1. ## simple simultaneous equations?

Hello, can someone help me sovle these, i know the gaussian and substitution methods but the coefficients in these equations are all proportional so when i multiply up to cancel out it removes all Xs and all Ys, any suggestions are welcome, it's probably really simple and i'm missing something, thanks guys.

10x+2y=8 and;
3y+15x=12

thanks again .

2. Originally Posted by captainlewis
Hello, can someone help me sovle these, i know the gaussian and substitution methods but the coefficients in these equations are all proportional so when i multiply up to cancel out it removes all Xs and all Ys, any suggestions are welcome, it's probably really simple and i'm missing something, thanks guys.

10x+2y=8 and;
3y+15x=12

thanks again .
10x+2y=8...(1)
15x+3y=12...(2)

Equation (1) and (2) are really the same. To see this multiply (1) by 15.

Work with equation (1).

10x+2y=8
Thus,
5x+y=4
Thus,
y=-5x+4

Let x=t i.e. and real number.
Then, y=-5x+4=-5t+4

Thus,
x=t
y=-5t+4

Are all solutions.

3. Hello, captainlewis!

You're not missing a thing.
. . The system does not have a unique solution.

10x + 2y .= .8 . [1]
15x + 3y .= 12 -[2]

Divide [1] by 2: .5x + y .= .4
Divide [2] by 3: .5x + y .= .4

You see, they gave us only one equation.

And one equation with two variables has an infinite number of solutions.

It says: "Find two numbers x and y so that 5 times x plus y equals 4"
. . and there are zillions of answers: .(1,-1), (0,4), (-1,9), . . .