Linear Equations in One Variable - 8th math

Question (6)

Solution

Given,

After transposing to LHS, we get

After cross multiplication, we get

5m – 1 = 6 × 1

⇒ 5m – 1 = 6

After transposing –1 to RHS, we get

5m = 6 + 1 =7

⇒ 5m = 7

After transposing 5 to RHS, we get

⇒ m = 7/5 Answer

Checking of Result

Given,

Now, LHS = to LHS, we get

After substituting the value of m = 7/5, we get

Thus, LHS =

Now, RHS =

After substituting the value of m = 7/5, we get

Thus, RHS

Thus, LHS = RHS Proved

Simplify and solve the following linear equations.

Question (7) 3(t – 3) = 5(2t + 1)

Solution

Given, 3(t – 3) = 5(2t + 1)

⇒ 3t – 3 × 3 = 5 × 2t + 1× 5

⇒ 3t – 9 = 10t + 5

After transposing –9 to RHS, we get

⇒ 3t = 10t + 5 + 9

⇒ 3t = 10t + 14

After transposing 10t to LHS, we get

⇒ 3t – 10t = 14

⇒ – 7t = 14

After transposing 7 to RHS, we get

⇒ –t = 14/7 = 2

⇒ t = –2 Answer

Checking of Result

Given, 3(t – 3) = 5(2t + 1)

Now, LHS = 3(t – 3)

After substituting the value of t = –2, we get

3(–2 – 3)

= 3 ( –5)

Thus, LHS = –15

Now, RHS = 5(2t + 1)

After substituting the value of t = –2, we get

5 [2×(–2) + 1]

= 5(–4 + 1)

= 5(–3) = –15

Thus, RHS = –15

Thus, LHS = RHS Proved

Question (8) 15(y –4) – 2(y – 9) + 5(y + 6) = 0

Solution

Given, 15(y –4) – 2(y – 9) + 5(y + 6) = 0

After opening of brackets

⇒ 15y – 60 –2y + 18 + 5y + 30 = 0

After rearranging the above expression

⇒ 15y – 2y + 5 y – 60 + 18 + 30 = 0

⇒ 13y + 5 y – 60 + 48 = 0

⇒ 18y – 12 = 0

After transposing –12 to RHS, we get

18y = 12

After transposing 18 to RHS, we get

Answer

Checking of Result

Given, 15(y –4) – 2(y – 9) + 5(y + 6) = 0

Now, LHS = 15(y –4) – 2(y – 9) + 5(y + 6)

Thus, after substituting the value of y = 2/3, we get

Thus, LHS = RHS Proved

Question (9) 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Solution

Given, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

⇒ 15z – 21 – 18z + 22 = 32z – 52 – 17

After rearranging the above expression, we get

⇒ 15z – 18z – 21 + 22 = 32z – 69

⇒ –3z + 1 = 32z – 69

Now after transposing 32z to RHS, we get

–3z – 32z = –69 – 1

⇒ –35z = –70

After transposing –35 to RHS, we get

⇒ z = 2 Answer

Checking of Result

Given, 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Now, LHS = 3(5z – 7) – 2(9z – 11)

After substituting the value of z = 2, we get

3(5 × 2 – 7) – 2(9 × 2 –11)

= 3(10 – 7) – 2(18 – 11)

= 3 (3) – 2(7)

= 9 – 14

⇒ LHS = – 5

Now, RHS = 4(8z – 13) – 17

After substituting the value of z = 2, we get

4(8 × 2 – 13) – 17

= 4(16 – 13) – 17

= 4(3) – 17

= 12 – 17

⇒ RHS = –5

Thus, LHS = RHS Proved

Question (10) 0.25(4f – 3) = 0.05(10f – 9)

Solution

Given, 0.25(4f – 3) = 0.05(10f – 9)

⇒ 1 × f – 0.75 = 0.5f – 0.45

⇒ f – 0.75 = 0.5f – 0.45

Now after transfosing 0.5f to LHS, we get

⇒ f – 0.75 – 0.5f = – 0.45

Again after transposing –0.75 to RHS, we get

⇒ f – 0.5f = –0.45 + 0.75

⇒ 0.5f = 0.3

Now, after transposing 0.5 to RHS, we get

⇒ f = 0.3/0.5

⇒ f = 3/5 = 0.6

⇒ f = 0.6 Answer

Checking of Result

Given, 0.25(4f – 3) = 0.05(10f – 9)

Now, LHS = 0.25(4f – 3)

After substituting the value of f = 0.6, we get

0.25 (4 × 0.6 – 3)

= 0.25(2.4 – 3)

= 0.25 (–0.6)

= – 0.150

⇒ LHS = – 0.15

Now, RHS = 0.05(10f – 9)

After substituting the value of f = 0.6, we get

0.05 (10 × 0.6 – 9)

= 0.05 (6 – 9)

= 0.05 (–3)

⇒ RHS = – 0.15

Thus, LHS = RHS Proved