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