Question

The sum of a number and 3 times a second number is represented by a+3b=39. The sum of half of the first number and 1/3 of the second number is represented by 12a+13b=9.

What is the first number (a)?
a =
What is the second number (b)?
b =
Please I need help I keep getting something unreasonable

Answer 1

you can set the first equation by a, that would be
a = 39 - 3b
then you plug in 39 - 3b into 'a' on the next equation
1/2(39 - 3b) + 1/3b = 9
simplify it
19.5 - 1.5b + 1/3b = 9
-10.5 = -7/6b
b = 9
plug in 9 in the first equation
a + 3 (9) = 39
a = 39 - 27
a = 12

FINAL
a = 12
b = 9

Answer 2

Answer: The first number is a = 12 and the second number is b = 9.

Step-by-step explanation: Given that the first number a and second number b is related to each other by the following system of equations :

`a+3b=39~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\\\(1)/(2)a+(1)/(3)b=9~~~~~~~~~~~~~~~~~~~~~~(ii)`

We are to find the first and second number, i.e., a and b.

From equation (i), we have

`a+3b=39\\\\\Rightarrow a=39-3b~~~~~~~~~~~~~~~~~~~~~`(iii)`

Substituting the value of a from equation (iii) in equation (ii), we get

`(1)/(2)*(39-3b)+(1)/(3)b=9\\\\\\\Rightarrow (3(39-3b)+2b)/(6)=9\\\\\Rightarrow 117-9b+2b=9*6\\\\\Rightarrow 117-7b=54\\\\\Rightarrow 7b=117-54\\\\\Rightarrow 7b=63\\\\\Rightarrow b=(63)/(7)\\\\\Rightarrow b=9.`

Putting the value of b in equation (iii) we get

`a=39-3*9=39-27=12.`

Thus, the first number is a = 12 and the second number is b = 9.