Simple Equations - Solutions

 CSBE Class –VII Mathematics

NCERT Solutions
Simple Equations (Ex. 4.1)

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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$m=6$

Answer:

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

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

Answer: (a) n+5=19(n=1)$n+5=19\left(n=1\right)$

Putting n=1$n=1$ in L.H.S.,

1 + 5 = 6

∵$\because$ L.H.S. ≠$\ne$ R.H.S.,

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

(b) 7n+5=19(n=−2)$7n+5=19\left(n=-2\right)$

Putting n=−2$n=-2$ in L.H.S.,

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

∵$\because$ L.H.S. ≠$\ne$ R.H.S.,

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

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

Putting n=2$n=2$ in L.H.S.,

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

∵$\because$ L.H.S. =$=$ R.H.S.,

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

(d) 4p−3=13(p=1)$4p-3=13\left(p=1\right)$

Putting p=1$p=1$ in L.H.S.,

4(1)−3=4−3=1$4\left(1\right)-3=4-3=1$

∵$\because$ L.H.S. ≠$\ne$ R.H.S.,

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

(e) 4p−3=13(p=−4)$4p-3=13\left(p=-4\right)$

Putting p=−4$p=-4$ in L.H.S.,

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

∵$\because$ L.H.S. ≠$\ne$ R.H.S.,

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

(f) 4p−3=13(p=0)$4p-3=13\left(p=0\right)$

Putting p=0$p=0$ in L.H.S.,

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

∵$\because$ L.H.S. ≠$\ne$ R.H.S.,

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

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

Answer: (i) 5p+2=17$5p+2=17$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(ii) 3m−14=4$3m-14=4$

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

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

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

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

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

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

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

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

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

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

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

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

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