#1

Ok, im finsihing off college and i did a test. I havnt done maths since highschool and i need help.

Could someone calculate.

1) 5x+y=9

8x+2y=30

1) 11m2 of the floor area will be taken up by work benches, storage shelves etc. What proportion of the room is this"

y=4x-1

y=-x+4

Could someone calculate.

__Simultaneous equations__1) 5x+y=9

8x+2y=30

__"The ICT support team will be based in a room measuring 4.5m long by 5.7 metres wide"__1) 11m2 of the floor area will be taken up by work benches, storage shelves etc. What proportion of the room is this"

__HARD MOTHER****ER - Simultaneous equations using graphs__y=4x-1

y=-x+4

#2

"The ICT support team will be based in a room measuring 4.5m long by 5.7 metres wide"

1) 11m2 of the floor area will be taken up by work benches,storage shelvesetc. What proportion of the room is this"

Shelves don't count as floor space. Being as you are working in 2 dimensions, I think you should just tell the teacher that it is far too ambiguous to possibly answer.

#3

EDIT: Soz i thought they were separate questions

*Last edited by jesus the strat at Jun 12, 2008,*

#4

College, eh?

#5

The first equation to solve for y is 5*(30/8 + 2/8*y)+y = 9, solve for y using simple algebra and x is easy to find after that, just plug the numerical value for y into any of the 2 equations.

#6

1) a. x=1 and y=4

b. x=3 and y=3

ill do the others in a sec.

Err...if I'm not wrong (which I may be) surely simultaneous equations have the same value for x and y?

#7

And I think x for the last one is 5/3.

#8

1) y = 9 - 5x

Sub it into the second equation: 8x + 18 - 10x = 30

So x = -6

sub it in to get y = 39

2) 4.5 x 5.7 = 25.65

11/25.65 x 100 = 43%

3) y = y

4x - 1 = x + 4

3x = 5

x = 5/3

Sub it into the second equation: 8x + 18 - 10x = 30

So x = -6

sub it in to get y = 39

2) 4.5 x 5.7 = 25.65

11/25.65 x 100 = 43%

3) y = y

4x - 1 = x + 4

3x = 5

x = 5/3

#9

3)(1,3)

last one is -x+4

1) y = 9 - 5x

Sub it into the second equation: 8x + 18 - 10x = 30

So x = -6

sub it in to get y = 39

2) 4.5 x 5.7 = 25.65

11/25.65 x 100 = 43%

3) y = y

4x - 1 = x + 4

3x = 5

x = 5/3

last one is -x+4

#10

1) a. x=1 and y=4

b. x=3 and y=3

the second one is 14.65m^2

1) actually should be x=-6 and y=39 because x and y have to be the same for both equations.

beat to it.

#11

Yeah x and y have to be the same.

#12

Err...if I'm not wrong (which I may be) surely simultaneous equations have the same value for x and y?

Correct. x= -6 and y= 39.

#13

Yeah x and y have to be the same.

I thought so. It's been years since I finished school, I'd forgotten how to do them.

By the way, to all those in school still:

I have NEVER found a use for these things out side of school. NEVER!

#14

3)(1,3)

last one is -x+4

oops, so it's 4x - 1 = -x +4

5x = 5

x = 1

#15

I thought so. It's been years since I finished school, I'd forgotten how to do them.

By the way, to all those in school still:

I have NEVER found a use for these things out side of school. NEVER!

well you're doing it in college so...

EDIT:nevermind thought you were TS

#16

I thought so. It's been years since I finished school, I'd forgotten how to do them.

By the way, to all those in school still:

I have NEVER found a use for these things out side of school. NEVER!

Hahaha, I use this stuff every day along with vector calculus and differential equations.

#17

"HARD MOTHER****ER - Simultaneous equations using graphs

y=4x-1

y=-x+4"

So does anyone know what Y is???

y=4x-1

y=-x+4"

So does anyone know what Y is???

#18

the do. The asnwer for number on is x= -6 and y=39.

it's not difficult,

5x+y=9, solve for y and we get y=9-5x

substitute for y in the second equation and it reads 8x+2(9-5x)=30

solve for x, then plug your solution into either equation to get y

EDIT: damn beaten to it

it's not difficult,

5x+y=9, solve for y and we get y=9-5x

substitute for y in the second equation and it reads 8x+2(9-5x)=30

solve for x, then plug your solution into either equation to get y

EDIT: damn beaten to it

#19

"HARD MOTHER****ER - Simultaneous equations using graphs

y=4x-1

y=-x+4"

So does anyone know what Y is???

3

I answered that one already

#20

"HARD MOTHER****ER - Simultaneous equations using graphs

y=4x-1

y=-x+4"

So does anyone know what Y is???

Do you not have a graphing calculator, TS? That'd make your life MUCH easier... but this is still quite easy, just set them equal to each other.

-x+4=4x-1

x=1

y=4(1)-1

y=3

...I'm guessing your major has nothing to do with math.

#21

This is year 9 work. I would have though common sense would have solved these...

#22

Hahaha, I use this stuff every day along with vector calculus and differential equations.

Looks at Title....

OK, so mech. engineers (wouldn't it be cool if it was actually a Mech engineer!) need it...

I'm an IT goon though, so I don't!

#23

this is what college math is like? wow im going to college lol soo easy

#24

This is year 9 work. I would have though common sense would have solved these...

yes I mean come on, this is basic maths you should really know.

#25

this is what college math is like? wow im going to college lol soo easy

Actually, university maths is damn tricky, and a lot of it is conceptual. School maths is learning techniques. Uni level is understanding them and teaching you to think in new ways about it.

I suspect TS is doing a non maths related subject, and a bit of maths has come up.

#26

Hey, yeh i was in a rush 2 get my maths done because i never went 2 the class and today was the last day to get everything done in every Unit, so i was alone there for 30minutes . Im doing an ICT course so there is a tiny bit of maths involved.

Anyway, THANKS

Anyway, THANKS