Linear equations with two variables is one of the interesting topics in linear algebra. We used linear equations many times in our real life. Suppose, Katrina hear that a shopkeeper has put on sale a dress for 20% off on the original price. Knowing this is a bargain, she drives to shop, quickly pay Rs. 1500 selling price for the dress and came back to home. Later, this question comes to mind, what was the original price of the dress? We look for linear equations to get answer for such questions. This section includes a pair of linear equations questions organized with advanced level of difficulty which provide learners sufficient opportunity to apply your skills.

Questions on Linear Equations in Two Variables

Our subject experts have designed below listed questions to make you master solving linear equations. Different methods to solve linear equations are explained below with the help of examples.

Example 1: Solve 2a + 3b = 10; 4a + 5b = 4
Solution:
Step 1
: 2a + 3b = 10  .....(1)
4a + 5b = 4   ......(2)
Step 2:  Using elimination method to solve this system of equations
Multiply equation (1) with 2 and subtract from equation (2), we get
2(2a + 3b = 10) => 4a + 6b = 20

4a + 5b = 4
4a + 6b = 20
-     -        -
___________
0 - b = -16
___________

or b = 16

Step 3:  Find the value of a
Substitute value of a in equation (2)
4a + 5(16) = 4
4a + 80 = 4
4a = 76
or a = -76/4 =  - 19

Answer:
a = -19
b = 16

Example 2: Solve: 5x + 3y = 10; -x + y = 17
Solution:
Step 1:
5x + 3y = 10  ...(1)
-x + y = 17     ...(2)
Step 2:  Using substitution method to solve this system of equations
Solve solve (2) for y and substitute the value in equation (2). Get value of x
y = 17 + x
Now, 
5x + 3(17 + x) = 10
Use distributive property, to simplify the above equation
5x + 51 + 3x = 10
8x + 51 = 10
8x = 10 - 51
8x = -41
or x = -41/8
Step 3: Find the value of y
use value of x in equation (2) and get the value of y
-(-41/8) + y = 17
41/8 + y = 17
y = 17 - 41/8 =  95/8

Answer: 
x = -41/8
y = 95/8

Example 3: Solve for x and y: x + y = 4, 2x + 2y = 9
Solution: 
x + y = 4        ...(1)
2x + 2y = 9    ...(2)
Multiplying (1) by number 2, we get, 
2 ( x+ y ) = 2(4)
2x + 2y = 8  ......(3)
2x + 2y = 9  ......(4)
Subtracting , (2) from  (3), we get,   
 
0 = 1, which is not true.
Hence the system of equations have no solution.


Example 4: Solve pair of linear equations using elimination method.  x + 10y = 105 ; x + 15y = 155

Solution:
Step 1
:
x + 10y = 105  ....(1)
x + 15y = 155  ...(2)

Subtract equation (1) from (2)

x + 15y = 155
x + 10y = 105
-  -       = -
_____________
0 + 5y = 50
_____________

This implies 5y = 50
or y = 50/5 = 10

Find the value of x:
x + 10y = 105  (Put value of y)
x + 10(10) = 105
x + 100 = 105
x = 105 - 100 = 5
Answer:
x = 5
y = 10

Example 5: Solve the set of simultaneous equation using elimination

x + 2y = 8
2x + y = 15

Solution:

x + 2y = 8  ------- (1)

2x + y = 15  ------(2)

We multiply equation (1) with 2

We get 2x + 4y = 16

Now we subtract the second equation from it

 2x + 4y = 16
-2x - y = -15
------------------------
 3y = 1
 
 y = $\frac{1}{3}$
 
Now substitute this value in one of the equations and solve.

x + 2y = 8

x + 2 $(\frac{1}{3})$ = 8

x + $\frac{2}{3}$ = 8

x = 8 - $\frac{2}{3}$

x = $\frac{22}{3}$

Solution set is (x, y) = $(\frac{22}{3}, \frac{1}{3})$ 

Practice Problems

Below listed problems are for practice. Try it and excel your skills.
Problem 1:
Solve x - y = 6 and -2x + 2y = 1
Problem 2: Find the solution set for the following pair of linear equations: 5a + 10b = 9; -4a + b = 3
Problem 3: Solve 10x + 5y = 4 ; 3x - 2y = -3
Problem 4: Solve x - y = 10 and x + y = 30
Problem 5: Use substitution method to solve: 3x - y = 10 and 2x - 3y = 1
Problem 6: 3x - y = 10 and y - 2x = 2. Solve the system using graphic method.