Hello. you shouldn't have people do your work full at a time.
The first number is x
The second number is y
So according to the given condition the sum of these two numbers is 84, so we can write
x+y=84 ————-(1)
Again one of them is more than 12 than the other that's why we can write,
x+12=y -------------(2)
Now putting the values of y in equation (1) we get,
x+x+12=84
=>2x+12=84
=>2x+12–12=84–12
=>2x=72
=>2x/2=72/2
=>x=36
Taking the value of x in equation(2) we get,
36+12=y
=>y=48
The two numbers are 48 and 36.
Check:
48+36=84
=>84=84
Also 48 =36+12; which is more than 12 of 36.
So
18*6=108
4*6=24
108-24=84
so the answer is 84
i hope it's correct
Answer:
![\huge\boxed{j=38}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7Bj%3D38%7D)
Step-by-step explanation:
![\dfrac{j}{-2}+7=-12\qquad\text{subtract 7 from both sides}\\\\\dfrac{j}{-2}+7-7=-12-7\\\\\dfrac{j}{-2}=-19\qquad\text{multiply both sides by (-2)}\\\\(-2)\!\!\!\!\!\!\!\diagup\cdot\dfrac{j}{-2\!\!\!\!\!\diagup}=(-2)(-19)\\\\j=38](https://tex.z-dn.net/?f=%5Cdfrac%7Bj%7D%7B-2%7D%2B7%3D-12%5Cqquad%5Ctext%7Bsubtract%207%20from%20both%20sides%7D%5C%5C%5C%5C%5Cdfrac%7Bj%7D%7B-2%7D%2B7-7%3D-12-7%5C%5C%5C%5C%5Cdfrac%7Bj%7D%7B-2%7D%3D-19%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%20%28-2%29%7D%5C%5C%5C%5C%28-2%29%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5Cdiagup%5Ccdot%5Cdfrac%7Bj%7D%7B-2%5C%21%5C%21%5C%21%5C%21%5C%21%5Cdiagup%7D%3D%28-2%29%28-19%29%5C%5C%5C%5Cj%3D38)