Question

Solve for a when b = 4 and c = 5
6(a + 2) + 2c = b ^ 2 - a + 27
Please tell that how it came the full process of it.​

Answer 1

$\large\sf\fbox\orange{Solution}$

6(a+2) + 2c = b² - a + 27

(Put the value of b = 4 and c = 5)

6(a+2) +2(5) = (4)² - a+27

6a + 12 + 10 = 16 - a + 27

6a + 22 = -a + 43

6a = -a + 43 - 22

6a = -a + 21

6a + a = 21

7a = 21

$a = \frac{21}{7}$

a = 3

$\huge\sf\fbox\red{Be~Brainly}$

Answer 2

Given :  6(a + 2) + 2c = b ² - a + 27

To Find :  Value of a  when b = 4 and c = 5

Solution:

 6(a + 2) + 2c = b ² - a + 27

Substitute b = 4 and c = 5 in the equation

=> 6*(a + 2) + 2(5) = 4² - a  + 27

=> 6a + 12 + 10  = 16 - a + 27

=> 6a + 22 = 43 - a

=> 6a + a = 43 - 22

=> 7a = 21

=> a= 3

Value of a is 3 when  b = 4 and c = 5

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