To do:We have to solve the given equations.Solution:(a) $2y+\frac{5}{2}=\frac{37}{2}$Transposing $\frac{5}{2}$ to R.H.S we get, $2y=\frac{37}{2}-\frac{5}{2}$$2y=\frac{37-5}{2}$$2y=\frac{32}{2}$$y=\frac{16}{2}$$y=8$(b) $5t+28 =10$Transposing 28 to R.H.S we get, $5t=10-28$$5t=-18$$t=\frac{-18}{5}$(c) $\frac{a}{5}+3=2$Transposing 3 to R.H.S we get, $\frac{a}{5}=2-3$$\frac{a}{5}=-1$Multiplying both side by 5, we get, $\frac{a}{5}\times5=-1\times5$$a=-5$(d) $\frac{q}{4}+7=5$Transposing 7 to R.H.S we get, $\frac{q}{4}=5-7$$\frac{q}{4}=-2$Multiplying both side by 5, we get, ... Read More

To do:We have to solve the given equations.Solution:(a) $10p=100$Dividing both the sides by 10 we get, $\frac{10p}{10}=\frac{100}{10}$$p=10$(b) $10p+10 =100$Subtracting 10 from both sides we get, $10p+10-10=100-10$$10p=90$Dividing both the sides by 10 we get, $\frac{10p}{10}=\frac{90}{10}$$p=9$(c) $\frac{p}{4}=5$Multiplying both the sides by 4 we get, $(\frac{p}{4})\times 4=5\times 4$$p=20$(d) $\frac{-p}{3}=5$Multiplying both the sides by ... Read More

To do:We have to give the steps you will use to separate the variable and then solve the equation.Solution:(a) $3n-2=46$Adding 2 to both sides of the equation, we get$\Rightarrow 3n-2+2=46+2$$\Rightarrow 3n=48$Dividing both the sides by 3 we get, $\Rightarrow \frac{3n}{3}=\frac{48}{3}$$\Rightarrow n=16$(b) $5m+7=17$Subtracting 7 from both sides of the equation, we ... Read More

To do:We have to give first the step we will use to separate the variable and then solve the equation.Solution: (a) $3l=42$Divide both the sides by 3 we get, $\frac{3l}{3}=\frac{42}{3}$$l=14$(b) $\frac{b}{2}=6$Multiplying both sides by 2 we get, $(\frac{b}{2})\times 2=6\times 2$$b=12$(c) $\frac{p}{7}=4$Multiplying both sides by 7 we get, $(\frac{p}{7})\times 7=4\times 7$$p=28$(d) $4x=25$Dividing ... Read More

To do:We have to give first the step we will use to separate the variable and then solve the equation.Solution:(a) $x-1=0$Adding 1 to both sides of the equation we get, $x-1+1=0+1$$x=1$(b) $x+1=0$Subtracting 1 from both sides of the equation we get, $x+1-1=0-1$$x=-1$(c) $x-1=5$Adding 1 to both sides of the equation ... Read More

To do:We have to set up an equation in the given cases.Solution:(i) Let Parmit have $m$ number of marblesThen, Irfan has $(5m+7)$ marblesTotal number of marbles Irfan has $=37$So, $5m+7=37$(ii) Let the age of Laxmi be $y$ yearsLaxmi’s father is four years older than three times Laxmi’s age.Then Laxmi's father's ... Read More

To do:We have to write the given equations in statement forms.Solution:(i) Given equation is $p+4=15$The given equation is written in the form of a statement as:"The sum of $p$ and 4 is 15."(ii) Given the equation is $m-7=3$.The given equation is written in the form of a statement as:"7 subtracted ... Read More

To do:We have to write equations for the given statements.Solution:(i) The sum of numbers $x$ and 4 is 9. This statement can be written in the form of an equation as:$\boxed{x+4=9}$(ii) 2 subtracted from $y$ is 8. This statement can be written in the form of an equation as:$\boxed{y-2=8}$(iii) Ten ... Read More

Given: Given equations are(i) $5p+2=17$(ii) $3m-14=4$To do: We have to solve the given equations by trial and error method.Solution:(i) $5p+2=17$$5p+2=$L.H.SBy putting, $p=0$, $5(0)+2=2$$≠17$By putting, $p=1$, $5(1)+2=7$$≠17$By putting, $p=2$, $5(2)+2=12$$≠$R.H.SBy putting, $p=3$, $5(3)+2=17$$=$R.H.STherefore, $p=3$ is a solution of the equation.(ii) $3m-14=4$$3m-14=$L.H.SBy putting, $m=5$, $3(5)-14=1$$≠6$By putting, $m=6$, $3(6)-14=4=$R.H.S.Therefore, $m=6$ is a solution ... Read More

To do:We have to check whether the value given in the brackets is a solution to the given equation or not.Solution:(a) Here, L.H.S $=n+5$R.H.S $=19$And $n=1$ (given)By putting, $n=1$, L.H.S $=1+5=6$$≠$R.H.SL.H.S $≠$ R.H.S, so $n=1$ is not a solution of the equation.(b) Here,  L.H.S $=7n+5$R.H.S $=19$And $n=-2$ (given)By putting, $n=-2$, L.H.S ... Read More