Question

Medals awarded if all completed and method shown :)
(Simultaneous Equations)

1.) 3x+2y=11
2x- y=-3

2.) 5x-7y=27
3x-4y=16

3.) 2x- y=5
\(\frac{ x }{ 4 } + \frac{ y }{ 3 }\)=2

4.) 0.4x + 3y = 2.6
x - 2y = 4.6

Answer 1

There are a two methods to solve this :

1) Substitution method
2) Elimination method


Which one would you prefer ?

Answer 2

Elimination method please :)

Answer 3

Well in the method of elimination we try to add and subract to get rid of one variable or find a particular variable.


Lets see the first one,

3x+2y=11---(1)
2x- y=-3---(2)


I would reccomend you to multiply eq(2) by 2

and then add those equations

You will get x.

Answer 4

i get x=5 but when i put it into the equation it does not work..... :(

Answer 5

3x+2y=11
4x-2y=-6
----------


I think you are doing it wrong :/


try doing it again :)

Answer 6

cant do it :( not good at this :/

Answer 7

You just have to add them nothing else


@Chloeee_Harris

Answer 8

oh ive got it now, how would i do the next question?

Answer 9

What did you get x as ?

Answer 10

5/7

Answer 11

x=5/7


Now put x in eq(1) and find y :)

Answer 12

4 3/7 :)

Answer 13

how would i do the next question?

Answer 14

5x-7y=27
3x-4y=16

Answer 15

@Chloeee_Harris

You got y wrong.

Answer 16

it says its correct on my answer sheet,

Answer 17

3x+2y=11

put x=5/7

3(5/7)+2y=11


@Chloeee_Harris

Answer 18

@Chloeee_Harris

43/7 ?

Answer 19

no \[4\frac{ 3 }{ 7 }\]

Answer 20

@Chloeee_Harris


Oh i got it


it is


4 3/7 or 31/7


Yeah right :)

Answer 21

sorry about that, thanks for helping me, could you do the second question?

Answer 22

5x-7y=27
3x-4y=16


Now lets see this

We can multiply eq(1) by 3

and eq(2) by 5

and then subract.

And get y :)

Answer 23

so what would be the answer to that ^

Answer 24

thanks :)

Answer 25

question 3?

Answer 26

or question 4?

Answer 27

2x- y=5
\[\frac{ x }{ 4 } + \frac{ y }{ 3 }=2 \]
If you want to use elimination, you could multiply the second by 8 and subtract it from the first.

Answer 28

\[8 \times ( \frac{ x }{ 4 }+\frac{ y }{ 3 })=8 \times 2\]

becomes
\[2x+\frac{ 8 y }{ 3 }=16\]\[2x-y=5 \]

Now subtract the second from first.