Simultaneous equations
Simultaneous equations are two or more equations with two or more variables. They are simultaneous because they can be solved to give values for the variables that are equal in each equation.
It may be useful to look at M4 Quadratic Equations, M4 Algebraic Fractions, M6 Simultaneous equations and M7 Simultaneous Equations.
The solution to the equations:
y = 2đ„ - 1 and y = 2đ„ - 1
Is when đ„ = 2 and đŠ = 3(2, 3)
Solving Simultaneous Equations (one linear and one non-linear)
In the paper, one of the equations will be linear and the other non-linear, e.g. quadratic.
There are 3 common styles of question.
- Both equations are given in the form đŠ =
đŠ = đ„ + 3
đŠ = đ„ÂČ + 3đ„ - One equation is given as đŠ = and the other is not
đŠ = 2đ„ - 3
2 = đ„ÂČ + đŠÂČ - Neither equation is given in the form đŠ =
đ„đŠ = 12
đŠ - 3đ„ + 9 = 0
The most common method for solving linear simultaneous equations is the elimination method. This method can be used for linear and non-linear but looks a little different.
- Eliminate one of the letters from both equations
- Calculate the values of the remaining letter - there will usually be two answers.
- Substitute each value back into one of the equations to find the value of the other letter - there will usually be two pairs of answers.
Example
Solve the following simultaneous equations.
đŠ = đ„ + 3
đŠ = đ„ÂČ + 3đ„
Solution
Eliminate one of the letters from both equations by equating the Right Hand Sides (RHS) of the equations.
Since both equations are in the form y = RHS, both Right Hand Sides can be equated or set equal to each other
đ + 3 = đÂČ + 3đCalculate the values of the remaining letter.
Simplify the quadratic by taking all the terms to one side and putting it equal to zero and solving to find both values of x - there will be two answers.
đ„ÂČ + 3đ„ - đ„ - 3 = 0
đÂČ + 2đ„ - 3 = 0
(đ„ + 3)(đ„ - 1) = 0
đ„ = -3 ââ đ„ = 1Substitute each value back into one of the equations to find the values of the other letter. It doesnât matter which equation you chose but the linear equation is a sensible choice!
đŠ = đ„ + 3
when đ„ = -3 ââ when đ„ = 1
đŠ = -3 + 3 = 0ââđŠ = 1 + 3 = 4
(-3,0) âââââ (1,4)
These can be checked by substituting the values into the other equation.
đŠ = đ„ÂČ + 3đ„
for (-3,0) ââââfor (1,4)
0 = (-3)ÂČ + 3(-3) â4 = (1)ÂČ + 3(1)
Both correct!
Example
Solve the simultaneous equations
đŠ = 2đ„ - 3
2 = đ„ÂČ + đŠÂČ
It is not possible to equate the equations since the second equation is not in the form y = __.
Instead one of the letters can be eliminated by substitution.
Solution
Eliminate one of the letters from both equations by substitution
The đŠ from the first equation đŠ = 2đ„ - 3 can be substituted for the đŠ in the second equation.
đŠ = (2đ„ - 3)
2 = đ„ÂČ + (2đ„ - 3)ÂČCalculate the values of the remaining letter
2 = đ„ÂČ + (2đ„ - 3)ÂČ
2 = đ„ÂČ + (2đ„ - 3)(2đ„ - 3)
2 = đ„ÂČ + 4đ„ÂČ - 6đ„ - 6đ„ + 9
5đ„ÂČ - 12đ„ + 7 = 0
\(x = \frac{7}{5}\)â\(x = 1\)Substitute these values back into one of the equations to find the values of the other letterIt doesnât matter which equation you chose but the linear equation is a sensible choice!
đŠ = (2đ„ - 3)
When \(x = \frac{7}{5}\)âââWhen \(x = 1\)
\(y = 2(\frac{7}{5}) - 3\)âââ\(y = 2(1) - 3\)
\(y = \frac{14}{5} - 3\)âââ\(y = -1\)
\(y = â\frac{1}{5}\)âââ\(y = -1\)
\((\frac{7}{5},-\frac{1}{5})\)âââââ\((1, -1)\)
These can be checked by substituting the values into the other equation.
2 = đ„ÂČ + đŠÂČ
\((\frac{7}{5},-\frac{1}{5})\)ââââââ\((1, -1)\)
\(2 = (\frac{7}{5})^2 + (\frac{1}{5})^2\)ââ \(2 = 1^{2} + (-1)^{2}\)
Both correct!
Example
Solve the simultaneous equations
đ„đŠ = 12
đŠ - 3đ„ + 9 = 0
In this case neither equation is in the form y = __.
In order to eliminate it is necessary to rearrange one of the equations and then substitute into the other equation.
Solution:
- Eliminate one of the letters from both equations by substitution
Rearranging the second equation to get it into the form đŠ = __
đŠ = 3đ„ - 9
This can be substituted for the y in the first equation.
đ„đŠ = 12
đ„(3đ„ - 9) = 12
- Calculate the values of the remaining letter đ„đŠ
đ„(3đ„ - 9) = 12
3đ„ÂČ - 9đ„ - 12 = 0 - divide all terms by 3
đ„ÂČ - 3đ„ - 4 = 0
(đ„ - 4)(đ„ + 1) = 0
đ„ = 4 ââ đ„ = -1
- Substitute these back into one of the equations to find the values of the other letter
đ„đŠ = 12
When đŠ = 4 â đŠ = -3
When đŠ = -1 âđŠ = -12
(4,3)ââââ (-1,-12)
These can be checked by substituting the values into the other equation.
đŠ = 3đ„ - 9
(4,3)ââââ (-1,-12)
3 = 3(4) - 9 â-12 = 3(-1) -9
Both correct!
Test yourself
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