Delve into solving simultaneous equations, where a solution works for multiple equations. This skill is essential for analysing systems with multiple variables and is widely used in fields such as engineering, economics, and science.
Simultaneous equations
Simultaneous equations are equations that share variables and must be solved at the same time. A pair of simultaneous equations therefore consists of two algebraic equations that share two variables, and must be solved together.
There are three methods for solving a pair of simultaneous equations using algebra:
elimination
substitution
graphical.
This resource shows you how to solve the same pair of simultaneous equations using the three different methods.
Elimination method
In the elimination method, we manipulate the equations so that one of the variables has coefficients that are equal or opposite in sign. To do this:
Multiply one or both of the equations by a constant to make the coefficents the same.
If the sign is opposite, add the equations to get rid of the variable.
If the sign is the same, subtract the equations to get rid of a variable.
Solve for the remaining variable, then substitute the value into an original equation to solve for the other variable.
Example 1 – solving simultaneous equations by elimination
Solve the simultaneous equations \(x+y=6\) and \(x-y=2\).
We can write the equations on top of each other, and call them \((1)\) and \((2)\).
Choose one variable and multiply one or both equations by a constant to make the coefficient in front of the variable equal or opposite in sign. In this case, \(y\) has a coefficient of \(1\) in both equations, and is opposite in sign: \(+\) in \((1)\) and \(-\) in \((2)\).
To eliminate the variable, we need to add the equations. That is, we do \((1)+(2)\). Add the terms on the LHS together, and add the terms on the RHS together.
Choose one variable and multiply one or both equations by a constant to make the coefficient in front of the variable equal or opposite in sign. In this case, we will choose \(x\). You can choose \(y\) but we will show \(x\) in this example.
\(x\) has a coefficient of \(3\) in equation \((1)\) and a coefficient of \(4\) in equation \((2)\). They are also equal in sign: \(+\) in both \((1)\) and \((2)\).
To eliminate the variable, we need to find the lowest common multiple of \(3\) and \(4\), which is \(12\), and multiply each equation to get \(12x\). This forms equations \((3)\) and \((4)\).
\[\begin{align*} 3x+2y & = 10\quad\textrm{multiply all terms by \(4\)}\\
12x+8y & = 40\quad(3)\\
4x+3y & = 13\quad\textrm{multiply all terms by \(3\)}\\
12x+9y & = 39\quad(4)
\end{align*}\]
Then, we subtract the equations. That is, we do \((1)-(2)\). Subtract the terms on the LHS, and subtract the terms on the RHS.
Choose the simpler equation and transpose it to make one of the variables the subject. These two are equally simple, so let's just go with \(x\) in equation \((1)\).
\[\begin{align*} x+y & = 6\quad(1)\\
x & = 6-y
\end{align*}\]
Substitute the transposed equation into the second equation. Solve for the remaining variable.
Choose the simpler equation and transpose it to make one of the variables the subject. Here, it doesn't really matter, so let's just go with \(x\) in equation \((1)\).
The final answer is \(x=4\) and \(y=-1\). We can write this in coordinate or ordered pair form: \((x,y)\), which would be \((4,-1)\).
As usual, you can substitute these values back into the original equations to check that your answers work.
Graphical method
In the graphical method, the solution is the coordinates of the intersection between the two linear graphs. To find this:
Plot the straight line graphs representing each equation on the same set of axes.
Find the intersection of the two lines.
The graphs must be plotted accurately for this method to give an accurate answer. For this reason it is not usually used unless specifically asked for.
Example – solving simultaneous equations using graphs
Solve the simultaneous equations \(x+y=6\) and \(x-y=2\) graphically.
Graphs representing each linear equation are drawn and the point of intersection is read from the graph.
The intersection between the graphs x+y=6 and y-x-2 is at x=4 and y=2.
The solution is \((4,2)\).
Again, you can check your answer by substituting the values into the equations.
Equations without solutions
If the solution to a pair of simultaneous equations is the intersection of their graphs, it would make sense that some pairs of equations do not have a solution. Graphically, this is when the two lines do not intersect.
Example – solving simultaneous equations
Solve the simultaneous equations \(2x-y=6\) and \(-4x+2y=-7\).
We can write the equations on top of each other, and call them \((1)\) and \((2)\).
You can use any of the three methods above to try and solve the equations. We will use substitution. Choose the simpler equation and transpose it to make one of the variables the subject. Let's go with \(y\) in equation \((1)\).
\[\begin{align*} 2x-y & = 6\quad(1)\\
y & = 2x-6
\end{align*}\]
Substitute the transposed equation into the second equation. Solve for the remaining variable.
The last line is not possible, so we can conclude that no solution exists.
We show see graphically that the two lines are parallel and never intersect, so no solution exists.
The two graphs are parallel to each other and do not intersect. Therefore, there is no solution to the simultaneous equations.
Equations with multiple solutions
Some pairs of simultaneous equations have multiple solutions. In this case, the equations actually have the same graph. That is, one line lies on the other line.
Example – solving simultaneous equations
Solve \(y=1-3x\) and \(6x=2(1-y)\).
Rearrange the equations so that \(x\) and \(y\) are on the same sides and number the equations.
These two lines are the same because \((2)=2\times(1)\). This means that they have the same graph and there are multiple values of \(x\) and \(y\) that are possible solutions.
