p-1)=34\\ 5\left(p-1\right)=30\\ \text{Divide both sides by 5 to get}\\ \text{p}-1=6\\ p=6+1=7\\ \text{Thus},\boxed{p=7}\\ \left(\text{h}\right)34-5(p-1)=4\\ -5\left(p-1\right)=-30\\ \text{Divide both sides by}-\text{5 to get}\\ p-1=6\\ p=6+1=7\\ \text{Thus,}\boxed{p=7}\end{array}

Q.3

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{equations}.\\ \left(\mathrm{a}\right)4=5(\mathrm{p}-2)\left(\mathrm{b}\right)-4=5(\mathrm{p}-2)\left(\mathrm{c}\right)-16=-5(2-\mathrm{p})\\ \left(\mathrm{d}\right)10=4+3(\mathrm{t}+2)\left(\mathrm{e}\right)28=4+3(\mathrm{t}+5)\\ \left(\mathrm{f}\right)0=16+4(\mathrm{m}-6)\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{4}=\text{5}(p-\text{2})\\ \text{Divide both sides by 5 to get}\\ \frac{4}{5}=p-2\\ p=\frac{4}{5}+2=\frac{4+10}{5}\\ \text{Thus},\boxed{p=\frac{14}{5}}\\ \left(\text{b}\right)-\text{4}=\text{5}(p-2)\\ \text{Divide both sides by 5 to get}\\ \frac{-4}{5}=p-2\\ p=\frac{-4}{5}+2=\frac{-4+10}{5}\\ \text{Thus},\boxed{p=\frac{6}{5}}\\ \left(\text{c}\right)-16=-5(2-p)\\ \text{Divide both sides by}-\text{5 to get}\\ \frac{16}{5}=2-p\\ \text{Multiply both sides by}-\text{1 to get}\\ \frac{-\text{16}}{5}=p-2\\ p=\frac{-16}{5}+2=\frac{-16+10}{5}\\ \text{Thus},\boxed{p=\frac{-6}{5}}\\ \left(\text{d}\right)\text{1}0=\text{4}+\text{3}(t+2)\\ 6=3\left(t+2\right)\\ \text{Divide both sides by 3 to get}\\ 2=t+2\\ \text{Thus,}\boxed{t=0}\\ \left(\text{e}\right)\text{28}=\text{4}+\text{3}(t+\text{5})\\ 24=3\left(t+5\right)\\ \text{Divide both sides by 3 to get}\\ 8=t+5\\ \text{Thus,}\boxed{t=3}\\ \left(\text{f}\right)0=\text{16}+\text{4}(m-\text{6})\\ -16=4\left(m-6\right)\\ \text{Divide both sides by 4 to get}\\ -4=m-6\\ \text{Thus,}\boxed{m=2}\end{array}

Q.4

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=2\\ \left(\mathrm{b}\right)\mathrm{Construct}3\mathrm{equations}\mathrm{starting}\mathrm{with}\mathrm{x}=-2\end{array}$

Ans.

\begin{array}{l}\text{(a) We have}\\ x=2\\ \text{Multiply both sides by 4 to get}\\ \boxed{4x=8}\dots \dots \dots \dots \left(\text{i}\right)\\ \text{Subtract 2 from both sides to get}\\ \boxed{4x-2=6}\dots \dots \left(\text{ii}\right)\\ \text{Divide both sides by 2 to get}\\ \boxed{2x-1=3}\dots \dots \text{(iii)}\\ \text{(b)}\text{We have}\\ x=-2\\ \text{Adding 4 to both sides to get}\\ \boxed{x+4=2}\dots \dots \dots \left(\text{i}\right)\\ \text{Multiply 2 from both sides to get}\\ \boxed{2x+8=4}\dots \dots \left(\text{ii}\right)\\ \text{Divide both sides by 2 to get}\\ \boxed{x+4=2}\dots \dots \left(\text{iii}\right)\end{array}