Question

Question #85f82

Answer 1

Let.... the first number be `x`
And the second number be `y`
A.T.Q.
`x+y=40rArr(1)`
`x-y=10rArr(2)`
Solve the first one first
`x+y=40`
Subtract y from both sides
`x+cancely-cancely=40-y`
You get
`x=40-y`
Substitute this value in `(2)` equation
`x-y=10`
`40-y-y=10`
`40-2y=10`
Subtract 40 from both sidess
`cancel40-2y-cancel40=10-40`
Which gives us
`-2y=-30`
Minuses cut each other out
`cancel- 2y= cancel- 30`
Which gives us
`2y=30`
Which is
`y=15`
`x=40-y`
`x=40-15`
`x=25`
`25+15=40`
`25-15=10`

Answer 2

15 and 25

Explanation:

`x =`number 1
`y =`number 2

(1) `x + y = 40`
(2) `x - y = 10`

Equation (2):
`x - y = 10`
`x = 10+y`

Put equation (2) in equation (1):
`x + y = 40`
`10+y + y = 40`
`10 + 2y = 40`
`2y = 30`
`y = 15`

Plug back in one of the initial equation to find x:
`x - y = 10`
`x = 10+y`
`x = 10+15`
`x = 25`

Answer 3

25 and 15

Explanation:

We can generalize the above statement into two equations:
`x+y=40`
and
`x-y=10`
Now we can solve this problem using simultaneous equations,
`x-y=10`
`x=y+10`
Substitution Time:
`x+y=40`
`(y+10)+y=40`
`2y=30`
`y=15`
To find x, simply substitute y back into any of the equations
`x-y=10`
`x-15=10`
`x=25`