1. ## simultaneous equation

could anyone please show me a detailed method you could use for solving these simultaneous equations

12x + 65y - 1 = 0 and

60y = 150x + 511

thanks for any help

2. Originally Posted by mark
could anyone please show me a detailed method you could use for solving these simultaneous equations

12x + 65y - 1 = 0 and

60y = 150x + 511

thanks for any help

Have you tried anything?

Start by taking 1 equation and isolating a single variable, it doesn't matter which.

3. i don't know what you mean by isolating a single variable. i've only started maths recently. i have tried things, like rearranging the top equation to match the bottom one in order. then make the x's or y's the same number and cancel them out. but i end up with 138 and 12/13 x = 511 and 12/13

4. Originally Posted by mark
i don't know what you mean by isolating a single variable. i've only started maths recently. i have tried things, like rearranging the top equation to match the bottom one in order. then make the x's or y's the same number and cancel them out. but i end up with 138 and 12/13 x = 511 and 12/13
It means make it something like x = f(y) or y = f(x)

For example: $12x +65y-1 = 0$

Add 1 to both sides and take 65y from both sides: $12x = 1 - 65y$

Divide by 12: $x = \frac{1}{12}(1-65y)$

Now that x is expressed in terms of y you can put this into equation 2 wherever you see x

5. thanks, the answer the book gave me was x = -3 and 1/6 and y= 3/5. could you show me how you would be able to express it that way please?

6. Not sure what/why you're asking, Mark.

From 12x + 65y - 1 = 0 we get x = (1 - 65y) / 12

We substitute that in 60y = 150x + 511, to get:
60y = 150(1 - 65y) / 12 + 511
Solving gives y = 1047/1745 = 3/5

Substituting y = 3/5 in 12x + 65y - 1 = 0 results in x = -19/6 or -3 1/6