How would I go about doing this question?

T345

T345

T345

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Hey guys, I need a bit of help

There is this question. A straight line passes through the points (3,7) and (21,09). Apply y = mx + c and use simultaneous equations to find the equation of the line.

I am not sure what to do when they say, "Apply y = mx + c"

Does it mean that I need to sub in X and Y into y = mx + c

Which will make 7 = 3m + c

And 9 = 21m + c?

I am only asking because I am not sure what to do, and the teacher told us to attempt this question, and I don't remember covering it last year, and we have definitely not covered it this year.
 
Simultaneous equations:
1) 7 = 3m + c
2) 9 = 21m + c

Times 1) by 7.....to give equation 3):
(This is done to make something in the equations match each other, so when subtracted/added together one unknown disappears - I matched m but it could be c).
3) 49 = 21m + 7c
2) 9 = 21m + c

1)-2) gives:
40 = 6c. Therefore c=6.67

Put this into an equation - Any will do. I will use 1):
7 = 3m + 6.67. So 3m=0.33. So m = 0.11.

Check by putting m=0.11 and c=0.67 into the equations:
1) 7 = 3(0.11) + 6.67. This balances so is correct.
2) 9 = 21m + c. This balances so is correct.

I used to find these hard in school when I could not be arsed but now I quite like maths like this!
The above is the first maths question like this I have done for 9 years so forgive me if I have made any errors! :p.

Regy53 said:
I have no idea. I just wonder why you need to know this? as in what will it be used for in life.
Click to expand...

I used to think most 'pure' maths was horse **** until I did engineering and realised a lot of it is useful in real life.
In all honesty Alevel Maths/Physics mechanics is extremely applicable to the real world, but I only realised this several years later.

OP: What level maths are you doing?
 
Last edited:
Hah thanks OpenToSuggestions for making sure I don't feel stupid. T345, you don't need to do that multiplication step.

y = mx + c

When x =3, y = 7, so: 7 = 3x + c.

When x = 21, y = 9, so: 9 = 21x + c

Rearrange so c is the subject, (or m, if you like):

c = 7 - 3m, and c = 9 - 21m

Make each equation equal to each other to eliminate c, leaving m:

7 - 3m = 9 - 21m and rearrange for m:

21m - 3m = 9 - 7
18m = 2
m = 2/18 = 1/9

Giving you y = x/9 + c

Now you obviously need to determine what c is, so choose one of the original equations:

7 = 3m + c

sub in m

7 = (3*1/9) + c

rearrange for c

c = 7 - (3*1/9)
c = 7 - 3/9
c = 7 - 1/3

Therefore c = 6 and 2/3 = 20/3

We also have m, worked out earlier, m = 1/9

Put that back into y = mx +c gives you your final equation, y = x/9 + 20/3
 
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