Question

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions as the one shown.
8x + 7y = 39
4x – 14y = –68

Answer 1

hi @justsmile531 <3

Answer 2

@Lyrae

Answer 3

multiply the first equation all the way across by \(2\)
what do you get?

Answer 4

16 + 14 ?

Answer 5

you are missing an \(x\) and a \(y\)

Answer 6

Oh. so 16x + 14y ?

Answer 7

better, but you are now missing the equal sign, and the number on the other side
don't forget to multiply it by 2 as well

Answer 8

16y + 14y = 78

Answer 9

whew now we got it

\[16x+14y=79\] is the same line as
\[8x+7y=39\] but now we have something better, the system looks like
\[16x+14y=79\\
4x-14y=-68\]

Answer 10

oops typo there, i meant \[16x+14y=78\\
4x-14y=-68\]

Answer 11

now what happens when you add these up in a column?

Answer 12

Would you add 16x and 4x together? and the same for the rest?

Answer 13

you would add \(16x+4x\) yes

Answer 14

that would 20x

Answer 15

yes it would

Answer 16

what about the rest?

Answer 17

Would the 14y + -14y = 0 ?

Answer 18

yes to that as well
so on the left we get just
\[20x\] and the \(y\) terms have added up to zero and are gone
how about on the right side of the equal sign?

Answer 19

20x = 10 . Then you divide by 20 on both sides to get x by itself. and you get 2

Answer 20

slow down there a bit
yes you do get
\[20x=10\]

Answer 21

but if you divide by \(20\) you do not get \(2\)
\[\frac{20}{10}=2\] but you are computing
\[\frac{10}{20}\]

Answer 22

You get .5 or 1/2

Answer 23

yes you do

Answer 24

And now I am done?

Answer 25

probably need to find \(y\) as well, but that is most of the work done
the rest it to replace \(x\) by \(\frac{1}{2}\) in either equation, say
\[8x + 7y = 39\] and find \(y\)

Answer 26

in other words, solve
\[8\times \frac{1}{2}+7y=39\] for \(y\)

Answer 27

y= 5

Answer 28

bingo

Answer 29

solution \[x=\frac{1}{2},y=5\] or sometimes written as the pair
\[(\frac{1}{2},5)\]

Answer 30

Wow. Thank you so much. You really helped alot. I appreciate it

Answer 31

yw, hope it was more or less clear

Answer 32

You helped a struggling Blonde succeed. Lol

Answer 33

lol struggling blond was the name of my first punk band

Answer 34

Nice .

Answer 35

just kidding though
wish it were true

Answer 36

Learning is struggling.

Answer 37

You got that right.

Answer 38

Great effort!

Answer 39

Thank you :)