Question

solve this questions

Answer 1

Answers:

1. t = 0
2. x = 11

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Given equation - 1:

10 = 4 + 3(t+2)

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Solution:

10 = 4 + 3(t+2)

$\implies$ 10 = 4 + 3t + 6

$\implies$ 10 = 10 + 3t

$\implies$ 10-10 = 3t

$\implies$ 0 = 3t

$\implies$ 0/3 = t

$\implies {\dag {\boxed {\bf {\purple { 0 = t}}}}}$

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Verification:

Substitute the value of t as 0 in the equation,

10 = 4 + 3(t+2)

$\implies$ 10 = 4 + 3(0+2)

$\implies$ 10 = 4 + 3×2

$\implies$ 10 = 4+6

$\implies$ 10 = 10

$\implies$ LHS = RHS

Hence Verified

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Value of t:

Value of t is 0.

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Given equation - 2:

$\sf \dfrac {x-5}{3} - 4 = -2$

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Solution:

$\sf \dfrac {x-5}{3} - 4 = -2$

$\implies {\sf {\dfrac {x-5}{3}}}$ = -2 + 4

$\implies {\sf {\dfrac {x-5}{3}}}$ = 2

$\implies$ x-5 = 2×3

$\implies$ x-5 = 6

$\implies$ x = 6+5

$\implies {\dag {\boxed {\bf {\pink {x = 11}}}}}$

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Verification:

Substitute the value of x as 11 in the equation,

$\sf \dfrac {x-5}{3} - 4 = -2$

$\implies {\sf \dfrac {11-5}{3} - 4 = -2}$

$\implies {\sf \dfrac {6}{3} - 4 = -2}$

$\implies$ 2 - 4 = -2

$\implies$ -2 = -2

$\implies$ LHS = RHS

Hence Verified

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Value of x:

Value of x is 11.