Question

A system of equations is shown below:

3x + 8y = 12
2x + 2y = 3

Part A: Create an equivalent system of equations by replacing one equation by the sum of that equation and a multiple of the other. Show the steps to do this. (6 points)


Part B: Show that the equivalent system has the same solution as the original system of equations. (4 points)

Answer 1

@SolomonZelman

Answer 2

Can you help me please

Answer 3

first add the equations together. What do you get after doing this?

Answer 4

5x+10y

Answer 5

yes and adding 12 +3 you get 15.
So the new equation would be

5x+10y=15.

Answer 6

Now, multiply the second equation times (lets say) 2.

Answer 7

Like, 5(2)+10(2)=15?

Answer 8

no, your second equation is:
`2x + 2y = 3`

multiply each term in it by 2.

Answer 9

Oh okay that's what I thought but i wasn't sure.
so then
4x+4y=6?
This probably isn't right but..

Answer 10

yes, it is right, actuallty.

Answer 11

I keep disconnecting

Answer 12

Oh okay

Answer 13

so you are replacing your 2 equations of the initial system, by the new ones.
We get a `NEW SYSTEM OF EQUATIONS`.

\(\large\color{slate}{ 5x+10y=15 }\)
\(\large\color{slate}{ 4x+4y=6 }\)

Answer 14

Alright

Answer 15

So you have 2 systems:

Initial system: New system:

\(\large\color{slate}{ 3x+8y=12 }\) \(\large\color{teal}{ 5x+10y=15 }\)
\(\large\color{slate}{ 2x+2y=3 }\) \(\large\color{teal}{ 4x+4y=6 }\)

Answer 16

you have to show that they are equivalent.

Answer 17

I believe that to do this, you need to solve each system.

Answer 18

Oh okay

Answer 19

I would (perhaps) use matrix for both.
Good practice.

Or you can do other methods.

Answer 20

I would go for matrix.

Answer 21

I don't know matrix

Answer 22

Oh, nvm about matrix.

Answer 23

Okay

Answer 24

lets start from system 1.
you got:

\(\large\color{slate}{ 3x+8y=12 }\)
\(\large\color{slate}{ 2x+2y=3 }\)



\(\large\color{slate}{ 2x+2y=3 }\) multiply this equation times -4

Answer 25

-2x+-2y=-1

Answer 26

that is not what you get when you multiply (the second equation) times -4.

Answer 27

lets go like this:

1. What is \(\large\color{slate}{ 2x }\) times \(\large\color{slate}{ -4 }\) ?
2. What is \(\large\color{slate}{ 2y }\) times \(\large\color{slate}{ -4 }\) ?
3. What is \(\large\color{slate}{ 3 }\) times \(\large\color{slate}{ -4 }\) ?

Answer 28

-8
-8
-12

Answer 29

you are forgetting the X's and Y's for the first two answers.
(the third one is correct)

Answer 30

-8x and -8y

Answer 31

yes.

Answer 32

\(\large\color{slate}{ 2x+2y=3 }\)


1. \(\large\color{slate}{ 2x }\) times \(\large\color{slate}{ -4 }\) is \(\large\color{slate}{ -8x }\).
2. \(\large\color{slate}{ 2y }\) times \(\large\color{slate}{ -4 }\) is \(\large\color{slate}{ -8y }\).
3. \(\large\color{slate}{ 3 }\) times \(\large\color{slate}{ -4 }\) is \(\large\color{slate}{ -12 }\).

So when you multiply the \(\large\color{slate}{ 2x+2y=3 }\) , times \(\large\color{slate}{ -4 }\), you get:

\(\large\color{slate}{ -8x-8y=-12 }\)

Answer 33

Right

Answer 34

This is the change we made to the second equation of the system.

So your system is now, NOT

\(\large\color{red}{ 3x+8y=12 }\)
\(\large\color{red}{ 2x+2y=3 }\)



BUT, it is:

\(\large\color{slate}{ 3x+8y=12 }\)
\(\large\color{slate}{ -8x-8y=-12 }\)

Answer 35

Now, add the equations together.

Answer 36

-5x+0y=0?

Answer 37

\(\large\color{slate}{ ~~~~~~~~~~~3x+8y=~~~12 }\)
\(\large\color{slate}{^{^{\color{blue}{\Huge^+}}}~~ -8x-8y=-12 }\)
\(\large\color{blue}{~~~~~~~~^{\text{______________________}} }\)
\(\large\color{slate}{ ~~~~~~~~-5x+0y=~~~0 }\) CORRECT!

Answer 38

That would be same as: \(\large\color{slate}{-5x=0 }\), and so, x= ?

Answer 39

0?

Answer 40

yes, x=0.

Answer 41

Can you sovle for y, knowing that x=0 ?

Answer 42

*solve.

Answer 43

5?

Answer 44

Lets see....

\(\large\color{slate}{3x + 8y = 12 \\ 2x + 2y = 3 }\)

you know that x=0. Plug that in.

\(\large\color{slate}{3(0) + 8y = 12 \\ 2(0) + 2y = 3 }\)

Answer 45

what will y be equivalent to?

Answer 46

10y? i don't really know

Answer 47

\(\large\color{slate}{ 3(0)+8y=12 \\ 2(0)+2y=3}\)

simplifies to,

\(\large\color{slate}{ 8y=12 \\ 2y=3}\)

Answer 48

so in each of the equations, the y is going to be ?

Answer 49

1.5?

Answer 50

Connection again, excuse me...
yes, 1.5 or to be mathematical, 3/2.

Answer 51

Okay

Answer 52

So, your solution to the first equation is (0, 3/2 )

Answer 53

Now, the second system:

\(\large\color{slate}{ 5x+10y=15 }\)
\(\large\color{slate}{ 4x+4y=6 }\)

\(\large\color{slate}{ 5x+10y=15 }\) divide this equation by 5.
\(\large\color{slate}{ 4x+4y=6 }\) divide this equation by -4.

Answer 54

Alright

Answer 55

so your new equations from this will be ...

Answer 56

1x+2y=3 and -1x+-1y=-1.5

Answer 57

yes, correct!

```
x +2y = 3
- x - y =-1.5
```

So, now add the equations to each other.

Answer 58

Uhm this probably isn't correct but, x+1y=4.5

Answer 59

when you add the equations, don't the X's cancel ?

Answer 60

Yes

Answer 61

also, are you sure that : `3 + (-1.5) = 4.5 ` ?

Answer 62

no

Answer 63

yes, so what will `3 + (-1.5) ` be equal to ?

Answer 64

1.5

Answer 65

yes.

Answer 66

So the addition would be giving you:

0x + 1y = 1.5

which is same as y=1.5

Answer 67

Now, can yo solve for x ?

Answer 68

the system was,
\(\large\color{slate}{ 5x+10y=15 }\)
\(\large\color{slate}{4x+4y=6 }\)

plug in y=1.5

\(\large\color{slate}{ 5x+10(1.5)=15 }\)
\(\large\color{slate}{4x+4(1.5)=6 }\)

Answer 69

Solve for x.

Answer 70

5x+15=15
4x+6=6
so x=0

Answer 71

yes, correct.

Answer 72

So you have showed that in both systems,

the solution is: \(\large\color{slate}{ (0~,~~1.5) }\)

Answer 73

I think you are done...

Answer 74

Okay so are we moving on to B?

Answer 75

Thank you btw.

Answer 76

we have made the 2 systems (for part A)

and we have showed that both systems have a same solution by solving each system (for part B).

Answer 77

I think you are done not only with part A, but also with part B.
You are done entirely.

Answer 78

Now that I read part B I realize I am done. So thank you for all your help i really appreciate it:)

Answer 79

Sure, yw!