Maths Question regarding linear interception

[acroviak]

hello.
Im never too sure about how one would go about finding the interception between two linear equations.

the question i have is.

1. Find where these two lines intersect : 3x + 4y = 26 and 7x - y = 9

thank you very much.

[BigDave44]

x = 2 y = 5

[BigDave44]

take 2nd equation and multiply each side by 4, giving

28x - 4y = 36 then add the first equation
3x + 4y = 26, giving

31x = 62

x = 2 and, by substitution, y = 5

[pauld1412]

3x + 4y =26 (1)
7x - y =9 (2)

Multiply equation 2 by 4 and add

3x + 4y =26
28x - 4y =36

31x =62
x =2

Put x = 2 into equation 1
3(2) + 4y = 26
6 +4y = 26
therefore 4y = 20
y = 5
Hence intersection point is (2,5)

[DustyR]

You need to find numbers for a and y which satisfy BOTH equations.
3x + 4y = 26
7x - y = 9

Multiply both sides of second by 4 :
28x - 4y = 36
Add
3x + 4y = 26
Hence 31x = 26 + 36 = 62
hence x=2
y = 7X2 - 9 =5

Thus the lines meet at (2,5)

[Buenchico]

(2-part post)

You're simply seeking a pair of values, for x and y, which satisfy both equations. So the standard techniques for solving simultaneous equations will work:

Number the equations as follows:
3x + 4y =26 . . . Equation [1]
7x-y = 9 . . . .Equation [2]

Now multiply one (or both) equations by a suitable multipliers (or multipliers) in order to make the coefficient of either x or y (ignoring whether it's positive or negative) the same.

With the equations we've got, simply multiplying [2} by 4 will make the coefficient of y equal to (minus) 4, which matches the coefficient in [1]. So we've now got:

3x + 4y =26 . . . [1] and
28x - 4y = 36 . . . [2a]

The reason for making the coefficient of one of the terms the same is so that the term will 'disappear' when we either add or subtract the two equations. Since the coefficient of y in [1] is +4 and the coefficient of y in [2] is -4, we need to add the two equations (in order to get zero, which is what we're after):

[Buenchico]

[1] + [2a] gives:
31x + 0 = 62

Solving that equation gives x = 2

Now we know x, so we can substitute it into one of the original equations. Substituting x = 2 into [1] gives
6 + 4y = 26
<=> 4y = 20
<=> y = 5

We now have the pair of values, x and y, which satisfy both equations. Therefore we know that (2,5) lies on both lines, which means that it must be at the intersection of those two lines.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

In that particular example it was only necessary to multiply one of the equations to make the coefficients of one of the terms the same. Sometimes it's necessary to multiply both equations by suitable values.

Also, in that example the 'new' equations needed to be added together (in order to get a zero coefficient). If the 'new' coefficients were both positive (or both negative) you'd have to subtract one equation from the other.

Chris

(PS: I see that other have been posting while I've been typing my answer. We've all agreed on the technique but I've posted anyway, in case you need further explanation of the reasons for the operations)