## Simultaneous Equations

As my tutoring students know, I detest calculators. I think the current math curriculum encourages laziness. Working with the numbers themselves gives students more of an idea of how numbers fit together. For example, I often can solve math problems in my head. The reason I can do that is that I did the vast majority of my math problems with pencil and paper or in my head, not with a calculator.

The reason I mention this is that my main tutee (he's in highschool and I've been working with him since he was in 2nd grade, so by now he knows how I feel about calculators) wanted to use his calculator , with a simultaneous equation function, to solve the problems. I told him if he wanted my help (which he did) he'd need to learn how to do it without the calculator.

The reason I mention this is that my main tutee (he's in high

**Simultaneous Equations**

## The Next Step

Multiply the

*entire*top equation by 4 and the*entire*bottom equation by 3 (so that box equations now have 12x as the first part of thee equation) to get the results in this picture.## After That:

Because out goal is to get this equation down to a solvable one variable equation (and one equation), we will want to subtract the second equation (because the "x" part of both equations will "drop out" when you take 12x from 12x). So we multiply the lower equation by -1 (or, simply, change the plus signs to minus and the minus signs to plus).

## Subtract and get this

## As you can see from this picture, you subtract each column (x from x, y from y, etc.) and the x drops out leaving only the y and the numbers. Then you solve for y by dividing by 17 and you end up with y = -2

##

Once you solve for y, you need to find x:

Substitute -2 for y and this is what you get:

## Do the same thing in the other equation:

You should get the same answer for x (this is a way to check your work -- if the answers are the same, then you probably did it right -- if not, you need to redo the problem)

## One More Way to do Simultaneous Equations

In this example, we take the second equation (x-y=0) and move the y to the other side by adding it so -- X - Y + Y = 0 + Y. the -Y +Y on the left side cancel out (-Y+Y = 0) and you are left with X=Y.

In the

**OTHER**equation (3x + 2y = -5) and substitute y for x (since y=x, as we just figured out) -- and we get 3y + 2y = -5. Add together and we get 5y=-5 or y = -1. Since x=y, then x=-1. So the answer it (-1, -1).

For the most part, it's harder to do this method, but occasionally, you'll see an equation that makes substitution easier. It's a matter of instinct. While this won't work well most of the time, it's nice to know how to do it when you need it

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