Simultaneous Equations
Introduction
After studying this article you will be able to:
- To be able to solve a pair of equations using the methods of elimination, balancing unknowns and the method of substitution.
- To be able to solve linear and non-linear simultaneous equations.
Simultaneous Equations - Introduction
Simultaneous equations are two equations with two variables that need to be solved.
Simultaneous Equations – The method of elimination
Suppose that we have the two equations as follows:
\begin{aligned} & 6 x+y=15 \\ & 4 x+y=11 \end{aligned}We want to find what the values of x and y.
One thing that is important is to label each equation such as 1 and 2 so we now have the following:
- 6 x+y=15
- 4 x+y=11
Now the method elimination requires us to “get rid” of one of the unknown variables. And if you look t the equations you can see that both equations have single y variable. Because of this we can subtract equation 2 from equation 1.
1 – 2 gives: 2 x=4 \text { and so } x=2
Remember that there are two unknown variables that we need to find. Because we have just found the value of x we now need to find the value of y and this can be done by substituting the value of x into either equation 1 or equation 2.
Using equation 1 we have:
\begin{gathered} 6(2)+y=12 \\ 12+y=15 \\ y=3 \end{gathered}The solutions are therefore x=2 \text { and } y=3
Simultaneous Equations - Balancing unknown variables
Not all equations that we will come across will have variables that have the same value as we have seen in the example above. What is needed in this process is to make one of those unknowns the same and to then continue with the process of elimination as seen above.
Example:
Solve the following pair of simultaneous equations
- 3 x+2 y=18
- 2 x-y=5
What you see here is that none of the coefficients are the same. So we need to make them the same for one of the unknowns.
Probably the easiest thing to do here is to multiply equation 1 by 1 and equation 2 by 2 to give new equations 3 and 4:
3. 3 x+2 y=18
4. 4 x-2 y=10
You can now see that the y terms have the same coefficient and because one is positive and the other is negative we can just add equation 3 and 4 together to give:
\begin{aligned} 7 x & =28 \\ x & =4 \end{aligned}Don’t forget that we know need to find the value of y and we can do this by substituting the value of x into any of the equations above (1, 2, 3 or 4).
Using equation 2:
\begin{gathered} 2(4)-y-5 \\ 8-5=y \\ y=3 \end{gathered}So the solutions are x=4 \text { and } y=3
Simultaneous Equations – Method of substitution
If you are given a pair of simultaneous equations to solve then to use the method of substitution means to make one of the unknowns from one of the equations the subject and to then substitute that into the other equation. Doing this will give a new equation in terms of one unknown only. The following example will illustrate this.
Example:
Solve the following simultaneous equations
- y=2 x+3
- 3 x+4 y=1
Now equation 1 is actually nice here because y is the subject. You could make either x or y the subject from equation 2 but this would introduce fractions and more work. But there could be instances where you might have to do this.
From above we will substitute equation 1 into equation 2 as follows:
3 x+4(2 x+3)=1We now have an equation involving x terms only. Expanding brackets gives us:
\begin{aligned} &\begin{gathered} 3 x+8 x+12=1 \\ 11 x=-11 \end{gathered}\\ &x=-1 \end{aligned}And the value of y using equation 1: y=2(-1)+3=-2+3=1
So the solutions are x=-1 \text { and } y=1
Simultaneous Equations – Solving linear and non-linear simultaneous equations
In order to solve these types of simultaneous equations we must use the method of substitution. The process is to use the easier equation and to make either x or y the subject and to substitute into the non-linear equation.
The following example will illustrate this.
Example:
Solve the following simultaneous equations
- x^2+y^2=5
- x+y=3
Equation 2 is the easier equation and the linear equation always will be. We need to make either x or y the subject.
From equation 2: x=3-y and we substitute this into equation 1.
\begin{gathered} (3-y)^2+y^2=5 \\ 9-6 y+y^2+y^2=5 \\ 2 y^2-6 y+4=0 \\ y^2-3 y+2=0 \\ (y-1)(y-2)=0 \end{gathered}So, y=1 or y=2
We now need the values of x for these values of y.
When y=1, x=2 and when y=2, x=1
And these are the solutions.
Notice we had to solve a quadratic equation here as well as.