# Simple question abt linear equations

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#1
Say I have two sets of coordinates (1,3) and (-2,-1) and I want to find what B is in y=Mx+b and I do the following workings out 3=(4/3)x+b, 3=(4/3)x1+b. My question is on that last workings out where I multiplied it by 1 is it only multipled by that because it’s the x coordinate from the set I chose? So say the coordinates were (5,8) would it be 8=(4/3)x5 instead or is it always times by one?
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1 month ago
#2
Is not x1. For the point (1,3) you
* Replace y by 3
* Replace x by 1
So the equation is
3 = 4/3 + b
The gradient m is multiplied by the x value. You may have meant x as multiplication? If so, that's right. You can obviously pick any point.
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
Is not x1, you
* Replace y by 3
* Replace x by 1
So the equation is
3 = 4/3 + b
The gradient m is multiplied by the x value. You may have meant x as multiplication? If so, that's right.
when you say 'the x value' does that mean the x value in the set of co-ordinates i chose, in this case I chose (1,3) and its multiplied by 1 because the 1 is the x value there?
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1 month ago
#4
(Original post by PinkOneAmong)
when you say 'the x value' does that mean the x value in the set of co-ordinates i chose, in this case I chose (1,3) and its multiplied by 1 because the 1 is the x value there?
Yes. The point (1,3) must satisfy the equation of the line. Just like any other point that lies on the line.
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#5
(Original post by mqb2766)
Yes. The point (1,3) must satisfy the equation of the line. Just like any other point that lies on the line.
Okay, thankyou!
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