3.15 Simultaneous Equation

Heya, can you please give me advice into how to prove that the coefficients of 2 equations representing two identical planes are proportional?

Welcome to the community!

It is usually quite obvious if you analyze your equations. For example:
you can see that all coefficients in equation (2) are double of those of equation (1) and all the coefficients of equation (3) are 4 times bigger than those of equation (1), including the answer. That means all three equations represent the same plane.

When you start solving your simultaneous equations, you first align all the variables, like you do for the calculator, first x’s, then y’s and then z’s (or whatever variables you have). If you put them to the calculator, it will give you an error:

so you can start looking what these equations are. First, you check if you can divide all the coefficients of any of the equations by the same number. In the example above, all coefficients of the equation (2) are divisible by 2 and of the equation (3) - by 4. You can divide/multiply both part of the equation by the same number. If you divide (2) by 2 and (3) by 4 you will get equation (1).

In your assessment you don’t need to proof this statement. You just inspect your equations and notice the pattern, then you explain what it means and provide a sketch of what the planes representing equations look like. That should be enough.

Can you give an in depth explanation and step by step method on finding multiple solutions for dependent equations using equations below:

Eq 1) 1200x + 1600y + 2000z = 200,000
Eq 2) 30x + 40y + 50z = 5,000
Eq 3) 2x + 2y + 2z = 220

First of all, examine each equation and see if you can simplify them. The basic rule is that you can divide/multiply both sides of an equation by the same number. So if you can divide all the coefficients and the constant by the same number - do it.

In the first equation all coefficients and the constant are divisible by 400. After dividing 1200, 1600, 2000 and 200,000 by 400 it will look like:

Eq 1) 3x + 4y + 5z = 500

When we examine the second equation we can notice that all coefficients are divisible by 10:

Eq 2) 3x + 4y + 5z = 500

What do we see now? They are the same equation! Which means both equations can be simplified into the same equation and they represent the same plane.

All coefficients and the constant in the third equation can be divided by 2:

Eq 3) x + y + z = 110

The resulting equation is different from what we had for equations 1) and 2). The coefficients are different and the constant is different. That means this equation represent a plane that is neither the same as the one 1) and 2) represent nor parallel to them (in that case you would have the same coefficients before x, y and z but the constant, the answer, would be different). The only option is that this plane intersects the plane described by equations 1) and 2)

In your assessment, start from simplifying equations. If you have a situation similar to the equations above, you just write your simplified version:

(1) 3x + 4y + 5z = 500
(2) 3x + 4y + 5z = 500
(3) x + y + z = 110

Then, make a statement: equations (1) and (2) are the same and (3) is different and provide a diagram:


Make a statement that two planes are the same and the third intersects them (along a line). Then you will have to explain, in context, that that means there are multiple solutions (more than one point of intersection) which are all along the line of intersection.

Up to here is usually what you need to do do get Merit. In order to get an Excellence, you will need to provide some possible solutions.

That is basically trial and error method, but there are ways to make it easier and faster. For that, start from solving your simultaneous equations (because (1) and (2) are the same, only write that equation once):

(2) 3x + 4y + 5z = 500
(3) x + y + z = 110

Next you can use the elimination method. For that, multiply equation (3) by -3 so that when you add the equations, the x’s are eliminated:

(2) 3x + 4y + 5z = 500
(3)*-3 -3x - 3y - 3z = -330

Then add them together:

(2) + (3)*-3: y + 2z = 170 (4)

This equation tells us that y and 2z added together are equal to 170. You also need to know the limitations of the value of x. For that, use Equation (3) and re-arrange it to make x the subject:

(3) x + y + z = 110
x = 110 - y - z (5)

That tells us that x will be equal to 110, minus z and y.

You don’t need to write all these explanations in your assessment, just your working. What you do have to do is clearly state that your answers can only be whole numbers (positive integers) as that is usually the case for your assessment as the equations represent practical situations: plants planted, days spent, items sold, etc. You can’t sell fraction of an item or a negative amount of items so your answer is limited by positive integers. Remember your three simultaneous equations have an infinite number of solutions, but for the practical situation that these equations represent you are usually limited by only those solutions which are positive integers.

So after that you implement the art of trial and error to find possible solutions:

y + 2z = 170 (4)
x = 110 - y - z (5)

We see that y+z can’t exceed 110, as that would make x negative. Also, y can’t be an odd number, as then 2z would equal an odd number which would make z not a whole number.

You can start from, for example, y = 50, then z = (170 - 50)/2 = 60 and x = 110 - 50 - 60 = 0. First possible solution: (0, 50, 60).

You can also try y = 40, then z = (170 - 40)/2 = 65 and x = 11 - 40 - 65 = 5 (5, 40, 65)

Other possible solutions are (10, 30, 70), (20, 10, 80)

You will only need to suggest 3-4 possible solutions and write an answer in context.

Thank you for the clarification, but with the statement would you be able to give me an example it is based on John’s small mask making business, there are 3 sizes of masks made small(x), medium(y), and large(z) and is made of 3 items; Material and amount required to make each size: 1200cm^2 - small, 1600cm^2 - medium and 2000cm^2 - large with a total of 200,000. Elastic: 30cm - small, 40cm - medium, 50cm - large with a total of 5,000. Toggles each mask requires 2 toggles in order to adjust the size with a total of 220.

For the situation described in the question, you can make a statement:

John can produce different combinations of different sizes of masks in order to use all material, elastic and toggles, for example:

No small masks, 50 medium size and 60 large masks, OR
5 small masks, 40 medium size and 65 large masks, OR
10 small masks, 30 medium size and 70 large masks

Can you please give me an example statement for the Merit part ?

Eq 1) 1200x + 1600y + 2000z = 200,000 | divide by 400
Eq 2) 30x + 40y + 50z = 5,000 | divide by 10
Eq 3) 2x + 2y + 2z = 220 | divide by 2

(1) 3x + 4y + 5z = 500
(2) 3x + 4y + 5z = 500
(3) x + y + z = 110

Equations (1) and (2) are the same and (3) is different which means two planes are the same and the third intersects them:


That means there are multiple solutions for this system of simultaneous equations.

If you’ve done up to here that would usually be an M. You created equations, solved them and interpreted the solution in the context.

Hey how do I do the excellence part my teacher explained something totally different, something like m = to something.

That is what you are expected to do for an E:

Develop a chain of logical reasoning by

  • Forming and solving all sets of equations (stating the variables clearly)
  • explaining answers in context by stating that the solutions must be both positive and integer
  • identifying and justifying the multiple nature of the solutions for the new situation
  • giving a general solutions with limits
  • listing at least TWO integer solutions
  • Communicating thinking using correct mathematical statements

As you see, for an Excellence, first of all you need to make comments, in context, about what you are doing. Start from defining the variables:
x - number of small masks,
y - number of medium masks
z - number of large masks

Answer the first question (unique solution) and provide an answer in context. Then solve the second question, with the adjustment, when you have multiple solutions.

As you see, you will need to clearly state that even though there is an infinite number of solutions for your three simultaneous equations, for this context, you only can consider positive integers. Then, you will need to start solving your new equations as was shown above so you can find limits for your variables. You may want to make some short statements, for example:

  • z and y, added together, can’t exceed 110
  • y plus 2z is equal to 170

Then provide two, three or even four solutions and write them in context.

I am not sure what your teacher was talking about. I assume that for some assessments you are asked to discuss when, for example, there will be no solutions. In your case, if you change the total amount of material or elastic available, that would make equations (1) and (2) have the same coefficients but different constants, so they represent two parallel planes and there would be no solutions.

(1) 3x + 4y + 5z = m
(2) 3x + 4y + 5z = k

when m = k there are multiple solutions, otherwise there are no solutions.

Equation 1: Divide through by 100, then divide through by 4 to obtain
3x +4y + 5z = 500 Call this equation 1a

Equation 2: Divide through by 10
3x + 4y + 5z = 500 Call this equation 2a

Equation 3
Divide through by 2
x + y + z = 110 Call this equation 3a

Eqation 1a minus equation 2a gives 0 = 0 , which implies an infinite number of solutions. A rather odd question.