MP Board Class 7 Maths Solutions For Chapter 4 Simple Equations
1. Give the first step you will unr to separate the variable and then solve the equation :
1) x-1 -0
Solution:
Given equation isi – 1- 0
Adding 1 on both sides we get
x -1 +1-0 +1
x=1
2) x +1 =0
Solution:
Given equation is x + 1 -0
Subtracting ‘1’ from both sides we get
x+1-1 -0-1
x =-l
3) x-1 = 5
Solution:
Given equation is x-1 =5
Adding 1 on both sides we get
x-1 +1=5+1
x = 6
4) x + 6 = 2
Solution:
Given equation is x + 6 = 2
Subtracting ‘6’ from both sides we get
x+6-6= 2-6
x = – 4
Mp Board Class 7 Maths Solutions
5) y- 4 = -7
Solution:
Given equation is y – 4 = – 7
Adding 4 on both sides we get
y-4+4=-7+4
y = – 3
6) y – 4 = 4
Solution:
Given equation is y- 4 = 4
Adding 4 on both sides we get
y-4 + 4 = 4+4
y = 8
7) y + 4 = 4
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y+4-4=4-4
y=0
8) y + 4 = – 4
Solution:
Given equation is y + 4 = – 4
Subtracting 4 from both sides we get
y+4-4 = -4-4
y=-8
Class 7 Maths Chapter 4 Solutions Mp Board
2. Give the first step you will use to separate the variable and then solve the equation:
1) 3l = 42
Solution:
The given equation is 3l = 42
Dividing both sides by 3 we get
\( \frac{3l}{3}=\frac{42}{3} \)l = 14
2) \( \frac{b}{2}=6 \)
Solution: Given equation is \( \frac{b}{2}=6 \)
Multiplying both sides by 2 we get
\( \frac{b}{2} \times 2=6 \times 2 \)b = 12
3) \( \frac{p}{7}=4 \)
Solution: Given equation is \( \frac{p}{7}=4 \)
Multiplying both sides by 7 we get
\( \frac{p}{7} \times 7=4 \times 7 \)p = 28
Mp Board Maths Chapter 4 Solutions
4) 4x = 25
Solution: Given equation is 4x = 25
Dividing both sides by 4 we get
\( \frac{4 x}{4}=\frac{25}{4} \) \( x=\frac{25}{4} \)5) 8y = 36
Solution:
Given equation is 8y 36
Dividing both sides by 8 we get
\( \frac{8 y}{8}=\frac{36}{8} \) \( y=\frac{36}{8}=\frac{36+4}{8+4}=\frac{9}{2} \) \( y=\frac{9}{2} \)6) \( \frac{z}{3}=\frac{5}{4} \)
Solution:
Given equation is \( \frac{z}{3}=\frac{5}{4} \)
Multiplying both sides by 3 we get
\( \frac{z}{3} \times 3=\frac{5}{4} \times 3 \Rightarrow z=\frac{5 \times 3}{4} \) \( z=\frac{15}{4} \)Mp Board Class 7 Book Solutions
7) \( \frac{a}{5}=\frac{7}{15}\)
Solution:
Given equation is \( \frac{a}{5}=\frac{7}{15} \)
\( \begin{aligned}& \frac{a}{5} \times 5=\frac{7}{15} \times 5 \\
& =\frac{35}{15}=\frac{35+5}{15+5}=\frac{7}{3}
\end{aligned} \) \( a=\frac{7}{3} \)
8) 20t = – 10
Solution:
Given equation is 20t = -10
Dividing both sides by 20 we get
\( \frac{20 \mathrm{t}}{20}=\frac{-10}{20} \Rightarrow t=\frac{-10}{20}=\frac{-1}{2} \) \( t=\frac{-1}{2} \)3. Give the steps you will use to separate the variable and then solve the equation:
1) 3n – 2 = 46
Solution: Given equation is 3n – 2 = 46
Adding 2 on both sides we get
3n -2 + 2 = 46 + 2
3n = 48
Dividing both sides by 3 we get
\( \frac{3 n}{3}=\frac{48}{3} \Rightarrow 9 n=144 \)n = 16
2) 5m + 7 = 17
Solution:
Given equation is 5m + 7 = 17
Subtracting 7 from both sides we get
5m + 7-7 = 17-7
=> 5m = 10
Dividing both sides by 5 we get
\( \frac{5 m}{5}=\frac{10}{5} \)25m = 50
\( \Rightarrow m=\frac{50}{25} \)m = 2
Class 7 Mp Board Maths Question Answers
3. \( \frac{20 p}{3}=40 \)
Solution:
Given equation is \( \frac{20 p}{3}=40 \)
Multiplying both sides by 3 we get
\( \frac{20 p}{3} \times 3=40 \times 3 \)20p = 120
Dividing both sides by 20 we get
\( \frac{20 p}{3}=\frac{120}{20} \Rightarrow 400 \mathrm{p}=2400 \)p = 6
4. \( \frac{3 p}{10}=6 \)
Solution:
Given equation is \( \frac{3 p}{10}=6 \)
Multiplying both sides by 10 we get
\( \frac{3 p}{10} \times 10=6 \times 10 \)3p = 60
Dividing both sides by 3 we get
\( \frac{3 p}{10}=\frac{60}{3} \Rightarrow 9 p=180 \)p = 20
4. Solve the following equations:
1)10p = 100
Solution:
Given equation is 10p = 100
Dividing both sides by 10 we get
\( \frac{10 p}{10}=\frac{100}{10} \quad \Rightarrow 100 p=1000 \) \( p=\frac{1000}{100} \)p = 10
2) 10p + 10 = 100
Solution:
Given equation is 10p+10 = 100
Subtracting 10 from both sides we get
10p + 10 – 10 = 100 – 10
10p = 90
Dividing both sides by 10 we get
\( \frac{10 p}{10}=\frac{90}{10} \)=> 100p =900
\( p=\frac{900}{100} \)p= 9
Mp Board Class 7 Maths Solutions
3) \( \frac{p}{4}=5 \)
Solution:
Given equation is \( \frac{p}{4}=5 \)
Multiplying both sides by 4 we get
\( \frac{p}{4} \times 4=5 \times 4 \)p =20
4) \( \frac{-p}{3}=5 \)
Solution:
Given equation is \( \frac{-p}{3}=5 \)
Multiplying both sides by -3 we get
\( \left(\frac{-p}{3}\right) \times(-3)=5(-3) \)p = – 15
5) \( \frac{3 p}{4}=6 \)
Solution:
Given equation is \( \frac{3 p}{4}=6 \)
Multiplying both sides by 4 we get
\( \frac{3 p}{4} \times 4=6 \times 4 \)3p = 24
Dividing both sides by 3 we get
\( \frac{3 p}{3}=\frac{24}{3} \)9p = 72
p = 8
6) 3s = – 9
Solution:
Given equation is 3s = -9
Dividing both sides by 3 we get
\( \frac{3 s}{3}=\frac{-9}{3} \)9s = – 27
s = – 3
Mp Board Maths Chapter 4 Solutions
7) 3s + 12 = 0
Solution:
Given equation is 3s + 12 = 0
Substracting 12 from both sides we get
3s + 12- 12 = 0 – 12
=> 3s =- 12
Dividing both sides by 3 we get
\( \frac{3 s}{3}=\frac{-12}{3} \)=> 9s = -36
s = – 4
8) 3s = 0
Solution:
Given equation is 3s = 0
Dividing both sides by 3 we get
\( \frac{3 s}{3}=\frac{0}{3} \)=> 9s = 0
s= 0
9) 2q = 6
Solution:
Given equation is 2q =6
Dividing both sides by 2 we get
\( \frac{2 q}{2}=\frac{6}{2} \)4q = 12
q = 3
10) 2q – 6 = 0
Solution:
Given equation is 2q- 6 = 0
Adding 6 on both sides we get
2q-6 + 6 = 0 + 6
2q = 6
Dividing both sides by 2 we get
\( \frac{2 q}{2}=\frac{6}{2} \)=4q = 12
q= 3
11) 2q + 6 = 0
Solution:
Given equation is 2q + 6 = 0
Subtracting 6 from both sides we get
2q +6-6 = 0-6
=> 2q= – 6
Dividing both sides by 2 we get
\( \frac{2 q}{2}=\frac{-6}{2} \)4q = -12
q = -3
Mp Board Maths Chapter 4 Solutions
12) 2q + 6 = 12
Solution:
Given equation is 2q + 6 = 12
Subtracting 6 from both sides we get
2q + 6-6 = 12 -6
=> 2q = 6
Dividing both sides by 2 we get
\( \frac{2 q}{2}=\frac{6}{2} \)4q = 12
q = 3