In mixed form, it would be equal to 1 2/7
In short, Your Answer would be Option B
Hope this helps!
Answer:
Step-by-step explanation:
Solution by substitution method
3x-4y=8
and 18x-5y=10
Suppose,
3x-4y=8→(1)
and 18x-5y=10→(2)
Taking equation (1), we have
3x-4y=8
⇒3x=4y+8
⇒x=(
4y+8)/
3 →(3)
Putting x=
(4y+8
)/3 in equation (2), we get
18x-5y=10
18(
(4y+8)
/3) -5y=10
⇒24y+48-5y=10
⇒19y+48=10
⇒19y=10-48
⇒19y= -38
⇒y=-
38
/19
⇒y= -2→(4)
Now, Putting y=-2 in equation (3), we get
x=4y+8
x=
(4(-2)+8)
/3
⇒x=
(-8+8)/
3
⇒x=
0/
3
⇒x=0
∴x=0 and y= -2
Let's call the two numbers
and
.
Given these variables, we can say:
, based on the first sentence in the problem.
Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

To solve, we can simply substitute
in for
in the second equation and solve.


- Get terms on the left side to a common denominator for easier addition


- Cross multiplication (
)


- Subtract
from both sides of the equation

- Factor left side of the equation

Now, notice that we have found two solutions, but the problem is only asking for one. This <em>likely </em>means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,
,
Since
is not positive in this case,
is not a solution.
Thus,
is our only solution. In this case,
,
which means that the smaller number is 14 and the larger number is 28.