Subtract as per given instruction from horizontal method of subtraction: (a) 3x from 5x (b) -7a from -3a (c) 5a + 6 from 4a + 3 (d) 3p + q + 3c from 5p + 4q + 7c
Answers
Answer:
The required difference =(3a–7b+4c)−(5a+7b−2c)
=3a–5a–7b–7b+4c+2c
=−2a–14b+6c
Step-by-step explanation:
Ncert solutions
Grade 7
Mathematics
Science
Chapters in NCERT Solutions - Mathematics, Class 7
Exercises in Simple Equations
Question 2
Q2) Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
(b) \frac{b}{2}=\ 6
2
b
= 6
(c) \frac{p}{7}=4
7
p
=4
(d) 4x = 25
(e) 8y = 36
(f) \frac{z}{3\ }=\ \frac{5}{4}
3
z
=
4
5
(g) \frac{a}{5\ }=\ \frac{7}{15}
5
a
=
15
7
(h) 20t = -10
Solution 2:
(a) 3l = 42
(dividing both sides by 3 )
\frac{3}{3}l\ =\ \frac{42}{3}
3
3
l =
3
42
l = 14
(b) \frac{b}{2}\ =\ 6\
2
b
= 6
(multipling both side by 2 )
\frac{b}{2}\times2\ =\ 6\times2
2
b
×2 = 6×2
b = 12
(c) \frac{p}{7}=4
7
p
=4
(multiplying both sides by 7)
p = 28
(d) 4x = 25
(dividing both sides by 4 )
\frac{4x}{4}=\ \frac{25}{4}
4
4x
=
4
25
x\ =\ \frac{25}{4}x =
4
25
(e) 8y = 36
(dividing both sides by 8 )
\frac{8}{8}y\ =\ \frac{36}{8}
8
8
y =
8
36
y\ =\ \frac{9}{2}y =
2
9
(f) \frac{z}{3\ }=\ \frac{5}{4}
3
z
=
4
5
(multipling by 3 on the both sides)
\frac{z}{3}\times3\ =\ \frac{5}{4}\times3
3
z
×3 =
4
5
×3
z\ =\ \frac{15}{4}z =
4
15
(g) \frac{a}{5\ }=\ \frac{7}{15}
5
a
=
15
7
(multiplying both sides by 5 )
\frac{a}{5}\times5\ =\ \frac{7}{15\ }\times3
5
a
×5 =
15
7
×3
a\ =\ \frac{7}{3}a =
3
7
(h) 20 t = -10
(dividing both sides by 20 )
\frac{20}{20}t\ =\ -\frac{10}{20}
20
20
t = −
20
10
t\ =\ -\frac{1}{2}t = −
2
1