# AP Board 7th Class Maths Solutions Chapter 3 Simple Equations Ex 2

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
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