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My mathematics teacher showed me a set of equations with more than one variable. She told me that the equations represent a linear system and she wanted me to find the solution. I kept staring at it because I didn't know what to do. It seemed impossible to find the solution to multiple variables at the same time. Well, it isn't and in this article, you will learn how to do just that.
Let's start by knowing what a linear system is.
A linear system is a mathematical model of a system of linear equations.
We recall that a linear equation in 3 variables takes the form of \(Ax+By+Cz=0.\)
The following are examples of a system of linear equations.
a. \(\left\{\begin{align} 3x+y&=6\\x+y&=2\end{align}\right.\)
b. \(\left\{\begin{align} -2x+y&=7\\x+y&=5\end{align}\right.\)
c. \(\left\{\begin{align} 3x+5y&=9\\2x+3y&=7\end{align}\right.\)
Solving linear systems is finding the set of values that will solve each of the equations of the system simultaneously.
Consider the system, \[\left\{\begin{align} x+y&=0\\x-y&=2\end{align}\right.\]
The solution to the above system is \((1,-1)\).
In fact, when substituting \(x=1\) and \(y=-1\), the two equations are satisfied.
We have \(1+(-1)=0\) and \(1-(-1)=1+1=2\).
Consider the linear system below, \[\left\{\begin{align} 2x+y&=0\\x-2y&=5\end{align}\right.\]
The solution to the linear system above is \((1,-2)\).In fact, when substituting \(x=1\) and \(y=-2\) in the above equations, we notice that they are verified simultaneously.
We have \(2(1)+(-2)=0\) and \(1-2(-2)=1+4=5.\)
How to solve linear systems?
There are various methods used in solving linear systems, we list the following,
If you follow the instructions for each method carefully, you'll be able to solve any linear system. Let's get a brief introduction to each of them.
As the name implies, the substitution method involves putting something in place of another. It is like a replacement. So what you will have to do is to solve for one of the variables in one of the equations and substitute them in the other equation.
You should follow the below steps when solving linear systems by substitution.
Step 1. Manipulate one of the equations to make one of the variables the subject of the equation.
Step 2. This is the substitution step. Substitute the isolated variable obtained in Step 1 in the other equation, to solve for the other variable.
Step 3. Now, substitute back the value of the variable you found in Step 2 in any of the equations of the system, to solve for the first isolated variable in Step 1.
Let's see how it's done with an example.
Solve the following linear system by substitution,
\[\left\{\begin{align} x+y&=6\\-3x+y&=2\end{align}\right.\]
Solution
Step 1 .
You should manipulate one of the equations so that one of the variables will be on one side and the other on the other side of the equation, by making one variable the subject of the equation. You can use any of the equations.
First, we will label our equations in the above system as,
\[x+y=6 \quad \text{as Equation 1} \quad \text{and}\quad -3x+y=2\quad \text{as Equation 2}.\]
Next, we choose one of the equations to manipulate, while noting that a different choice of equations will lead to the same solution.
We choose Equation 1, \[x+y=6.\] We isolate one of the variables in terms of the other. Here we are isolating \(x\) to get, \[x=6-y\]
Step 2.
This is the substitution step. You will now substitute the value of \(x\) in Equation 2.
You now have,
\[\begin{align} -3(6-y)+y&=2\\ -18+3y+y&=2\\-18+4y&=2\\4y&=20\\y&=\frac{20}{4}\\y&=5\end{align}\]
You now have the value of \(y\) as 5.
Step 3.
Since you now know the value of \(y\), you can now substitute that in any of the equations to get the value of \(x.\) But we know that \(x=6-y\) thus we get \[x=6-y=6-5=1.\] Therefore, the solution of the system is \[x=1\quad \text{and}\quad y=5.\] We can verify that \(x=1\) and \(y=5\) verify the equations of the system simultaneously.
In step 1, \(x\) was the subject of the formula. Let's see what happens if \(y\) is made the subject of the formula.
We recall Equation 1, \[x+y=6\]
Making \(y\) the subject of the formula, you will have the following, \[y=6-x\]
Now, you will follow step 2 and substitute \(y\) in Equation 2, to get
\[\begin{align} -3x+(6-x)&=2\\-3x+6-x&=2\\-4x&=2-6\\-4x&=-4\\x&=1\end{align}\]
Now, we calculate \(y\),\[y=6-x=6-1=5\]
We see that the solution of the system is \(x=1\) and \(y=5\).
The elimination method is another method of solving linear systems. It is called the elimination method because it requires you to eliminate or cancel out one of the variables from both equations.
It is also called the addition method because it involves the addition of both equations.
You should follow the following steps when solving linear systems by elimination,
Step 1. Make sure to write the equations in standard form \(ax+by=c\).
Step 2. Make sure that the variable to be eliminated has opposite coefficients in both equations, this can be done by multiplying corresponding equations with constants.
Step 3. Add both equations.
Step 4. Solve for the remaining variable.
Step 5. Substitute the value of the variable you just got in Step 4 in any of the equations to get the value of the other one.
Let's see how this is done in the below example.
Solve the following linear system by elimination,
\[\left\{\begin{align} x+y&=6\\-3x+y&=2\end{align}\right.\]
Solution
Step 1 .
We ensure that both equations are written in standard form.
Step 2.
You have to think of a way to eliminate one of the variables, so you will be able to get the value of the other. We notice that the coefficients of the variable \(y\) are equal in both equations. And thus, to have opposite coefficients of the variable \(y\) in both equations, we have to multiply one of the equations by \((-1)\).
We first label our equations as:
\[x+y=6\quad \text{as Equation 1}\quad \text{and}\quad -3x+y=-2\quad \text{as Equation 2} .\]
We multiply Equation 2 by (-1) to get, \(3x-y=-2\).
Now, we have the following system, \[\left\{\begin{align} x+y&=6\\3x-y&=-2\end{align}\right.\]
Step 3.
Now we are ready to add both equations, to get \(4x+0y=4\).
Step 4.
The resulting equation is \(4x=4\). We solve for \(x\) to get, \(x=1.\)
Step 5.
We now substitute the value of \(x\) in any of the equations to find the value of \(y\).
From Equation 1, we have \[\begin{align}x+y&=6\\1+y&=6\\y&=6-1\\y&=5\end{align}\]
Therefore, the solution of the system is\[x=1 \quad \text{and}\quad y=5. \]
In order to solve linear systems by graphing, we follow the below steps.
Step 1. Write each equation in its slope intercept form, that is \(y=mx+c\).
Step 2. Plot the two lines.
Step 3. The solution of the linear system of equations can be determined depending on the position of the straight lines in the coordinate system.
We encounter three cases:
We first start with Case 1, where the two lines intersect once.
The solution of the linear system can be found graphically by reading the coordinates of the point of intersection.
We move to Case 2, where the two lines are parallel.
Since the two lines have no point of intersection as they are parallel, the solution of the system does not exist. We say that the system has no solution, or is inconsistent.
We move finally to Case 3, where the two lines are congruent.
Since the two lines are congruent, they are in the same position. This means that there are infinite solutions to that linear system.
In the figure below, the first graph shows a point of intersection meaning it has a solution. The second shows that the lines are parallel which means there are no solutions at all, and the third graph shows both lines on top of each other meaning that there are multiple solutions for that system.
Let's see some examples.
Solve the following linear system by graphing,
\[\left\{\begin{align} x+y&=4\\2x+y&=5\end{align}\right.\]
Solution
Step 1.
We put the equations in the slope-intercept form to get,
\[\left\{\begin{align}y&=-x+4\\y&=-2x+5\end{align}\right.\]
Step 2.
We plot the lines in the graph. We notice that these lines have a single point of intersection. This is the solution we are looking for.
From the figure above, you can see that the point of intersection is \((1, 3)\) meaning that our solution is \(x=1\) and \(y=3\).
Let's take another example.
Solve the following linear system graphically,
\[\left\{\begin{align} -2x+3y&=-9\\4x-6y&=18\end{align}\right.\]
Solution
Step 1. We put the equations in slope intercept form, to get
\[\begin{align} y&=\frac{2}{3}x-3\\ y&=\frac{2}{3}x-3\end{align}\]
Step 2. We plot the two lines.
On the graph, it seems like there's only one line. But that's not true. What is happening is that the from both equations are lying on top each other, which means that there is an infinite number of solutions to this system.
Let's see another example.
Solve the following linear system graphically,
\[\left\{\begin{align} -2x+5y&=-15\\-4x+10y&=10\end{align}\right.\]
Solution
Step 1. Write each equation in its slope intercept form, we get
\[\begin{align}y&=\frac{2}{5}x-3\\y&=\frac{2}{5}x+1\end{align}\]
Step 2. We plot the two lines in the graph
The graph shows that the lines are parallel which means that there are no solutions to this linear system.
The solution of a set of linear equations can also be obtained by using a table of values.
The first step to do is to write each equation in its slope intercept form, and then to fill the table of values and notice where we have the same values for \(x\) and \(y\). Let's see how this is done.
Find the solution of the linear system below using tables,
\[\left\{\begin{align} x+y&=4\\2x+y&=5\end{align}\right.\]
Solution
Step 1. We write each equation in its slope-intercept form,
\[\begin{align}y&=-x+4\\y&=-2x+5\end{align}\]
Step 2. We change the variable \(y\) in both equations to \(y_1\) and \(y_2\) respectively, to get
\[\begin{align} y_1&=-x+4\\y_2&=-2x+5\end{align}\]
Now, we draw the table. We do this by substituting different values of \(x\) to get values for \(y_1\) and \(y_2\)_{.}
x | y_{1} | y_{2 } |
1 | 3 | 3 |
2 | 2 | 1 |
3 | 1 | -1 |
From the table above, we notice the values of \(y_1\) and \(y_2\)_{ } that were obtained when \(x\) was set to 1, 2 and 3. We can go on to use more numbers and even negative numbers but in this case, we can stop at the points we have because we can already see the solution.
So how do we obtain your solution from the table?
We notice that when \(x\) is 1, \(y_1\) and \(y_2\) are both 3. This means that the solution of the linear system is \((1, 3)\).
We can use the table of values to plot a graph and we will get the same plot and answer as in the previous example.
One thing to note is that when making the table of values, we may not get equal values for \(y_1\) and \(y_2\) as quickly as we did in our example. Sometimes, we'll have to put in some more values of \(x\) before we can obtain that, but we just need a few numbers of \(x\) to get the solution by graphing.
To solve linear systems by graphing, you plot the graphs of the set of equations given and find the solutions from the graph. If the lines on the graph intersect, it means there is a solution. If they are parallel, it means there is no solution but if they are directly on top of each other, it means there is an infinite number of solutions.
Step 1. Manipulate one of the equations to make one of the variables the subject of the equation.
Step 2. This is the substitution step. Substitute the isolated variable obtained in Step 1 in the other equation, to solve for the other variable.
Step 3. Now, substitute back the value of the variable you found in Step 2 in any of the equations of the system, to solve for the first isolated variable in Step 1.
Step 1. Make sure to write the equations in standard form \(ax+by=c\).
Step 2. Make sure that the variable to be eliminated has opposite coefficients in both equations, this can be done by multiplying corresponding equations with constants.
Step 3. Add both equations.
Step 4. Solve for the remaining variable.
Step 5. Substitute the value of the variable you just got in Step 4 in any of the equations to get the value of the other one.
We write the system as in the matric representation method, and test if the square matrix is invertible. If the matrix is not invertible, or not a square matrix, we use the Gaussian Elimination method.
There is no such obligation when solving a system of linear equations unless it is stated in the question to solve it algebraically.
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