How do you solve a system algebraically

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Solve systems of linear equations

An equation that has only one unknown can (in all cases known to you) be solved for this unknown and thus the set of solutions can be determined. The solution set means all numbers that can be substituted for the unknown, so that the equation is true, i.e. "true".

However, some questions contain two or more unknowns, but it is also possible to set up two or more independent equations. For example a little word problem:

Christina buys ten items from item A and item B twelve times. Daniel, on the other hand, buys fifteen items from A, but only two from B. Christina pays 38 euros, Daniel 19.40 euros.

The individual prices of A and B are unknown. Since the individual quantities and the total price are known for both buyers, two equations can be set up that describe how the respective total price is made up. The unit price of A is described by the variable a and the unit price of B by the variable b:

10a + 12b = 38 (Christina's purchase) 15a + 2b = 19.4 (Daniel's purchase)

Unfortunately, here you cannot solve any of the individual equations for a variable in such a way that you can read off the unit price, because you cannot get rid of the other variable. But we know that the solutions to be found for a and b for both Equations must apply at the same time. You have one here system two equations with two unknowns.
It is customary to number the equations with Roman numerals and to put the system of equations between parallel vertical bars:

I: 10a + 12b = 38  
II:15a + 2b = 19.4

All methods of cracking the problem are based on making one equation with one unknown out of the n equations with n unknowns (where n is the number of equations and variables). In addition to guessing and the graphical solution method, there are essentially four algebraic methods:

→ Equation procedure
→ Appointment procedure
→ Addition method
→ Elimination procedure (also → here, with interactive examples with calculation method)

If you have more than two equations, every single step in every process always leads to an equation that contains one less variable.

 

The equation procedure

Solving equation II from the above example for b, we get b = -7.5a + 9.7. We appropriately call this transformed equation II '. Since the a still occurs on the right-hand side, b depends on a. After all, one can immediately calculate an associated b for every value of a.

For example, if a were EUR 0.50, then b could be calculated with -7.5 · 0.5 + 9.7 = 5.95.
Perhaps a = 0.5 is already the solution. If so, then it should also satisfy the first equation (10a + 12b = 38) with b = 5.95. Unfortunately, however, 10 · 0.5 + 12 · 5.95 = 76.4 and not 38.

You will (hopefully) notice immediately that b = -7.5a + 9.7 describes a linear function with the slope -7.5 and the y-axis intercept 9.7. A graph can be drawn for this function, which is a straight line.

The same can be done with the first equation: Solve for b and draw the corresponding graph b = -5/6a + 19/6 (II ').

The point of intersection of both graphs is the point of the sought solution pair (a | b), because it lies on both graphs, and its coordinates (a | b) therefore "fit" into both equations. If you draw reasonably precisely, you can read the coordinates and thus the prices, possibly down to 5-10 cents. Of course, generosity ends with money at the latest: we have to know exactly!

Now remember how to calculate the intersection of two linear functions: You set the function terms equal. In our example, the function terms are -7.5a + 9.7 and -5/6a + 19/6. So you equate them and you get an equation that contains only one unknown. So you can use it to determine the solution for a. This is the equation procedure:

II '= I' -7.5a + 9.7 = -5 / 6 * a + 19/6 | · 6 -45a + 58.2 = -5a + 19 | + 45a 58.2 = 40a + 19 | - 19 39.2 = 40a | : 40 0.98 = a

With this value one can easily calculate b: One only has to insert the value 0.98 for a into one of the two equations transformed into b:

Insertion in I ': b = -5 / 6 a + 19/6 = -5 / 6 0.98 + 19/6 = 2.35 Insertion in II': b = -7.5 a + 9, 7 = -7.5 * 0.98 + 9.7 = 2.35

It makes sense to choose the more pleasant equation, which is certainly II 'here.

The equation procedure
 
1.)  Solve both equations for the same variable.
2.)Put the other sides of the equations equal.
3.) Solve the resulting equation for the contained variable.
4.) Plug the solution into one of the transformed equations from step 1 to find the other variable.

Attention:
The system of equations I: 3x - 4y = 17; II: 2x + 3y = 17 could lead to the approach I = II: 3x - 4y = 2x + 3y. That would be correct, because 17 is equal to 17, but it does not help because this is still an equation with two unknowns. A variable is only dropped if you solve both equations for this one and, so to speak, "short-circuit" them by equating them.

 

The installation process

A new example. The system of equations is given

  5 - 4x = y  
7x - 3y = 51.5

If you have solved one of the two equations for a variable, you know its value as a function of the other variable. In our example, equation I has already been solved for y. All y that want to be the solution of the system of equations must be equal to 5 - 4x.

If you are now in the other Equation replaced all y by this term, which is equal to y according to the first equation, one obtains an equation that only contains x:

II: 7x - 3y = 51.5 For the y (5 - 4x) is inserted: I in II: 7x - 3 (5 - 4x) = 51.5

Attention: You have to put the term in brackets, because otherwise you would not take the "complete" y (i.e.) times -3, but only the 5.

Now you can solve the equation for x as above and insert the result in I to calculate y:

7x - 3 (5 - 4x) = 51.5 | Dissolve bracket 7x - 15 + 12x = 51.5 | To summarize 19x - 15 = 51.5 | + 15 19x = 66.5 | : 19 x = 3.5 In I: y = 5 - 4x = 5 - 4 * 3.5 = -9
The installation process
 
1.)  Solve an equation for a variable.
(It is possible that a given equation already exists. Otherwise proceed in such a way that you get no or at least "simple" fractions if possible.)
2.)Plug the term for that variable into the other equation.
3.) Solve the resulting equation for the contained variable.
4.) Plug the solution into the transformed equation from step 1 to find the other variable.

 

Other procedures

The addition method is described → here and the Gaussian elimination method → ​​here.

 

Examples

On → this page I have put together three examples of linear systems of equations (two equations each with two unknowns), which can be solved in all possibilities and in both methods. Understanding the solution steps should answer almost all questions.

 

Solve systems of equations of degree 3

The degree indicates how many equations and how many unknowns the system of equations has; in case 3 there are three equations with three unknowns.

Example:

  3a - b = c  
a + 2c = 4 - 4b
2b + c = 1

It is quite common that not all variables appear in all equations. Here, for example, the a is missing in III. This 3rd equation can now be used to eliminate c in I and II (eliminate = extinguish). To do this, we first solve them for c in order to then insert them in I and II:

III: 2b + c = 1 | - 2b III ': c = 1 - 2b in I: 3a - b = c 3a - b = 1 - 2b | + 2b 3a + b = 1 in II: a + 2c = 4 - 4b a + 2 (1 - 2b) = 4 - 4b a + 2 - 4b = 4 - 4b | + 4b -2 a = 2

This resulted in two equations with a total of two unknowns (a and b), i.e. this system of equations:

 3a - b = 1 - 2b  
a = 2

Fortunately, in the second equation there is no longer any b, so the solution for a is already known and can be inserted into the first to calculate b:

3a + b = 1 3 * 2 + b = 1 | -6 b = -5

With the now known values ​​for a and b, c can be calculated. b alone is sufficient for this, since there is no a in III ':

in III ': c = 1 - 2b c = 1 - 2 * (-5) = 1 + 10 = 11

 

Systems of equations with an infinite number of solutions or without a solution

The following system of equations has no solution:

 x - z = 2y - 5 
y - 4x + z = 6
2x + 3y = 3 - e.g.
I: x - z = 2y - 5 | + z I ': x = 2y + z - 5 in II: y - 4 (2y + z - 5) + z = 6 -7y - 3z + 20 = 6 | -20 II ': -7y - 3z = -14 in III: 2 (2y + z - 5) + 3y = 3 - z 7y + 2z - 10 = 3 - z | + z 7y + 3z - 10 = 3 | · (-1) -7y - 3z + 10 = -3 | -10 III '-7y - 3z = -13

Compare II 'with III'. The left sides are identical, but the right sides are not. Further equation leads to the wrong statement -14 = -13. In such cases there is no solution.

If, on the other hand, the system of equations were given as:

 x - z = 2y - 5 
y - 4x + z = 7
2x + 3y = 3 - e.g.

... then you get with analog steps

... II ': -7y - 3z = -14 ... III' -7y - 3z = -14

So two identical equations. In such a case it is said that the equations are linearly dependent. In fact, III is obtained by taking I + 2 · II. So you actually only have two independent equations with three unknowns and cannot find a unique solution.

In these cases one starts with a free variable, e.g. z, and describes the others as a function of it:

 

Solve a system of equations with four unknowns
 I: 6p - q + m = 12n - 5  
II:-2q - 8 = -6p + 8n - 2m
III:2m = 4n - 3p + 5
IV:3p = 9 + 4n + q
Solve equation IV for q: 3p = 9 + 4n + q IV ': q = 3p - 4n - 9 IV' in I: 6p - q + m = 12n - 5 6p - (3p - 4n - 9) + m = 12n - 5 6p - 3p + 4n + 9 + m = 12n - 5 | +5 -4n I ': 3p + 14 + m = 8n IV' in II: -2q - 8 = -6p + 8n - 2m -2 (3p - 4n - 9) - 8 = -6p + 8n - 2m -6p + 8n + 18 - 8 = -6p + 8n - 2m | -8n + 6p 10 = -2m | : (- 2) -5 = m very nice! in I ': 3p + 14 + (-5) = 8n 3p + 9 = 8n p = 8 / 3n - 3 III: 2m = 4n - 3p + 5 2 (-5) = 4n - 3 (8 / 3n - 3 ) + 5 -10 = 4n - 8n + 9 + 5 -10 = -4n + 14 | -14 -24 = -4n | : (- 4) 6 = n p = 8 / 3n - 3 (see above) p = 8/3 6 - 3 p = 13 IV ': q = 3p - 4n - 9 = 3 * 13 - 4 * 6 - 9 = 6

Such systems of equations can be solved more systematically with the → Gaussian method


© Arndt Brünner, October 3, 2003
eMail: [email protected]
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Version: October 4, 2003