Simple Equations - Solutions

CSBE Class –VII Mathematics

NCERT Solutions
Simple Equations (Ex. 4.1)

Question 1. Complete the last column of the table:

 S. No. Equation Value Say, whether the Equation is satisfied. (Yes / No) 1 x+3=0$x+3=0$ x=3$x=3$ 2 x+3=0$x+3=0$ x=0$x=0$ 3 x+3=0$x+3=0$ x=−3$x=-3$ 4 x−7=1$x-7=1$ x=7$x=7$ 5 x−7=1$x-7=1$ x=8$x=8$ 6 5x=25$5x=25$ x=0$x=0$ 7 5x=25$5x=25$ x=5$x=5$ 8 5x=25$5x=25$ x=−5$x=-5$ 9 m3=2$\frac{m}{3}=2$ m=−6$m=-6$ 10 m3=2$\frac{m}{3}=2$ m=0$m=0$ 11 m3=2$\frac{m}{3}=2$ m=6$m=6$

 S. No. Equation Value Say, whether the Equation is satisfied. (Yes / No) 1 x+3=0$x+3=0$ x=3$x=3$ No 2 x+3=0$x+3=0$ x=0$x=0$ No 3 x+3=0$x+3=0$ x=−3$x=-3$ Yes 4 x−7=1$x-7=1$ x=7$x=7$ No 5 x−7=1$x-7=1$ x=8$x=8$ Yes 6 5x=25$5x=25$ x=0$x=0$ No 7 5x=25$5x=25$ x=5$x=5$ Yes 8 5x=25$5x=25$ x=−5$x=-5$ No 9 m3=2$\frac{m}{3}=2$ m=−6$m=-6$ No 10 m3=2$\frac{m}{3}=2$ m=0$m=0$ No 11 m3=2$\frac{m}{3}=2$ m=6$m=6$ Yes

Question 2.Check whether the value given in the brackets is a solution to the given equation or not:
(a) $n+5=19\left(n=1\right)$
(b) $7n+5=19\left(n=-2\right)$
(c) $7n+5=19\left(n=2\right)$
(d) $4p-3=13\left(p=1\right)$
(e) $4p-3=13\left(p=-4\right)$
(f) $4p-3=13\left(p=0\right)$

Answer: (a) $n+5=19\left(n=1\right)$

Putting $n=1$ in L.H.S.,

1 + 5 = 6

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $n=1$ is not the solution of given equation.

(b) $7n+5=19\left(n=-2\right)$

Putting $n=-2$ in L.H.S.,

$7\left(-2\right)+5=-14+5=-9$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $n=-2$ is not the solution of given equation.

(c) $7n+5=19\left(n=2\right)$

Putting $n=2$ in L.H.S.,

$7\left(2\right)+5=14+5=19$

$\because$ L.H.S. $=$ R.H.S.,

$\therefore$ $n=2$ is the solution of given equation.

(d) $4p-3=13\left(p=1\right)$

Putting $p=1$ in L.H.S.,

$4\left(1\right)-3=4-3=1$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p=1$ is not the solution of given equation.

(e) $4p-3=13\left(p=-4\right)$

Putting $p=-4$ in L.H.S.,

$4\left(-4\right)-3=-16-3=-19$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p=-4$ is not the solution of given equation.

(f) $4p-3=13\left(p=0\right)$

Putting $p=0$ in L.H.S.,

$4\left(0\right)-3=0-3=-3$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p=0$ is not the solution of given equation.

Question3.Solve the following equations by trial and error method:
(i) $5p+2=17$
(ii) $3m-14=4$

Answer: (i) $5p+2=17$

Putting $p=-3$ in L.H.S. $5\left(-3\right)+2$ = $-15+2=-13$

$\because$$-13\ne 17$ Therefore, $p=-3$ is not the solution.

Putting $p=-2$ in L.H.S. $5\left(-2\right)+2=$$-10+2=-8$

$\because$$-8\ne 17$ Therefore, $p=-2$ is not the solution.

Putting $p=-1$ in L.H.S. $5\left(-1\right)+2=$$-5+2=-3$

$\because$$-3\ne 17$ Therefore, $p=-1$ is not the solution.

Putting $p=0$ in L.H.S. $5\left(0\right)+2=$$0+2=2$

$\because$$2\ne 17$ Therefore, $p=0$ is not the solution.

Putting $p=1$ in L.H.S. $5\left(1\right)+2=$$5+2=7$

$\because$$7\ne 17$ Therefore, $p=1$ is not the solution.

Putting $p=2$ in L.H.S. $5\left(2\right)+2=$$10+2=12$

$\because$$12\ne 17$ Therefore, $p=2$ is not the solution.

Putting $p=3$ in L.H.S. $5\left(3\right)+2=$$15+2=17$

$\because$$17=17$ Therefore, $p=3$ is the solution.

(ii) $3m-14=4$

Putting $m=-2$ in L.H.S. $3\left(-2\right)-14=-6-14=-20$

$\because$$-20\ne 4$ Therefore, $m=-2$ is not the solution.

Putting $m=-1$ in L.H.S. $3\left(-1\right)-14=-3-14=-17$

$\because$$-17\ne 4$ Therefore, $m=-1$ is not the solution.

Putting $m=0$ in L.H.S. $3\left(0\right)-14=0-14=-14$

$\because$$-14\ne 4$ Therefore, $m=0$ is not the solution.

Putting $m=1$ in L.H.S. $3\left(1\right)-14=3-14=-11$

$\because$$-11\ne 4$ Therefore, $m=1$ is not the solution.

Putting $m=2$ in L.H.S. $3\left(2\right)-14=6-14=-8$

$\because$$-8\ne 4$ Therefore, $m=2$ is not the solution.

Putting $m=3$ in L.H.S. $3\left(3\right)-14=9-14=-5$

$\because$$-5\ne 4$ Therefore, $m=3$ is not the solution.

Putting $m=4$ in L.H.S. $3\left(4\right)-14=12-14=-2$