Question

How do you do simultaneous equations?

Answer 1

There are two main ways - see below

Explanation:

The first way is to solve one equation for one of the variable and then substitute that solution into the second equation. Solve for the remaining variable and then substitute that answer into the first to find the other variable.

The second way is to find the LCM for one of the variables between the two equations. Multiply the entire equation(s) by the number that will find the product as the LCM.
Subtract the two equations leaving only one variable. Solve for the variable then substitute into one of the equations to find the other variables.

Both of these ways work so use the one that fits your problem the best.

Answer 2

Couple of ways

Explanation:

At first you are taught to try to equate the coefficients of x or y in order to eliminate them e.g.

`" "color(red)"x + 2y = 7"`
`" "color(blue)"2x + 2y = 8"`

In this case, the `y` coefficient (number in front of `y`) is `2` in both equations. You can eliminate `y` by subtracting the two equations. As the blue one is bigger, you would probably do blue - red but it works the other way around too.

`color(blue)"2x"-color(red)"x"=color(blue)8-color(red)7`

This gives you `x=1`

You can use this to now find `y` by substituting `x=1` into the red or blue equation. I'll choose `color(red)"red"`:

`color(red)(1+2y=7)`
`2y=6`
`y=3`

Finally it is good practice to check your answer by substituting your new values into the other equation that you didn't use to check if it works as well:

`color(blue)"2(1)+2(3)=8"` as required.

If your coefficients are not equal like in this case, you have to multiply the whole equation to get them equal. Alternatively you can use a substitution method to get the same answer. For example, using the above equations:

`color(red)"x = 7 - 2y"`
`color(blue)2color(red)(("7-2y")color(blue)" + 2y = 8"`
`14-4y+2y=8`
`-2y =-6`
`y=3`

Now you know y, put that back into your `x` equation to get `x=1`.