# Simultaneous Equations

Contents

## Motivation

The aim of this section is to understand what are simultaneous equations and how we can solve them? After reading this section we will be able to write down a word problem in the form of simultaneous equations and be able to find out the solution.

## Content

• Introduction
• Definition
• Types of Simultaneous Equations
• Methods for solving Simultaneous Equations.
• Who to verify the solution?
• Problems on Simultaneous Equations

## Introduction

Mathematics plays a vital role in our life; without mathematics many situations go wrong. There is no problem in physical science that can be solved without converting it into mathematics. Whenever a problem is converted to mathematics it gives an equation in some variable form. Like this situation if a person buys two cupcakes in £3, then what will be the price of one cupcake?

Here the problem in mathematical form is

2x = 3

Where x represents the price of cupcake. We can find the value of x by dividing 2 on both sides, but sometimes problems give the two or more equations. These equations involve two or more unknown variables, as x is an unknown value in above equation, which we have to determine.  These types of equations are called simultaneous equations.

Definition

Word simultaneous represents “at a same time”. So simultaneous equations are those equations which are correct for the certain values of unknown variables at a same time.

For examples

2x + y = 6

x + 2y = 9

Above equations are simultaneous equations in unknown variables ‘x’  and ‘y’. Both equations are true for x = 1 and y = 4. Let’s check that these equations are true or not?

2x + y = 6

Put x = 1 and y = 4

21 + 4 = 6

Now the other equation

x + 2y = 9

Now put x = 1 and y = 4

1 + 24 = 9

See both the equations are true at a same time for single value of x and y.

## Types of Simultaneous Equations

There are two types of simultaneous equations which we will see in this section.

1) Linear simultaneous equations

Linear simultaneous equations are called those equations in which power of each unknown variable is one. i.e.

x = 2y        5x – y = 15

2) Nonlinear simultaneous equations

Nonlinear simultaneous equations are those equations in which power of at least one unknown variable must be greater than one. i.e.

$x^{2}+y = 5$       x + y = 5

In above set first equation comes with second degree so this set will be called nonlinear simultaneous equations.

Note: An equation involves a variable with second degree is known as quadratic equation. In addition, above nonlinear equation is also quadratic simultaneous equations.

## Methods for solving Simultaneous Equations.

There are well known three methods we use to solve simultaneous equations, as are listed below.

1) Elimination Method: In this method we eliminate one variable to find the value of other variable.

In this method first we multiply both equations with different numbers to make coefficient same of any one variable and then subtract these equations, after subtraction one variable vanishes out so that we can find the value of another unknown variable easily. After finding out the value of one unknown variable we put this in any one equation and find out the other equations. We will see this method in examples.

Example 1: Solve the simultaneous equations 2x + 3y = 8   and 3x + 2y = 7

Solution:

First give the name to both equations.

2x + 3y = 8       (1)

3x + 2y = 7       (2)

We will solve these equations by elimination method. To eliminate the one unknown variable, we make the coefficient same of one variable, here we are going to eliminate the unknown variable ‘x’ first multiply equation (1) by 3 and equation (2) by 2. (multiplying by coefficient of ‘x’ in equation (1) with equation (2) and coefficient of ‘x’ in equation (2) by equation (1) is an easy way to make coefficient same

3×1⇒  6x+9y=24

2×2⇒  6x+4y=14

Now subtract equation (2) from equation (1)

In this step ‘x’ eliminates, we get the equation in term of ‘y’ only. Now divide by ‘5’ on both side.

$\frac{5y}{5}=\frac{10}{5}$

Its gives

y = 2

Now put the value of ‘y’ in any equation, we get the same results, let’s put the value of ‘y’ in equation (1)

2x + 3(2) = 8

⇒  2x + 6 = 8

Subtracting ‘6’ on both sides

2x + 6 – 6 = 8 – 6

2x = 2

Dividing by ‘2’ on both sides of the above equation, we get:

$\frac{2x}{2}=\frac{2}{2}$

x = 1

The solution of simultaneous equation is x = 1  and y = 2.

2) Substitution Method: In this method first we write any one unknown variable in terms of second unknown variable from one equation. Then substitute the value of this variable in the other equation. Then one variable vanishes out and we find out the value of other one. After it, the procedure is same as discussed in elimination method. We will see this method in the following example.

Example 2: Solve the simultaneous equations $x^{2}+y^{2}=10$ and x + y = 4

Solution:

Since it is nonlinear simultaneous equations, we first list them by numbering.

$x^{2}+y^{2}=10$       (1)

x + y = 4          (2)

We solve them by substitution method, firstly write down the equation two in term of ‘x’ only,

x = 4 – y     (3)

Now substitute the value of ‘x’ from equation (1) to equation (2)

$(4-y)^{2}+y^{2}=10$

By using the formula $(a-b)^{2}=a^{2}-2ab+b^{2}$

$16-8y+y^{2}+y^{2}=10$

Adding same terms and rearranging above equation

$2y^{2}-8y+16=10$

Subtracting ’10’ on both side

$2y^{2}-8y+16-10=10-10$

$2y^{2}-8y+6=0$

Dividing by two ‘2’ on both sides

$\frac{2y^{2}}{2}-\frac{8y}{2}+\frac{6}{2}=0$

$y^{2}-4y+3=0$

Since it is a quadratic equation in term of ‘y’, we can solve it by factorization.

$y^{2}-3y-y+3=0$

Taking common similar terms

$y(y-3)-1(y-3)=0$

$(y-3)(y-1)=0$

From above equation we can write as

y – 3 = 0   or   y-  1 = 0

Rearranging above equations we get

y = 3 or y = 1

So, here is the values of ‘y’, now put these values in equation (3) one by one

Put y = 3

x = 4 – 3 = 1

Now put y = 1

x = 4 – 1 = 3

So, there are two different answers, one is x = 1, when y = 3 and the other is x = 3, when y = 1.

3) Graphical Method: In this method we draw the graph of each line and trace out the intersection of these lines. Basically, this intersection is the solution of these equation. Normally we use graphically method for linear. In nonlinear simultaneous equations graphically, method is not so effective because its solution out in surd form.

Example 3: Solve the simultaneous equations by graphical method. 6x + y = 40;   4x + 3y = 36

Solution:

First of all, we draw the graph of both equation one by one and then trace out the intersection of lines, which will be the our required solution.

In this graph point A representing the point of intersection. At point A the value of x-axis is 6 and y-axis is 4, so point of intersection is 6,4, which is the required solution.

Who to verify the solution?

Many times, we find the solution but forget to check that even it is true or false. In mathematics it is very important to find out the collect solution for carrying good grades. So, for checking the solution that it is true of false we put the answer or values of unknown variables in both equation and see that either both sides are same or not, if same then our solution is correct, if that then we have to check our calculation again.

For example, in the last section we find out the solution of 6x + y = 40;   4x + 3y = 36, which is x = 6 and y = 4. Put the values in both equation one by one and see that either it is correct or not.

Put in first equation.

66 + 4 = 36 + 4 = 40

Now in second equation

46 + 34 = 24 + 12 = 36

Since values of x and y satisfy both equations, so our solution is correct.

## Word Problems for Simultaneous Equations

Problem 1: If Jon bought three packets of chips and 2 packets of biscuits in £29, and Charlie bought one packet of chips and seven packets of biscuits in £54. Then what is the price of each packet of chips and biscuits?

Solution:

Here we first give the name to the objects for which we have to determine the price.

Let ‘x’ is the price of one packets of chips and ‘y’ is the price of one packet of biscuit.

Now write down the statement in mathematical form step by step.

Mathematical expression for Jon:

Jon bought three packets of chips so, amount for chips will be ‘3x’ and he bought two packets of biscuits, amount for biscuits will be ‘2y’. According to statement he spends £29 on chips and biscuits. Mathematical expression for Jon is written below.

3x + 2y = 29      (1)

Mathematical expression for Charlie:

Jon bought one packets of chips so, amount for chips will be ‘x’ and he bought seven packets of biscuits, amount for biscuits will be ‘7y’. According to statement he spends £54 on chips and biscuits. Mathematical expression for Charlie is written below.

x + 7y = 54       (2)

From equation (1) and equation (2) we will determine the value of x and y. Since these are the simultaneous equations. So, we can solve them elimination method.

According to method first make the coefficient same of a one variable, here we make the same coefficient of x.

3×(2)⇒   3x + 21y = 162

Subtract the equation (1) from the above equation.

Dividing on both sides by ’19’

$\frac{19y}{19}=\frac{133}{19}$

y = 7

Put the value of y = 7, in equation (2)

x + 7(7) = 54

x + 49 = 54

x = 54 – 49

x = 5

So, the price of one packet of chips is £5 and price of one packet of biscuit is £7.

Problem 2: If the sum money in the pocket of two person A and B is $8 and sum of square their amount is$34, then how much amount each person has?

Solution:

Let consider Person A have x and Person B have y in their pockets.

Then mathematical form for first condition, which is the sum of their amount is 8

x + y = 8      (1)

Then mathematical form for first condition, which is the sum of square their amount is 34

$x^{2}+y^{2}=34$       (2)

Rearranging equation (1), we get

y = 8 – x

Put this value of y in equation (2)

$x^{2}+(8-x)^{2}=34$

$x^{2}+64+x^{2}-16x=34$

Collecting same terms

$2x^{2}-16x+64-34=0$

$2x^{2}-16x+30=0$

Dividing by 2 on both side

$x^{2}-8x+15=0$

$x^{2}-5x-3x+15=0$

$x(x-5)-3(x-5)=0$

$(x-5)(x-3)=0$

x – 5 = 0    or     x – 3 = 0

⇒   x=5   or    x=3

Put the value of x, in y

For x = 5

y = 8 – 5 = 3

For x = 3

y = 8 – 3 = 5

Here is the solution x = 5 , when y = 3 and x = 3, when y = 5.

Its mean one of them have $3 and one of them have$5.

Problem 3: Solve the following Nonlinear Systems of Equations.

x + y = 2         (1)

$6x^{2}+3y^{2}=12$     (2)

Solution:

Rearrange the equation (1), we will get

y = 2 – x

Put this value in equation (2)

$6x^{2}+3(2-x)^{2}=12$

$6x^{2}+3(4-4x+x^{2})=12$

$6x^{2}+12-12x+3x^{2}-12=0$

$9x^{2}-12x=0$

$3x(3x-4)=0$

3x = 0     or       3x – 4 = 0

x = 0        or      x = 43

Now put the values of x into y = 2 – x, one by one

x = 0  ⇒ y = 2 – 0 = 2

$x=\frac{4}{3}$   ⇒  $y=2-\frac{4}{3}=\frac{6-4}{3}=\frac{2}{3}$

When x = 0, then y = 2 and when $x=\frac{4}{3}$ then $y=\frac{2}{3}$.

Sources:

1. Pure mathematics 1 by Sophie Goldie
2. Advance Level Mathematics (Pure Mathematics 1) by Hugh Neill and Douglas Quadling