In these tutorials you are shown how to solve a simultaneous equation by the substitution method. You are also shown how it relates to the intersection of two graphs and why there are two sets of solutions.

## Example 1

#### Example in the video

Solve the simultaneous equations:

## Example 2

#### Example in the video

Solve the simultaneous equations:

AlastairJanuary 11, 2017 at 4:35 pmHi

At 9.09 in the second video you added 2x^2 both sides and subtracted an x both sides.

I can see why you added the 2x^2 but doesn’t it appear on the other side (I am still getting a hang of my algebra)?

Why did you make the plus x a minus? Should that appear also on the other side?

Thank you

Stuart the ExamSolutions GuyJanuary 12, 2017 at 8:03 pmAdd 2x^2 to both sides

-2x^2 + x + 2x^2 = 2x^2 -10

gives + x = 2x^2 -10

Now subtract x to both sides

+x -x = 2x^2 -x -10

to give 0 = 2x^2 -x -10

which is the same as

2x^2 -x -10 = 0

Hope that helps

AlastairJanuary 13, 2017 at 8:43 amBrilliant, thanks very much for the reply and explanation.

AlastairJanuary 13, 2017 at 10:29 amDid you mean take away an x both sides to get 0 on the left side which leaves 0=2x^2 -x-10 (It’s the 0 we want by clearing terms).

2x^-x-10 =0 swapping the 0 round with the equation.

Sorry to be so pedantic but it’s the method that is so useful.

Alastair

AlastairJanuary 13, 2017 at 10:36 amCould you not simply multiply

-2x^2 +x=-10

equivalent to -2x^2 +x +10 =0 by -1 to get the 2x^2 -x-10 ?