If 2a + b = 13 and 3c-6a = 5, what is the value of b + c?

Step-by-step explanation:

Solve the first eqn for b, b=13–2*a, and the 2nd for c, c=(6*a+5)/3. Then

b+c=13–2*a+(6*a+5)/3=13–2*a+2*a+5/3=44/3. We see that b+c has a definite value because the terms in a cancel.

44/3 (a fraction)

Step-by-step explanation:

If 2a + b = 13. Equation 1 and

3c – 6a = 5. Equation 2.

From Equation 1.

2a + b = 13 so

2a = 13 - b so

a = (1/2)(13 - b). Equation 3.

Substitute a from Equation 3 into Equation 2:

From Equation 2:

3c - 6a = 5 or

3c - 6[ (1/2)(13 - b) ] = 5 or

3[ c - 2(1/2)( 13 - b ) = 5 so

3[ c - ( 13 - b ) ] = 5

Expanding:

c + b - 13 = ( 5/3)

Therefore...

b + c = ( 5/3 ) + 13 or

b + c = [ (13)(3) + 5 ]/( 3 ) so

b + c = ( 44 )/( 3 ) = ( 44/3 )