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.

Our subject experts have designed below listed questions to make you master solving linear equations. Different <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> methods to solve linear equations are explained below with the help of examples.

**Example 1:** Solve 2a + 3b = 10; 4a + 5b = 4

Step 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

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

Substitute value of a in equation (2)

4a + 5(16) = 4

4a + 80 = 4

4a = 76

or a = -76/4 = - 19

4a + 80 = 4

4a = 76

or a = -76/4 = - 19

a = -19

b = 16

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

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

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

x = -41/8

y = 95/8

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.

Step 1

x + 10y = 105 ....(1)

x + 15y = 155 ...(2)

Subtract equation (1) from (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

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

x = 5

y = 10

x + 2y = 8

2x + y = 15

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) = $(\frac{22}{3}, \frac{1}{3})$

Problem 1: