Maths GCSE – Simultaneous Equations Notes

Simultaneous equations – Linear (Elimination method)
At the intersection point of two straight lines, the x values will be the same for BOTH  equations and so will the y values. To find the coordinates, we need to eliminate either x or y by adding or subtracting the equations from each other:

1) First write the equations one above the other. You may need to rearrange terms so that the letters match vertically. For example:
2x – y  =  17
x   – 3y = 1

2) Then multiply one (or both) equations so that you ‘match’ one of the letters. For example, we would need to multiply the second equation through by 2 in order to match the ‘x’ terms:
2x – y  =  17
2x – 6y = 2

3) Now add both equations together (if the matched terms are the different sign), or subtract the equations (if the matched terms are the same signs, remembering that “Same Signs Subtract”).
Here, the x terms are both positive, so we subtract the equations:
2x – y  =  17
2x – 6y = 2
————-

0  + 5y = 15

Note that a double minus means add, so -y – -6y = -y + 6y = 5y

4) Then solve to find the value of y (in this example).
y = 15/5 = 3

5) put this value back into one of the original equations (choose the simplest) and solve. Here the first equation is probably the simplest:
2x -(3) = 17
2x – 3 = 17
2x = 20
x = 10

… It’s then usually a good plan to show your answer in coordinate format, e.g. (10 , 3)

… If you need to ‘zap’ minus signs on both sides of an equation, you can multiply every term through by -1. For example:
-y     =    -5
(×-1)   (×-1)
y      =     5


Simultaneous Equations – non-linear (squared terms)

At the intersection coordinates of a straight line with a curve, the x values will be the same for BOTH curve equations and so will the y values. To find the coordinates,  we need to substitute one equation into the other.

So if the two equations are
y – 6x = 2      y = x² + 10
We can substitute the y of the second equation into the first equation:
(x² + 10) – 6x = 2   It’s useful to substitute using brackets
x² + 10 – 6x = 2   Now subtract 2 from both sides
x² + 8 – 6x = 0   Now rearrange in the usual quadratic format
x² – 6x + 8 = 0   Then factorise to solve
(x – 2)(x – 4) = 0
so x = 2 or x = 4 are the two solutions here.

Now sub these values of x into the easiest equation to find the corresponding y values.
y = 2 + 6x    Here we have rearranged the first equation to make y the subject
x = 2, so y = 2 + 6(2) = 14
x = 4, so y = 2 + 6(4) = 26
So the solutions to the simultaneous equation are:
(2, 14) and (4, 26)

… If the more complicated second equation features x and y in the denominator, first multiply both sides so that x and y become numerators. For example:
1/x + 1/y = 5     (× by x)
1 + x/y = 5x     (× by y)
y + x = 5xy
Now it is easier to substitute the first equation into this to form a quadratic equation.

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