Question

Let a, d, c, d, e, and f be real numbers.
Consider the equations: A. ax+by=c
B. dx+ey=f
C. (a+7d)x+(b+7e)y= (c+7f)
a) Given that x and y satisfy both (A) and (B), demonstrate that (the same) x and y also satisfy (C).

Answer 1

Start with the left hand side of C

Answer 2

(a+7d)x+(b+7e)y

distributing you get
ax + 7dx + by + 7ey

Answer 3

Next you need to group terms in useful way

Answer 4

group first and third terms,
group second and last terms

Answer 5

What would the final answer be?

Answer 6

this is a written question, you need to show work and explain stuff

Answer 7

Yes, I know that I just want to know what will be the final answer and then I will show the work.

Answer 8

"Finally" you need to show that expression yields the right hand side which is c+7f

Answer 9

How will I do that?

Answer 10

(a+7d)x+(b+7e)y

distributing you get
ax + 7dx + by + 7ey

Answer 11

group first and third terms,
group second and last terms

Answer 12

(ax + by) + (7dx + 7ey)

factor out 7 from second group
(ax + by) + 7(dx + ey)

Answer 13

Now look at A and B equations and try to conclude.

Answer 14

What do you mean by "factor out 7 from second group -> (ax+by+7(dx+ey)"?

Answer 15

(ax + by) + ( `7`dx + `7`ey)

`7` is common in that group, yes ?

Answer 16

pull that out :

(ax + by) + `7`( dx + ey)

Answer 17

So, if I followed the steps you told me the final answer would be "ax + by) + (dx + ey)"?

Answer 18

what do you mean by a final answer, this question is like an essay. there is no final answer OK

Answer 19

Start with left hand side of equation C
```
(a+7d)x+(b+7e)y
```

distributing you get
```
ax + 7dx + by + 7ey
```


group first and third terms,
group second and last terms gives
```
(ax + by) + ( 7dx + 7ey)
```


factor out `7` from second group
```
(ax + by) + 7(dx + ey)
```


From equations A and B this becomes
```
(c) + 7(f)
```


which is same as the right hand side of equation C.
So we're done.

Answer 20

thats the complete work, see if it makes sense

Answer 21

Yes and thank you.