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

Exercise: 4.3

1. Solve the following equations:

a) \[2y+\frac{5}{2}=\frac{37}{2}\]

Ans:     

       \[\Rightarrow 2y=\frac{37}{2}-\frac{5}{2}=\frac{37-5}{2}=\frac{32}{2}=16\]

       \[\Rightarrow 2y=16\]

       \[\Rightarrow y=\frac{16}{2}=8\]

       \[\Rightarrow y=8\]

b) \[5t+28=10\]

Ans:

       \[\Rightarrow 5t=10-28=-18\]

       \[\Rightarrow 5t=-18\]

       \[\Rightarrow t=\frac{-18}{5}\]

c) \[\frac{a}{5}+3=2\]

Ans:

      \[\Rightarrow \frac{a}{5}=2-3=-1\]

      \[\Rightarrow \frac{a}{5}=-1\]

     \[\Rightarrow a=-1\times 5=-5\]

     \[\Rightarrow a=-5\]

d) \[\frac{q}{4}+7=5\]

Ans:

     \[\Rightarrow \frac{q}{4}=5-7=-2\]

     \[\Rightarrow \frac{q}{4}=-2\]

     \[\Rightarrow q=-2\times 4=-8\]

     \[\Rightarrow q=-8\]

e) \[\frac{5}{2}x= -10\]

Ans:     

        \[\Rightarrow x= -10\times \frac{2}{5}=-4\]

         \[\Rightarrow x=-4\]

f) \[\frac{5}{2}x=\frac{25}{4}\]

Ans:   

     \[\Rightarrow x=\frac{25}{4}\times \frac{2}{5}=\frac{5}{2}\]

      \[\Rightarrow x=\frac{5}{2}\]

g) \[7m+\frac{19}{2}=13\]

Ans:    

      \[\Rightarrow 7m=13-\frac{19}{2}=\frac{13\times     2-19}{2}=\frac{26-19}{2}=\frac{7}{2}\]

      \[\Rightarrow 7m=\frac{7}{2}\]

      \[\Rightarrow m=\frac{7}{2}\times \frac{1}{7}=\frac{1}{2}\]

h) \[6z+10=-2\]

Ans:    

      \[\Rightarrow 6z=-2-10=-12\]

      \[\Rightarrow 6z=-12\]

      \[\Rightarrow z=\frac{-12}{6}=-2\]

      \[\Rightarrow z=-2\]

i) \[\frac{3l}{2}=\frac{2}{3}\]

Ans:    

      \[\Rightarrow 3l=\frac{2\times 2}{3}=\frac{4}{3}\]

      \[\Rightarrow 3l=\frac{4}{3}\]

      \[\Rightarrow l=\frac{4}{3\times 3}=\frac{4}{9}\]

      \[\Rightarrow l=\frac{4}{9}\]

j) \[\frac{2b}{3}-5=3\]

Ans:    

      \[\Rightarrow \frac{2b}{3}=3+5=8\]

      \[\Rightarrow \frac{2b}{3}=8\]

      \[\Rightarrow 2b=8\times 3=24\]

      \[\Rightarrow 2b=24\ \ \ \ \ \]

      \[\Rightarrow b=\frac{24}{2}=12\]

      \[\Rightarrow b=12\]

2. Solve the following equations:

a) \[2\left( x+\text{ }4 \right)=12~\]

Ans:    

\[\Rightarrow x+4=\frac{12}{2}=6\]

\[\Rightarrow x=6-4=2\]

\[\Rightarrow x=2\]

b) \[3\left( n-5 \right)=21\]

Ans:    

\[\Rightarrow n-5=\frac{21}{3}=7\]

\[\Rightarrow n=7+5=12\]

c) \[3\left( n-5 \right)=-21\text{ }\]

Ans: 

\[\Rightarrow n-5=\frac{21}{3}=-7\]

\[\Rightarrow n=-7+5=-2\]

\[\Rightarrow n=-2\]

d) \[-4\left( 2+x \right)=8\]

Ans:    

\[\Rightarrow -4\times 2-\left(  4\times x \right)\text{=8}\]

\[\Rightarrow -8-4x=8\]

\[\Rightarrow -4x=8+8=16\]

 \[\Rightarrow x=\frac{-16}{4}=-4\]

e) \[4\left( 2-x \right)=8\]

Ans:    

 \[\Rightarrow 4\times 2-\left(  4\times x \right)\text{=8}\]

\[\Rightarrow 8-4x=8\]

\[\Rightarrow -4x=8-8\]

\[\Rightarrow x=0\]

3. Solve the following equations:

a) \[4=5\left( p-2 \right)\]

Ans:    

      \[\Rightarrow 4=5\times p-5\times 2\]

      \[\Rightarrow 4=5p-10\]

      \[\Rightarrow 5p=4+10\]

      \[\Rightarrow 5p=14\]

      \[\Rightarrow p=\frac{14}{5}\]

b) \[-4=5\left( p-2 \right)\]

Ans:    

       \[\Rightarrow -4=5\times p-5\times 2\]

       \[\Rightarrow -4=5p-10\]

       \[\Rightarrow 5p=-4+10\]

       \[\Rightarrow 5p=6\]

       \[\Rightarrow p=\frac{6}{5}\]

c) \[16=4+3\left( t+2 \right)\]

Ans:    

       \[\Rightarrow 16-4=3\left( t+2 \right)\]    

       \[\Rightarrow 12=3\left( t+2 \right)\]

       \[\Rightarrow \frac{12}{3}=\left( t+2 \right)\]

       \[\Rightarrow 4-2=t\]

       \[\Rightarrow t=2\]

d) \[4+5\left( p-1 \right)=34\]

Ans:    

       \[\Rightarrow 4+5\times p-5\times 1 = 34\]

       \[\Rightarrow 4+5p-5=34\]

       \[\Rightarrow 5p=34+1\]

       \[\Rightarrow 5p=35\]

       \[\Rightarrow p=\frac{35}{5}\]

       \[\Rightarrow p=7\]

e) \[0=16+4\left( m-6 \right)\]

Ans:    

     \[\Rightarrow 0=16+4\left( m-6 \right)\]

     \[\Rightarrow \frac{-16}{4}=\left( m-6 \right)\]

     \[\Rightarrow -4=m-6\]

     \[\Rightarrow -4+6=m\]

      \[\Rightarrow m=2\]

4. 

a) Construct 3 equations starting with \[x=2\].

Ans: The following equation can be solving in following steps;

Step I: 

\[x=2\]  

Multiplying both sides by 10,

\[10x=20\]                  

Adding 2 both sides

\[10x+2=20+2=22\]

\[10x+2=22\]

Step II:

Multiplying both sides by 5

\[x=2\]

\[5x=10\]

Subtracting 3 from both sides

\[5x-3=10-3=7\]

\[5x-3=7\]

Step III:

Dividing both sides by 5

\[x=2\]

\[\frac{x}{5}=\frac{2}{5}\]

b) Construct 3 equations starting with \[x=-2\].

Ans: The following equation can be solving in following steps;

Step I: 

\[x=-2\]

Multiplying both sides by 3

\[3x=-6\]

Step II:

Multiplying both sides by 3

\[x=-2\] 

\[3x=-2\times 3=-6\]

\[3x=-6\]

Adding 7 to both sides

\[3x+7=-6+7=1\]

\[3x+7=1\]

Step III:

Multiplying both sides by 3 

\[x=-2\]

\[3x=-2\times 3=-6\]

\[3x=-6\]

Adding 10 to both sides

\[3x+10=-6+10\]

\[3x+10=4\]

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

Opting for the NCERT solutions for Ex 4.3 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.3 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 4 Exercise 4.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 4 Exercise 4.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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