**Simultaneous equations** are equations containing two unknown quantities whose values are to be determined in two equations at the same time. Solved Examples and Shortcut Tricks of simultaneous equations are well explained here.

I will try to bring this lesson down to a lay man’s understanding such that after reading this post, you will never find it difficult to solve simultaneous equations again.

Contents

## Terminologies Used in Simultaneous Equations

There are some important terms one needs to understand before moving on to solve simultaneous equations. The terms are listed here under:

- Coefficient
- Term
- Variable

Let’s quickly explain the terms listed above.

**Coefficient:** This is a constant by which an algebraic term is multiplied. For instance, If we have 2x, 2 is the coefficient of x.

**Variable: **A variable has no fixed quantitative value. Example: From the example above, x is a variable. Any alphabet can be regarded as a variable. It changes. That is to say, its values are not constant.

**Term:** Any value (variable or constant) or expression separated from another term by a space or an appropriate character, in an overall expression or table. Put more aptly, a term is a combination of coefficient and variable. Example: 2x is a term.

Having explained the necessary terms used in solving simultaneous equations, we can progress to stating the difference between simultaneous equations and linear equation.

## Difference between Simultaneous equations and Linear equations

Linear equation has only one equation and only one unknown. Example: 2x+1 = 1

While Simultaneous equations have at least two equations and two unknown variables. Example:

2x+3y = 5 ————————————— equation 1

3x+3y = 2 —————————————- equation 2

Good! I believe you can proudly tell the difference between linear and simultaneous equations now.

## Methods of solving simultaneous equations

There are three most commonly used methods. Namely:

- Substitution method
- Elimination method
- Matrix method

For the scope of this lesson, I will emphasize on substitution and elimination methods.

### Substitution Method Solving Simultaneous Equation

In this method, one of the two unknowns is made the subject of the formula. This is later substituted for in the second equation to have a simple equation with one unknown variable. Then the equation is solved linearly to obtain one value and the value obtained is therefore substituted for to get the other unknown.

### Steps to solve simultaneous equations using substitution method:

- Call the first equation ‘equation 1 and second one ‘equation 2’
- Make either x or y the subject of the formula in any of the two equations.
- Substitute for the variable made the subject of the formula in the other equation.
- Collect like terms and solve linearly for the one unknown in the simple equation.
- Substitute for the solved unknown in either equation 1 or 2 to get the second unknown

**Example 1**

3x+2y = 10

x+2y = 2

**Step 1**

3x+2y = 10 ————————————— equation 1

x+2y = 2 —————————————- equation 2

Step 2: Let’s make x the subject of the formula in equation 2

x = 2 – 2y ————————————————– equation 3

**Step 3:** Substitute for x in equation 1 ( this means, put 2 – 2y in equation 1 wherever you see x)

This implies;

3(2 – 2y) + 2y = 10

open the bracket

6 – 6y + 2y = 10

**Step 4**

Collecting like terms and solving for the unknown

-6y + 2y = 10 – 6

-4y = 4

y = -4/4

y = -1

**Step 5**: To get x, substitute y in equation 2

x+2y = 2 —————————————- equation 2

x + 2(-1) = 2

x – 2 = 2

x = 4

Therefore, X = 4, y = -1

That’s how simple it is.

Read Also: **Divide 1423 by 24 in base 5**

### Elimination Method of Solving Simultaneous Equation

Here, one of the two unknowns is made to be removed either by addition or subtraction being performed on the two equations.

Note: Any unknown that is to be eliminated have the same coefficient so as to enable addition or subtraction that is to be performed.

### Steps to solve simultaneous equations using elimination method:

- Call the first equation ‘equation 1 and second one ‘equation 2’
- Check if any of the unknowns in the two equations have the value
- If they have, subtract or add to eliminate one of the unknown variables
- If they are not, use either the coefficient of x or y in equation 1 to multiply everything in equation 2 and call it equation 4. Also use either the coefficient of x or y in equation 2 to multiply everything in equation one to make one the unknowns equal and call it equation 3
- Check if addition or subtraction will eliminate the variables that are equal and carry out the one that eliminates one of the unknowns.
- Collect like terms and solve linearly for the one unknown in the simple equation.
- Substitute for the solved unknown in either equation 1 or 2 to get the second unknown.

Let’s pick up the **example 1** above and solve using elimination method

3x+2y = 10

x+2y = 2

**Step 1**

3x+2y = 10 ————————————— equation 1

x+2y = 2 —————————————- equation 2

**Step 2**: the coefficient of x in equation 1 is not the same with the coefficient of x in equation 2. The same thing applies to y. Skip step 3 since none is equal.

**Step 4:** multiply equation 1 by 1 (coefficient of x in equation 2) and multiply equation 2 by 3 (coefficient of x in equation 1)

This implies;

3x + 2y = 10 ——————————————— equation 3

3x + 6y = 6 ——————————————— equation 4

**Step 5**

subtracting equation 4 from equation 3

-4y = 4

y = -1

**Step 5**

To find x, substitute for y in equation

x+2y = 2 —————————————- equation 2

x + 2(-1) = 2

x – 2 = 2

x = 4

That’s it.

### Shortcut to solve simultaneous equations

Suppose you have set of simultaneous equations as shown below:

a_{1}x+b_{1}y = c_{1} ————————————- 1

a_{2}x+b_{2}y = c_{2} ————————————- 2

X = ( c_{1} b_{2} – c_{2} b_{1} ) / ( a_{1} b_{2} – a_{2} b_{1} )

Using this to solve for x from the example 1 above

3x+2y = 10 ————————————— equation 1

x+2y = 2 —————————————- equation 2

a_{1} = 3

a_{2} = 1

b_{1} = 2

b_{2} = 2

c_{1} = 10

c_{2} = 2

putting it in the above formula

x = (10*2 – 2*2) / (3*2 – 1*2)

x = 20 – 4 / (6 – 2)

x = 16 / 4

x = 4.

You can now substitute for x in either equation 1 or 2 to get y.

Please share this lesson with your friends by clicking the Facebook share button or other social media button at the left corner of this post. thanks for reading and keep coming back for more mathematics lessons.

**See: **

**Subtract 265 from 317 in base eight**

**Evaluate 2115 base seven Divided by 12 base seven**

Last Updated on July 6, 2019 by Admin