# How to Solve Simultaneous Equations with Multiple Unknowns

This is the third in our series of brief articles discussing important topics relevant to electronics and electromechanical technicians and technician students preparing for today’s workforce. In this series, we will be discussing some everyday skills and topics for practicing technicians, as well as some areas that have been identified as “difficult to understand” by our technician students while performing general circuit analysis. Topics of discussion will include circuit reduction techniques, transient responses, as well as areas of difficulty when working with linear dc network theorems.

**How to Solve Simultaneous Equations with Multiple Unknowns**

Many technicians encounter difficulty in solving node or loop equations containing multiple unknown quantities. In this third installment of The Practicing Technicians Series, we will review a means of solving such equations to get loop currents or node voltages when performing linear DC network analysis. The two technician level methods for solving simultaneous equations having multiple unknowns used when dealing with two or three equations are “substitution” and “elimination”. In order to solve for a given number of unknowns, we require that the same number of equations be provided. For instance, we would require two equations to solve for two unknown quantities. We would require three equations to solve for three unknown quantities, and so on.

**Using the Substitution Approach To Solve Simultaneous Equations**

Solve for x, and y given these two equations containing the two unknown quantities.

Eq.1 3x + 2 = 2y

Eq.2 8x – 4 = 4y

We want to isolate the y term in one equation. By this we mean, we need an equation that states the value of a single y in terms of x.

Eq.1 | 3x + 2 = 2y |

y = (3x + 2) / 2 |

Since we now have an expression for the value of y in terms of x, we will replace/substitute the y term in Eq.2 with this new expression we obtained above. This will leave us with only one unknown quantity, x, to solve for in Eq.2 instead of the two unknowns we had before.

Eq.2 | 8x – 4 = 4y |

8x – 4 = 4(3x + 2) / 2 | |

8x – 4 = (12x + 8) / 2 | |

8x – 4 = 6x + 4 | |

8x = 6x + 4 + 4 | |

8x – 6x = 4 + 4 | |

2x = 8 | |

x = 8/2 | |

x = 4 |

Having the value of x, we can use it in Eq.1 or Eq.2 to find the value of y. (Using Eq.1)

3(4) + 2 = 2y

12 + 2 = 2y

14 = 2y

y = 14 / 2

**y = 7**

**Using the Elimination Approach To Solve Simultaneous Equations**

This approach can also be used to solve for the two unknowns in the same two equations.

Eq.1 3x + 2 = 2y

Eq.2 8x – 4 = 4y

This time, our objective is to find a factor to multiply one of the equations by which will allow us to sum the two equations and eliminate one of the unknowns. If we multiply both sides of Eq.1 by a factor of -2 and then sum the two equations, we will be left with an expression containing only one unknown. We selected -2 as the factor to multiply Eq.1 by so that the right hand side of Eq.1 when added to the right hand side of Eq.2 will result in the y term being eliminated from the resulting equation.

Eq.1 | 3x + 2 = 2y |

-2 (3x + 2) = -2 (2y) | |

-6x – 4 = - 4y |

Now we will sum this new expression for Eq.1 with our original Eq.2. Adding the -4y from the right side of our new expression for Eq.1 to the 4y on the right hand side of Eq.2 will result in 0y, which effectively eliminates the y term from the resulting equation.

Eq.1 -6x – 4 = - 4y (modified Eq.1 by factor of -2)

Eq.2 8x – 4 = 4y

To sum these equations, add the terms from the left hand side of each equation together, and add the terms from the right hand side of the equations together as follows…..

(-6x – 4) + (8x – 4) = (-4y) + (4y)

2x – 8 = 0

2x = 8

x = 8/2

**x = 4**

We would now use the value of 4 for x in either of the two original equations to solve for the value of y. (Using Eq.2)

8(4) – 4 = 4y

32 – 4 = 4y

28 = 4y

y = 28/4

**y = 7**

In this last example, we eliminated the y term from the equations because it was an easily recognizable means of reducing the equation to a single unknown. We could have decided to eliminate the x variable instead, leaving an equation with y as the only unknown quantity.

Eq.1 3x + 2 = 2y

Eq.2 8x – 4 = 4y

In order to clear the x terms from the above equations, it will be necessary to apply a factor to both equations in order to create the situation we desire. If we multiply Eq.1 by 8 and Eq.2 by -3, the x terms become 24x and -24x respectively. These will cancel each other out when the two equations are summed.

Eq.1 8(3x + 2) = 8(2y)

Eq.1 24x + 16 = 16y

Eq.2 -3(8x – 4) = -3(4y)

Eq.2 -24x + 12 = -12y

Now we sum our two new equations to obtain an equation with a single variable. If we write the equations one on top of the other, we can simply sum vertically to arrive at our single variable equation.

Eq.1 | 24x + 16 = 16y |

Eq.2 | -24x + 12 = -12y |

----------------------- | |

0x + 28 = 4y | |

4y = 28 | |

y = 28/4 | |

y = 7 |

We would now use this known value of y in one of the original equations to solve for the value of x. (Use Eq.1)

3x + 2 = 2(7)

3x + 2 = 14

3x = 14 – 2

3x = 12

x = 12/3

**x = 4**

A practical example of how these techniques are applied is provided in the video animation on Loop Analysis below. This example uses the Substitution and Elimination techniques to solve simultaneous KVL loop equations and is illustrated step by step.

We hope this has been helpful to you as a practicing or student technician. If you liked this post check out our previous article in The Practicing Technician Series;

Using the Natural Log or “ln” function in circuit analysis.

How to Create Correct Ohm's Law KCL Branch Equations for Nodal Analysis

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