Simultaneous linear equations
What are simultaneous equations?
Simultaneous equations are a set of two or more equations, each containing two or more variables whose values can simultaneously satisfy both or all the equations in the set, the number of variables being equal to or less than the number of equations in the set.
Simultaneous linear equations
Simultaneous linear equations are a set of two or more linear equations with 2 or more variables. The solution of system of simultaneous linear equations is the ordered pair (x,y) if the set has two linear equations and (x,y,z….) if it has more linear equations.
Simultaneous linear equations with 2 variables
Example 1: $\displaystyle \left\{ \begin{array}{l}3x-y=1\\x+y=3\end{array} \right.$
Methods for solving simultaneous linear equation
- Graphical Method
- Substitution Method
- Elimination Method
Graphical Method
Simultaneous means at the same time with simultaneous linear equations you are trying to find the point where two lines cross, where the value of x and y are the same for both equations. There is only one point where the value of x and y are the same for both equations, this is where the lines cross ( the intersection). This is the simultaneous graphical solution.
Example 2: Solve $\displaystyle \left\{ \begin{array}{l}3x-y=1\\x+y=3\end{array} \right.$ graphically.
Solution: Draw each of the straight lines on the coordinative plan.
Firstly find some helping points (x,y) of our first equation 3x-y=1.
Draw them on the coordinative plan and join each with a straight line d1.
Secondly find some helping points (x,y) of our second equation x+y=3.
Draw each of them separately on the coordinative plan and join each with a straight line d2.
The point (1,2) where those two straight lines intersect is the graphical solution of our simultaneous linear equation with two variables.
Substitution Method
The substitution method consists on using one of the equations to get an expression of the form “y=” or “x=” and then substituting this into the other equation.This gives an equation with just one unknown. After solving this equation the value we find we substitute it at one of the equations of our initial system. This is how we find our pair (x,y) that is the solution.
Example 3: Solve $\displaystyle \left\{ \begin{array}{l}3x-y=1\\x+y=3\end{array} \right.$ by using substitution method.
Firstly we manipulate the equations in a way that one variable is defined in the terms of the other. You can choose whichever equation or variable you want, but to make it simple you have to choose the one that looks easier to rearrange. In our case we choose the equation x+y=3 and define y since in our first equation y is a simple variable with the coefficient 1.
x + y = 3
y = 3 – x
Secondly we substitute the variable y that we defined at our other equation, in our case at the first equation.
y = 3 – x
3x – y = 1
3x – (3-x) = 1
3x – 3 +x = 1
4x – 3 = 1
4x = 1 + 3 = 4
x = 1
Thirdly the value x that we found we substitute it to the other equation to find y. This is how we find our pair (x, y) that is the solution of the system.
y = 3 – x
x = 1
y = 3 – 1
y = 2
The solution is the pair (1, 2)
In the end we try the solution we found at the other equation to prove it that is the correct one.
(x, y) = (1, 2)
3x – y = 1
(3 × 1) – 2 = 1
3 – 2 = 1
Elimination Method
The elimination method is also called the addition method. It consists in adding or subtracting the equations in order to get one equation in one variable. If the coefficients of one variable are opposite then we add the equations and if the coefficients are equal you subtract the equations to eliminate the variables.
Tip: “Make sure that all the like terms and equal signs are in the same columns”
Be careful!
If you don’t have equations where you can eliminate one variable by adding or subtracting then you have to multiply one or both the equations to fix the coefficients in a proper way to obtain an equivalent system for eliminating one variable by adding or subtracting the equations.
Example 4: Solve $\displaystyle \left\{ \begin{array}{l}3x-y=1\\x+y=3\end{array} \right.$ by using the elimination method.
Solution: $ \displaystyle \left\{ \begin{array}{l}3x-y=1\\x+y=3\end{array} \right.$
Firstly since the coefficients of the variable y are opposite then we add the equations in order to get one equation with one variable.
$\displaystyle \frac{{~\begin{array}{*{20}{l}} {~~~3x-y=1} \\ {~~~x+y=3} \end{array}}}{{4x=4}}$
So x = 1
Secondly the value of x can be substituted into one of the equations of our system.
x = 1
x + y = 3
1 + y = 3
y = 3 – 1
y = 2
The solution is the pair (1,2)
Example 5: Solve $\displaystyle \left\{ \begin{array}{l}x+2y=2\\3x+y=1\end{array} \right.$ by using the elimination method.
Solution: In this case we can´t eliminate one variable byusing the addition or subtraction in order to get one equation with one variable. What we have to do is multiply one or both the equations of our system to fix the coefficients so we can eliminate one variable. We have to see which variable it´s easier to eliminate. In our system both of the variables can be eliminated by fixing the coefficients. We can eliminate the variable x by multiplying the first equation with -3 or we can eliminate the variable y by multiplying the second equation with -2.
Eliminating the variable x by multiplying the first equation with -3.
$ \displaystyle x+2y=2\Leftrightarrow -3x-6y=-6$
We obtain equivalent systems.
$ \displaystyle \left\{ {\begin{array}{*{20}{l}} {x+2y=2} \\ {3x+y=1} \end{array}} \right.\Leftrightarrow \left\{ \begin{array}{l}-3x-6y=-6\\3x+y=1\end{array} \right.$
$\displaystyle \frac{\begin{array}{l}-3x-6y=-6\\~~~3x+~~y~~=~~1\end{array}}{{~~~~~~-5y=-5}}$
Now we solve the equation with one variable.
-5y = -5
y = 1
Substitute y in either of the initial equations to obtain x.
x + 2y = 2
$\displaystyle x+\left( {2\cdot 1} \right)=2$
x=0
So the solution of our system is the pair (0,1).
