# Word Problems on simultaneous linear equations

In this chapter we will discuss some problems related to the concept of simultaneous linear equations.

To solve the problems, you have to comprehend the statement carefully and then frame the linear equation.

After getting two linear equations, you then have to get the value of variables to get the final solution.

## Simultaneous linear equations word problems

Example 01
Twice one number minus three times a second is equal to 2, and the sum of these numbers is 11. Find the numbers.

Solution
Let the first number be x and second number be y.

(A) Let’s form the equation using the statements

First statement:
Twice the first number – three times second number equals two.

Forming the linear equation;
2x – 3y = 2.

Second statement
Sum of the two numbers is 11.

Linear equation becomes;
x + y = 11

(B) Solving the equations

Now we have two linear equations;
2x – 3y = 2
x + y = 11

Using substitution method.

Taking second equation and finding value of y.
x + y = 11

y = 11- x

Now putting this value of y in first equation.
2x – 3y = 2

2x – 3 (11 – x) = 2

2x – 33 + 3x = 2

5x = 2 + 33

5x = 35

x = 35 / 5

x = 7

Here we have got exact value of variable x.
Putting this value in any given equation.

Taking second equation
x + y = 11

7 + y = 11

y = 4

Hence, 7 & 4 are the two required numbers

Example 02
Six years hence a man’s age will be three times his son’s age, and three years ago he was nine times as old as his son. Find their present ages

Solution
Let the man’s present age be x and son’s present age be y.

(A) Forming the equations

First statement
After six years, man’s age will be three times the son age.

After 6 years;
Man age will be x + 6
Son age will be y + 6

At this point man’s age will be 6 times that of son.

(x + 6) = 3 (y + 6)

x + 6 = 3y + 18

x = 3y + 12

This is the first equation.

Second Statement
Three years ago men was nine times as old as his son

Age 3 years back;
Man age = x – 3
Son age = y – 3

At this time, men was 9 times as old as son.

x – 3 = 9(y – 3)

x – 3 = 9y -27

x – 9y = -24

This is our second equation

(B) Solving the equations

The two equations are;
x = 3y + 12
x – 9y = -24

Using substitution method.
Substitute the value x = 3y + 12 in second equation.

x – 9y = -24

3y + 12 – 9y = -24

-6y = -24 -12

-6y = -36

y = 36 / 6

y = 6 years.

Here we have got exact value of y. Put this value in first equation.

x = 3y + 12

x = 3 (6) + 12

x = 30 years

Hence, age of man is 30 years and age of son is 6 years.

Example 03
Two times a number plus ten times a second number equals twenty. Thirty times the second number plus three times the first number equals 45. Find the two numbers.

Solution
Let the first number be x and second number be y.

(A) Framing the equation

First statement
Two times a number plus ten times a second number equals twenty.

2x + 10y = 20

Second statement
Thirty times the second number plus three times the first number equals 45

3x+ 30y = 45

(B) Solving the equations

Here we have got two equations.
2x + 10y = 20
3x+ 30y = 45

We will solve the equations using substitution method.

Find the value of x in first equation.

\mathtt{2x+\ 10y\ =\ 20}\\\ \\ \mathtt{x=\frac{20-10y}{2}}\\\ \\ \mathtt{x\ =\ \frac{20}{2} -\frac{10y}{2}}\\\ \\ \mathtt{x=\ 10\ -\ 5y}

Put the value of x in 2nd equation.

3x + 30y= 45

3 (10 – 5y) + 30y = 45

30 – 15y + 30y = 45

30 + 15y = 45

15y = 15

y = 1

Here we have got exact value of y. Put this value in first equation to get the value of x.

2x + 10y = 20

2x+ 10 = 20

2x = 10

x = 5

Hence, the two numbers are 5 and 1.

Example 04
At a certain time in a deer park, the number of heads and the number of legs of deer and human visitors were counted and it was found that there were 41 heads and 136 legs. Find the number of deer and human visitors in the park.

Solution
Let number of deer be x and number of humans be y.

We know that deer has 4 legs and 1 head. So total number of deer legs and head for one deer will be 4x and x.

Total deer legs = 4x

We know that humans have 2 legs and 1 head. So count of heads and legs is given as;

Total human legs = 2y

(A) Forming equations

Statement 01
There were total of 41 heads.
x + y = 41

Statement 02
There are 136 legs
4x + 2y = 136

(B) Solving equations

Here we have got two equations;
x + y = 41
4x + 2y = 136

Using substitution method for solving linear equations.

Find value of x in first equation.
x + y = 41

x = 41 – y

Substitute this value of x in 2nd equation.

4x + 2y = 136

4 (41 – y) + 2y = 136

164 -4y + 2y = 126

-2y = 136 – 164

-2y = -28

y = 14

Now we have got exact value of y. Put this value in first equation to get the value of x.

x + y = 41

x + 14 = 41

x = 27

Hence, there are 27 deer and 14 humans.

Example 5
2 shirts and 3 trousers cost $425. 3 Shirts and 2 Trousers cost$ 350. Find the price of one short and trousers.

Solution
Let price of one shirt be x and price of trouser be y.

(A) Framing equations

First statement
2 shirts and 3 trousers cost $425 2x + 3y = 425 Second statement 3 Shirts and 2 Trousers cost$ 350

3x + 2y = 350

(B) Solving equations

Here we have got two equations.

2x + 3y = 425
3x + 2y = 350

Here we will use elimination method to solve equations.

Multiply first equation by 3 and second equation by 2.

First equation
3 ( 2x + 3y = 425 )

6x + 9y = 1275

Second equation
2 ( 3x + 2y = 350)

6x + 4y = 700

Sligh both equation vertically and subtract.

Solving the equation further.

5y = 575

y = 575 / 5

y = 115

Here we got exact value of y. Put this value in first equation to get value of x.

2x + 3y = 425

2x + 3 (115) = 425

2x + 345 = 425

2x = 80

x = 40

Hence, price of shirt is 40$and price of trouser is 115$