Question

solve this question please​

Answer 1

$\huge{\mathcal{\underline{\pink{Solution:-}}}}$

$12. \: \: 3(t - 3) = 5(2t - 1) \\ \\ = > 3t - 9 = 10t - 5 \\ \\ = > 7t = - 4 \\ \\ = > t = \frac{ - 4}{7} \\ \\ \\ 13. \: \: \: \frac{5(1 - x) + 3(1 + x)}{1 - x} = 8 \\ \\ = > 5 - 5x + 3 + 3x = 8 - 8x \\ \\ = > 8 - 2x = 8 - 8x \\ \\ = >6 x = 0 \\ \\ = > x = 0$

Answer 2

12) Solve 3(t - 3) = 5(2t - 1)

⇒ 3 × (t) - 3 × (3) = 5 × (2t) - 5 × (1)

⇒ 3t - 9 = 10t - 5

⇒ 3t - 10t =  - 5 + 9

[Taking 10t to LHS and (-9) to RHS]

⇒ -7t = 4

⇒ t = $\boxed{ \frac{-4}{7}}$

Verification:

LHS = 3(t - 3)

$= 3(\frac{-4}{7} - 3)$

$= 3(\frac{-4-21}{7})\ [ \therefore Taking\ LCM = 7]$

$= \boxed{\frac{-75}{7}}$

RHS = 5(2t - 1)

$= 5(2 \times \frac{-4}{7} - 1)$

$= 5( \frac{-8}{7} - 1)$

$= 5( \frac{-8-7}{7})\ [ \therefore Taking\ LCM = 7]$

$= 5( \frac{-15}{7})$

$=\boxed{ \frac{-75}{7}}$

Hence, LHS = RHS (verified)

.

13) Solve $\bf \frac{5(1-x)+3(1+x)}{1-x}=8$

$\implies \frac{(5 \times 1 - 5 \times x) + (3 \times 1 + 3 \times x)}{1-x} = 8$

$\implies \frac{(5 - 5x) + (3 + 3x)}{1-x} = 8$

$\implies (5 - 5x) + (3 + 3x) = 8(1-x)$

[Taking (1-x) to RHS]

$\implies 5+3 -5x+3x=8-8x$

$\implies 8 -2x=8-8x$

$\implies 8x -2x=8-8$

$\implies 6x =0$

$\implies \boxed{x =0}$

Verification:

LHS = $\frac{5(1-0)+3(1+0)}{1-0}$

$= \frac{5(1)+3(1)}{1}$

$= 5+3 = 8 = RHS$

Hence, RHS = LHS

Verifired.