Math help from the Learning Centre

A system of linear equations consists of two or more equations with the same set of unknowns. A solution to the system is a pair of values that satisfies all equations at the same time. If we graphed each equation, we can see three possible solutions to a system of two linear equations. 

Solutions to systems of equations come in three types:

1. No solutions. When this happens, we say the system is inconsistent. Below is an example of an inconsistent system of equations:

   \[ x + y = 2 \] \[ x + y = 4 \]​ There is no solution to this system of equations because you can not find two numbers that add to make \(2\) and \(4\) simultaneously. If you were to proceed with attempting to solve the system, you end up with a contradictory statement - for example, \( 0 = 2 \), or \(4 = 2\).

2. Unique solutions. When this happens, we say the system is consistent. Below is an example of a consistent system of equations with a unique solution: \[ x + y = 4\] \[ x - y = 2\]. Only \(x = 3\) and \(y = 1\) satisfy both equations so this is the unique solution. 
3. Infinitely many solutions. When this happens, we also say the system is consistent. Below is an example of a consistent system of equations with a infinitely many solutions: \[ 2x - 2y = 8\] \[ -x + y = 4\]. In this case both equations are essentially the same equation (you can see this by multiplying across the second equation by \(-2\)).                  

These three cases can be represented graphically as follows:

• No Solution (inconsistent system): Parallel lines with no point of intersection (never meet).
• One Unique Solution (consistent system): Lines meet at only one point.
• Infinitely many solutions (consistent system): Coincident lines which meet at all points.

We can solve a system of linear equations, by graphing both equations and identifying the intersection point.

Example: \begin{align} y &= 2x+1\\ y &=-x+4\end{align}

Graph both lines on the same graph

The intersction point of both graphs is (1,3). which is the solution.

We can also solve systems of linear equations algebraically by substitution. We use the following system of equations as an example:

\(2x+4y=12\)    \((1)\)

\(6x-2y=8\)       \((2)\)

We first isolate for \(x\) in equation \((1)\) so we can use it in the substitution step:

\(2x+4y=12\)

\(\implies2x=12-4y\)

\(\implies x=\frac{12-4y}{2}\)

\(\implies x=\frac{2(6-2y)}{2}\)

       \(\implies x=6-2y\)     \((*)\)

Then, we can substitute the equation \((*)\) into equation \((2)\) to find \(y\):

\(6x-2y=8\)

\(\implies 6(6-2y)-2y=8\)

\(\implies 36-12y-2y=8\)

\(\implies 36-8=12y+2y\)

\(\implies 28=14y\)

\(\implies y=2\)

Now, we can substitute \(y=2\) into the equation \((*)\) to solve for \(x\):

\(x=6-2y\)

\(\implies x=6-2(2)\)

\(\implies x=6-4\)

\(\implies x=2\)

So the unique solution to this system is:

\(x=2\)

\(y=2\)

We can also solve a system of linear equations by aligning variables and eliminate one variable.

Example: \begin{align} 2x+3y &=7 \\ 4x-3y &=5 \end{align}

When we add both equations together to eliminate the y variable:

\begin{align} (2x + 3y) + (4x-3y) &= 7+5 \\ (2x+4x) + (3y-3y) &= 12 \\ 6x &=12 \\ x&=2 \end{align}

Substitute \(x=2\) into one of the equations to solve for y.

\begin{align} 2(2) + 3y &=7 \\ 4 + 3y &= 7 \\ 3y &= 3 \\ y&=1 \end{align}

Thus, the solution is \((2,1)\).

This method uses determinants and applies only when:

• The system has two linear equations.
• Coefficients are known and constants are on the right side of the equals sign.

The following is an example for a system of two linear equations with two variables.

Format:

\begin{align} a_{1}x+b_{1}y &= c_{1} \\ a_{2}x+b_{2}y &=c_{2} \end{align}

Then solve for x and y by finding the determinants D of the matrices:

\[D=\begin{array}{|cc|} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\]

\[D_{x}=\begin{array}{|cc|} c_{1} & b_{1} \\ c_{2} & b_{2} \end{array}\]

\[D_{y}=\begin{array}{|cc|} a_{1} & c_{1} \\ a_{2} & c_{2} \end{array}\]

\[x=\frac{D_{x}}{D}\]

\[y=\frac{D_{y}}{D}\]

Example:

\begin{align} 2x+3y &= 13 \\ 4x-y &=5 \end{align}

\[D=\begin{array}{|cc|} 2 & 3 \\ 4 & -1 \end{array} = (2)(-1)-(3)(4)=-2-12=-14 \]

\[D_{x}=\begin{array}{|cc|} 13 & 3 \\ 5 & -1 \end{array} = (13)(-1)-(3)(5) = -13-15=-28 \]

\[D_{y}=\begin{array}{|cc|} 2 & 13 \\ 4 & 5 \end{array}= (2)(5) -(13)(4)=10-52=-42 \]

\[x=\frac{-28}{-14}=2\]

\[y=\frac{-42}{-14}=3\]

Solution: \(x=2,y=3\)

Mixture problems involve combining two or more parts, each with known properties (such as concetraion, cost, or value), to form a final mixture with a desired property. These problems naturally lead to systems of equations because:

• One equation models the total amount,
• ANother models the total value or concentration

Approach

1. Define variables for the unknown quantities (amounts of each component)
2. Set up two equations:
   • One for total amount.
   • One for total concentration, value, or other key property.
3. Solve the system using substitution, elimination, or determinants.

Example:

You need 10 litres of 40% acid solution. You have 30% and 60% solutions. How much of each should you use?

Solution:

Let x = litres of 30% solution

Let y = litres of 60% solution

Set up equations:

Total volume: \(x+y=10\)

Total acid content: \(0.3x+0.6y=0.4(10)=4\)

Solve (e.g., using subsitution)

Rearrange first equation \(y=10-x\)

Substitute into second:

\[0.3x+0.6(10-x)=4\]

\[0.3x+6-0.6x=4\]

\[-0.3x=-2\]

\[x=\frac{20}{3}\approx 6.67\]

Then \(y=10-\frac{20}{3}=\frac{10}{3}\approx 3..33\)

There is approximately 6.67 L of 30% and 3.33 L of 60% solutions.

For systems with more variables and equations, Gaussian elimination is a useful method.

In discussing the use of Gaussian elimination in solving a system of equations, we will use the following system of equations and as example:

\[ 2x-3y+3z=6 \]

\[ x+2y-z=3 \]

\[ x-y+z=2 \]

In order to solve a system of equations using Gaussian Elimination, the system must first be transformed into an augmented matrix: 
$$ \left[ \begin{array}{ccc|c} 2 & -3 & 3 & 6 \\ 1 & 2 & -1 & 3 \\ 1 & -1 & 1 & 2\\ \end{array} \right] $$
There are three operations you are allowed to perform on an augmented matrix when using Gaussian elimination:

Elementary Row Operations:

1. Interchange the rows
2. Multiply a row by a nonzero constant
3. Add a nonzero multiple of a row to another row.

These operations mirror operations you can use on the system of equations closely; take a look at the side-by-side comparisons below:

When performing Gaussian elimination, the goal is to convert the augmented matrix into row echelon form. Below are the conditions required for a matrix to be in row echelon form:

Row Echelon Form: 

1. The leftmost nonzero entry of every row is \(1\) 
2. The leading 1 in each row is to the left of the leading 1 in every row below it.
3. The entries directly below a leading \(1\) are all zero. 

4. Any row of zeros are below the rows of leading \(1\)'s

Notice that the operations above that we have chosen have led us toward row echelon form. Below are the rest of the operations required to do so. Read them carefully and make sure you understand how and why each operation was taken.

Now that the matrix is in row echelon form, it can be converted back into a system of equations and solved.

$$ \begin{align*} x + 2y - z &= 3 \\ y - z &= -2 \\ z &= 7 \end{align*} $$

\(z = 7\) has been solved, so subbing this into the second equation gives

\[ y - 7 = -2 \qquad \implies \qquad y = 5 \]

Now that we have \(z = 7\) and \(y = 5\), both can be subbed into the first equation to solve for \(x\).

\[ x + 2(5) - 7 = 3 \qquad \implies \qquad x = 0 \]

Now the system has been solved -- the unique solution to the system is 

$$ \begin{align*} x &= 0\\ y &= 5 \\ z &= 7 \end{align*} $$

See the video below for another example of Gaussian elimination.