AP State Syllabus AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 2 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 2

Question 1.

Solve the following equations without transposing and check your result.

(i) x + 5 = 9

(ii) y – 12 = -5

(iii) 3x + 4 = 19

(iv) 9z = 81

(v) 3x + 8 = 5x + 2

(vi) 5y + 10 = 4y – 10

Solution:

(i) x + 5 = 9

Solution:

i) x + 5 = 9

x + 5 – 5 9 – 5 (subtract 5 from both sides)

x = 4

Check

LHS = x + 5

(substituting x = 4)

= 4 + 5 = 9

RHS = 9

∴ L.H.S = R.H.S

ii) y – 12 = – 5

y – 12 = – 5

y – 12 + 12= – 5 + 12 (add l2onbothsides)

y = 7

Check

LHS = y – 12

= 7 – 12= – 5

RHS = -5

∴ L.H.S = R.H.S

iii) 3x+4= 19

3x + 4 = 19

3x + 4 – 4 = 19 – 4

(subtract 4 from both sides)

3x = 15

\(\frac{3 x}{3}=\frac{15}{3}\) (Divide both sides by3)

x = 5

Check

LHS = 3x + 4

= 3 x 5 + 4

= 15 + 4 = 19

RHS = 19

∴ L.,H.S = R.H.S

iv) 9z = 81

\(\frac{9 z}{9}=\frac{81}{9}\) (Divide both sides by 9)

z = 9

Check

LHS = 9z = 9 x 9 = 81

RHS = 81

∴ LHS = RHS

v) 3x + 8 = 5x + 2

3x + 8 = 5x + 2

3x + 8 – 8 = 5x + 2 – 8

(Adding -8 on both sides)

3x = 5x – 6

3x – 5x = 5x – 6 – 5x

(Subtract 5x from both sides)

-2x = -6

\(\frac{-2 x}{-2}=\frac{-6}{-2}\)(Divide both sides by -2)

x = 3

Check

LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17

RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17

∴ LHS = RHS

(vi) 5y + 10 = 4y – 10

5y + 10 = 4y – 10

5y + 10 – 1o = 4y – 10 – 10

(Subtract 10 from both sides)

5y =4y – 20

5y – 4y = 4y – 20 – 4y

(Substract ty from both sides)

y = – 20

Check

LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90

RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90

∴ LHS = RHS

Question 2.

Solve the following equations by transposing the terms and check your result.

(i) 2 + y = 7

Solution:

y= 7 – 2 (transposlng+2)

y = 5

Check:

LHS = 2 + y = 2 + 5 = 7

RHS = 7

∴ L.H.S = R.H.S

(ii) 2a – 3 = 5

2a – 3 = 5

2a = 5 + 3 (transposing – 3)

2a = 8

(transposing x 2)

a = 4

Check

LHS = 2a – 3 = 2 x 4 – 3 = 8 – 3 = 5

RHS =5

∴ L.H.S = R.H.S

(iii) 10 – q = 6

10 – q = 6

– q = 6 – 10(transposing + 10)

– q – 4

q = \(\frac{-4}{-1}\) = 4 (transposing x ( – 1)

Check

LHS= 10 – q= 10 – 4= 6

RHS = 6

∴ L.H.S = R.H.S

(iv) 2t – 5 = 3

2t – 5 = 3

2t – 5 = 3 (transposing – 5)

2t = 3 + 5

2t = 8

t = \(\frac{8}{2}\) (transposing x (2))

Check

LHS=2t – 5= 2 x 4 – 5 = 8 – 5 = 3

RHS = 3

∴ L.H.S = R.H.S

(v) 14 = 27 – x

14 = 27 – x

0 = 27 – x – 14 (transposing + 14)

0 = 13 – x (transposIng – x)

x = 13

Check

LHS = 14

RHS = 27 – x = 27 – 13 = 14

∴ L.H.S = R.FIS

(vi) 5(x + 4) = 35

5(x + 4) = 35

x + 4 = \(\frac{35}{5}\) (lransposingx5)

x + 4 = 7

x = 7 – 4 (transposing + 4)

x = 3

Check

LHS = 5(3 + 4) = 5 x 7 = 35

RHS = 35

∴ L.H.S = R.H.S

(vii) -3x = 15

– 3x= 15

x = \(\frac{15}{-3}\) (transposingx( – 3))

x= – 5

Check

LHS = – 3x = -3x( – 5)= 15

RHS= 15

∴ L.H.S = R.H.S

(viii) 5x – 3 = 3x – 5

5x – 3 = 3x – 5

5x = 3x – 5 + 3(transposing – 3)

5x = 3x – 2

5x – 3x = – 2(trarisposing + 3x)

2x = – 2

x = \(\frac{-2}{2}\) (transposingx2)

x= – 1

C heck

LHS = 5x – 3 = 5x( – 1) – 3 = – 5 – 3 = – 8

RHS = 3x – 5 = 3x( – 1) – 5 = – 3 – 5 = – 8

∴ L.H.S = R.H.S

(ix) 3y + 4 = 5y – 4

3y + 4 = 5y – 4

3y = 5y – 4 – 4 (transposing + 4)

3y – 5y = – 8 (transposing + 5y)

– 2y = – 8

y = \(\frac{-8}{-2}\) =4 (transposingx( – 2)

y = 4

Check

LHS = 3y + 4 = 3 x (4) + 4 = 12 + 4 = 16

RHS = 5y – 4 = 5 x (4) – 4 = 20 – 4= 16

∴ L.H.S = R.H.S

(x) 3(x-3)=5(2x + 1)

3(x – 3)=5(2x+ 1)

3(x – 3)= 5(2x ÷ 1)

3x – 9= 10x+5

3x = 10x + 5 ÷ 9 (transposing – 9)

3x = 10x + 14

3x – 10x = 14(transposing+ lOx)

– 7x =14

x = \(\frac{14}{-7}\)(transposing x ( – 7))

x = – 2

Check

LHS = 3(x – 3) = 3[( – 2) – 3] = 3x( – 5) = – 15

RHS = 5(2x + 1) = 5 x [2( – 2) + 1] = 5 x [ – 4 + 1]

= 5 x ( – 3)= – 15

∴ LH.S = R.H.S