Home > Algebra > Simultaneous Equations

# Simultaneous Equations

When there are a set of two equations, we try to solve them together to get a solution. In other words, we try to find a common solution when two conditions occur together. For example, in a farm containing cows and chickens, there are 70 eyes and 110 legs. Find out how many cows and chickens each are in the farm.

This problem can be solved in four different ways:

1. Guess and Check method
2. Graphical method
3. Substitution method
4. Elimination method

The substitution and elimination methods are algebraic methods of solution. Let us look at each of the 4 methods to solve the problem.

## 1. Guess and Check method

Given there are 70 eyes, and both the cow and chicken have 2 eyes each, hence, there should be 70 ÷ 2 = 35 animals.

Let ‘x’ represent the number of cows and ‘y’ represent the number of chickens. So

x + y = 35

Cows have 4 legs and chickens have 2 legs. That means 4x is the number of legs of x cows, and 2y is the number of legs of y chickens. So

4x + 2y = 110

We now have two conditions (sets of equations) –

x + y = 35

4x + 2y = 110

To solve these equations, let us make a guess of number of cows = 15. So the number of chickens = 35 – 15 = 20.

The number of legs now will be 15×4 + 20×2 = 60 + 40 = 100, which is less than 110. We need an extra 10 legs.

So let us make another guess of 20 cows and 15 chickens. Now the number of legs will be 20×4 + 15×2 = 80 + 30 = 110 legs.

Hence there are 20 cows and 15 chickens in the farm.

Clearly this process can be a lengthy one, since it will involve guessing and checking the answers repeatedly. The other 3 methods are more efficient.

## 2. Graphical method

Here we draw the lines represented by the two simultaneous equations, and we try to find the meeting point of the lines.

Graph the lines y = 35 – x, and y =  =  55 – 2x on the same number plane. The intersecting points 20 and 15 are the answers to our problem.

When using the graphical method, sometimes, you will get a rough estimate rather than accurate values.

## 3. Substitution method

In this method, one pro numeral is replaced by an equivalent expression involving the other pro numeral.

Here we have the two equations –

1. x + y = 35
2. 4x + 2y = 110

Equation 1 can be rearranged to get y = 35 – x. We will substitute y = 35 – x in equation 2 to get –

4x + 2(35 – x) = 110

4x + 70 – 2x = 110

2x = 110 – 70

2x = 40

x = 20

Now put x = 20 in y = 35 – x to get y = 35 – 20 = 15.

So the solution is x = 20 and y = 15, meaning there are 20 cows and 15 chickens.

Note: it is always a good idea to check your answer with one of the equations.

## 4. Elimination method

In this method, one of the pronumerals is eliminated by making one of the coefficients of the pronumerals same. Then we add or subtract the equations.

1. x + y = 35
2. 4x + 2y = 110

Multiply equation 1 by 4 to get 4x + 4y = 140

Now subtract the two equations

 4x + 4y = 140 – 4x + 2y = 110 0 + 2y = 30

In the above step, we have eliminated ‘x’. So

2y = 30

y = 15.

Now putting y = 15 in either of the two equations, we get

x + y = 35

x + 15 = 35

x = 35 – 15  = 20

So x = 20 and y = 15 i.e. there are 20 cows and 15 chickens.

Note: To eliminate a pronumeral, the size of the co-efficient in each equation must be made the same by multiplying one or both equations by a constant.