First, let us restate the given conditions
c is 5 more than variable a ( c = a + 5)
c is also three less than variable a (c = a - 3)
Now, lets look at the answer choices and or given
c = a − 5
c = a + 3
Here, c is 5 less than "a"...so automatically disqualified
a = c + 5
a = 3c − 3
Here, we have to get "C" by itself in both top and bottom equation.
So,
Simplified version :
c = a - 5
Here, c is 5 less than "a"...so automatically disqualified
a = c − 5
a = 3c + 3
Here also, we have to get "C" by itself in both top and bottom equation.
So,
simplified version:
c = a + 5
Here, c is 5 more than "a"...so we continue
c = (a - 3) / 3
Here, c is 3 less than "a" <u>divided by 3</u><u /> . So, this is not correct
c = a + 5
c = a − 3
Here, c is 5 more than "A"
Also, c is 3 less than "a"
Which satisfies the given.
So, our answer is going to be the last one:
c = a + 5
c = a - 3
The greatest common factor is 2
We first have to look for the largest number that goes into both equations. The factors of 12 are 1, 2, 3, 4, 6 and 12. None of 3, 4, 6, or 12 go into 26 evenly. So 2 is the largest number you can take out.
With the variables, we take out as many as the lowest number will let us. Since the smallest number of n's is 2 in the second term, we take that many.
X = 4y
3x - y = 70
substitute the x in the second equation with the other half of the first equation.
3(4y) - y = 70
12y - y = 70
11y = 70
y = 6.36 repeating, or 6 4/11
now use this to solve first equation
x = 4 (6 4/11)
x = 25.45 repeating
done
