Solving simultaneous linear equations

Worked Example

Solve the following pair of simultaneous equations graphically:
\[\begin{align}y &= 3.8 - 0.6x  \\ y &= 1.3 + 0.3x\end{align}\]

Step 1: Plot each equation on the same set of axes.

Step 2: Identify the point of the intersection

Reading the coordinates from the graph, the approximate point of intersection of the two equations is \((2.79,2.15)\).

Worked Example

Both equations in the form \(Ax+By=C\)

Solve the following pair of simultaneous equations:
\[\begin{align} 3x + 4y &= 23\\ 5x - 2y &= 10 \end{align}\]

Step 1: Equalise the coefficients of one variable
\[\begin{align}3x + 4y &=23 \quad\quad [1] \\ 5x - 2y &= 10 \quad\quad [2]\end{align}\]

Multiply the second equation by \(2\) to align the coefficients of \(y\):

\[10x - 4y = 20 \quad\quad [3]\]

Step 2: Add or subtract the equations to eliminate one variable
Add equations \([1]\) and \([3]\) to eliminate \(y\):

\[\begin{align}3x + 4y &= 23 \quad\quad [1] \\  10x - 4y &= 20 \quad\quad [3]\end{align}\]

\([1] + [3]\) gives:

\[(3x + 4y) + (10x - 4y) = 23 + 20\]

Step 3: Solve the equation for the one variable
Solving the equation for \(x\):

\[\begin{align}(3x + 4y) + (10x - 4y) &= 23 + 20 \\ 3x + 10x + 4y - 4y &= 43 \\ 13x &= 43 \\ x &= \frac{43}{13}\end{align}\]

Step 4: Substitute the value into one of the original equations
Substitute \(x = \frac{43}{13}\) into equation \([1]\) and solve for \(y\):

\[\begin{align}3\left(\frac{43}{13}\right) + 4y &= 23 \\ \left(\frac{129}{13}\right) + 4y &= \frac{299}{13} \\ 4y &= \frac{299}{13} - \frac{129}{13} \\ 4y &= \frac{170}{13} \\ y &= \frac{170}{13 \times 4} \\ y &= \frac{85}{26}\end{align}\]

The graphs for the pair of simultaneous equations intersect at \(\left(\dfrac{43}{13},\dfrac{85}{26}\right)\).

Worked Example

Consider the following pair of simultaneous equations:
\[\begin{align}kx + 4y &= 10 \\ 3x - 5y &= 15\end{align}\]

Determine all values of k that results in a unique solution.

To determine the values of \(k\) for each case, start by rearranging both equations to express \(y\) in terms of \(x\).

• Rearrange \(kx + 4y = 10\) to \(y=\frac{10}{4} - \frac{k}{4}x\)
• Rearrange \(3x - 5y = 15\) to \(y = -3 + \frac{3}{5}x\)

A unique solution occurs when the two lines have different gradients, meaning they intersect at exactly one point. Therefore, the gradients of the equations must not be equal:

\[-\dfrac{k}{4} \neq \dfrac{3}{5}\]

Hence:

\[k \neq -\dfrac{12}{5}\]

This means that for all values \(k\), except for \(-\frac{12}{5}\) there will be a unique solution, this is when \(k \in \mathbb{R}\setminus \left\{-\dfrac{12}{5}\right\}\).

In this example, when \(k = -\frac{12}{5}\) the equations will have the same gradient and different \(y\)-intercepts. As a result, the lines are parallel, and there are no solutions. For infinite solutions, the equations must represent the same line, which would require matching both the gradient and the \(y\)-intercept.

Worked Example

Yuli has the choice of two printing plans. Plan A offers a rate of 25 cents for each page printed or copied. Plan B offers fee of $10 per month plus 15 cents per page printed or copied. Determine the break-even point for the two plans.

Step 1: Read the problem and write down the key information

Yuli has two printing plans to choose from:

• Plan A: 25 cents ($0.25) per page.
• Plan B: $10 per month plus 15 cents ($0.15) per page.

Need to determine the break-even point, i.e., the number of pages for which the total cost of both plans is the same.

Step 2: Define the variables

• Let \(x\) represent the number of pages printed in a month.
• Let \(C_A\) and \(C_B\) represent the total monthly cost of Plan A and Plan B, respectively.

Step 3: Translate into equations

• Plan A: \(C_A = 0.25x\)
• Plan B: \(C_B = 10 + 0.15x\)

Step 4: Solve the problem
Need to determine the value of x for which both plans cost the same, i.e. \(C_A = C_B\).

\[\begin{align}0.25x &= 10 + 0.15x \\
0.10x &= 10 \\  
x &= 100\end{align}\]

The break-even point occurs when Yuli prints 100 pages in a month. At this point, both plans will cost the same.

Step 5: Verify your solution
Substitute \(x = 100\) into both equations to verify they both result in the same cost.

• Plan A: \(C_A = 0.25\left(100\right) = $25\)
• Plan B: \(C_B = 10 +0.15\left(100\right) = 10 + 15= $25\)

Both plans cost $25, confirming the solution.

Worked Example

A sporting club has custom jackets made for $30 each and sells them to supporters for $80. The club incurs fixed yearly costs of $25,000 for marketing and equipment.

How many jackets must be sold each year to cover the marketing and equipment costs?

Step 1: Read the problem and write down the key information

The sporting club:

• Produces custom jackets at a cost of $30 each.
• Sells the jackets for $80 each.
• Has fixed yearly costs of $25,000 for marketing and equipment.

Need to determine the number of jackets to be sold each year to cover the fixed costs (break-even point).

Step 2: Define the variables

• Let \(x\) represent the number of jackets sold in a year.
• Let \(C_C\) represent the total yearly costs (jacket production, marketing and equipment)
• Let \(C_R\) represent the total yearly revenue from jacket sales

Step 3: Translate into equations

• Total costs for: \(C_C = 25000 + 30x\)
• Total revenue: \(C_R = 80x\)

Step 4: Solve the problem
Need to determine the value of x for which annual revenue and costs are equal, i.e. \(C_C = C_R\).

\[\begin{align}80x &= 25000 + 30x \\  
50x &= 25000 \\    
x &= 500\end{align}\]

The break-even point occurs when 500 jackets are sold annually.

Step 5: Verify your solution
Substitute \(x = 500\) into both equations to verify they both result in the same cost.

• Plan A: \(C_C = 25000 + 30(500) = $40000\)
• Plan B: \(C_R = 80\left(500\right) = $40000\)

Both costs and revenue are equal, confirming the solution.