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 = 2饾懃 - 1 and y = 2饾懃 - 1
Is when 饾懃 = 2 and 饾懄 = 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 饾懄 =
饾懄 = 饾懃 + 3
饾懄 = 饾懃虏 + 3饾懃 - One equation is given as 饾懄 = and the other is not
饾懄 = 2饾懃 - 3
2 = 饾懃虏 + 饾懄虏 - Neither equation is given in the form 饾懄 =
饾懃饾懄 = 12
饾懄 - 3饾懃 + 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.
饾懄 = 饾懃 + 3
饾懄 = 饾懃虏 + 3饾懃
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
饾挋 + 3 = 饾挋虏 + 3饾挋Calculate 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.
饾懃虏 + 3饾懃 - 饾懃 - 3 = 0
饾挋虏 + 2饾懃 - 3 = 0
(饾懃 + 3)(饾懃 - 1) = 0
饾懃 = -3 鈥冣 饾懃 = 1Substitute each value back into one of the equations to find the values of the other letter. It doesn鈥檛 matter which equation you chose but the linear equation is a sensible choice!
饾懄 = 饾懃 + 3
when 饾懃 = -3 鈥冣 when 饾懃 = 1
饾懄 = -3 + 3 = 0鈥冣凁潙 = 1 + 3 = 4
(-3,0) 鈥冣冣冣冣 (1,4)
These can be checked by substituting the values into the other equation.
饾懄 = 饾懃虏 + 3饾懃
for (-3,0) 鈥冣冣冣僨or (1,4)
0 = (-3)虏 + 3(-3) 鈥4 = (1)虏 + 3(1)
Both correct!
Example
Solve the simultaneous equations
饾懄 = 2饾懃 - 3
2 = 饾懃虏 + 饾懄虏
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 饾懄 from the first equation 饾懄 = 2饾懃 - 3 can be substituted for the 饾懄 in the second equation.
饾懄 = (2饾懃 - 3)
2 = 饾懃虏 + (2饾懃 - 3)虏Calculate the values of the remaining letter
2 = 饾懃虏 + (2饾懃 - 3)虏
2 = 饾懃虏 + (2饾懃 - 3)(2饾懃 - 3)
2 = 饾懃虏 + 4饾懃虏 - 6饾懃 - 6饾懃 + 9
5饾懃虏 - 12饾懃 + 7 = 0
\(x = \frac{7}{5}\)鈥僜(x = 1\)Substitute these values back into one of the equations to find the values of the other letterIt doesn鈥檛 matter which equation you chose but the linear equation is a sensible choice!
饾懄 = (2饾懃 - 3)
When \(x = \frac{7}{5}\)鈥冣冣僕hen \(x = 1\)
\(y = 2(\frac{7}{5}) - 3\)鈥冣冣僜(y = 2(1) - 3\)
\(y = \frac{14}{5} - 3\)鈥冣冣僜(y = -1\)
\(y = 鈥揬frac{1}{5}\)鈥冣冣僜(y = -1\)
\((\frac{7}{5},-\frac{1}{5})\)鈥冣冣冣冣僜((1, -1)\)
These can be checked by substituting the values into the other equation.
2 = 饾懃虏 + 饾懄虏
\((\frac{7}{5},-\frac{1}{5})\)鈥冣冣冣冣冣僜((1, -1)\)
\(2 = (\frac{7}{5})^2 + (\frac{1}{5})^2\)鈥冣 \(2 = 1^{2} + (-1)^{2}\)
Both correct!
Example
Solve the simultaneous equations
饾懃饾懄 = 12
饾懄 - 3饾懃 + 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 饾懄 = __
饾懄 = 3饾懃 - 9
This can be substituted for the y in the first equation.
饾懃饾懄 = 12
饾懃(3饾懃 - 9) = 12
- Calculate the values of the remaining letter 饾懃饾懄
饾懃(3饾懃 - 9) = 12
3饾懃虏 - 9饾懃 - 12 = 0 - divide all terms by 3
饾懃虏 - 3饾懃 - 4 = 0
(饾懃 - 4)(饾懃 + 1) = 0
饾懃 = 4 鈥冣 饾懃 = -1
- Substitute these back into one of the equations to find the values of the other letter
饾懃饾懄 = 12
When 饾懄 = 4 鈥 饾懄 = -3
When 饾懄 = -1 鈥凁潙 = -12
(4,3)鈥冣冣冣 (-1,-12)
These can be checked by substituting the values into the other equation.
饾懄 = 3饾懃 - 9
(4,3)鈥冣冣冣 (-1,-12)
3 = 3(4) - 9 鈥-12 = 3(-1) -9
Both correct!
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