Ncert solutions

# Question 1

Q1) Solve the following equations:

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

(b) 5t + 28 = 10

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

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

(e) \frac{5}{2}x=-5

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

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

(h) 6z + 10 = –2

(i) \frac{3l}{2}=\ \frac{2}{3}

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

Solution 1:

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

2y\ =\ \frac{37}{2}-\ \frac{5}{2}

2y\ =\ \frac{37-5}{2}

2y\ =\ \frac{32}{2}

2y = 16

y = 8

(b) 5t+28 =10

5t = 10-28

5t = -18

t\ =\ -\frac{18}{5}

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

\frac{a}{5}=\ 2-3

\frac{a}{5}=\ -1

a = -5

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

\frac{q}{4}=\ 5-7

\frac{q}{4}=\ -2

q\ =\ -2\times4

q= -8

(e) \frac{5}{2}x=-5

5x\ =\ -5\times2

5x = -10

x= -2

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

5x\ =\ \frac{25}{4}\ \times\ 2\

5x\ =\ \frac{25}{2}

x=\ \frac{25}{2\times5}

x=\ \frac{5}{2}

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

7m\ =\ 13-\ \frac{9}{2}

7m\ =\ \frac{26\ -\ 19}{2}

7m\ =\ \frac{7}{2}

m\ =\ \frac{7}{2\times7}

m\ =\ \frac{1}{2}

(h) 6z + 10 = -2

6z= -2-10

6z = -12

z\ =\ -\frac{12}{6}

6z = -2

(i) \frac{3l}{2}=\ \frac{2}{3}

3l\ =\ \frac{2}{3}\times2

l\ =\ \frac{4}{3\times3}

l\ =\ \frac{4}{9\ }

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

\frac{2b}{3}=\ 3+5

\frac{2b}{3}=\ 8

2b\ =\ 8\times3

b\ =\ \frac{24}{2}

b= 12

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