**Solving Linear Equations Using Elimination Method :**

In this section, we will learn, how to solve linear equations using elimination method.

**Step 1 :**

By taking any one equations from the given two, first multiply by some suitable non-zero constant to make the co-efficient of one variable (either x or y) numerically equal.

**Step 2 :**

If both coefficients which are numerically equal of same sign, then we may eliminate them by subtracting those equations.

If they have different signs, then we may add both the equations and eliminate them.

**Step 3 :**

After eliminating one variable, we may get the value of one variable.

**Step 4 :**

The remaining variable is then found by substituting in any one of the given equations.

**Example 1 :**

Solve the following pairs of equations by reducing them to a pair of linear equations

(1/2x) + (1/3y) = 2 and (1/3x) + (1/2y) = 13/6

**Solution :**

**Let 1/x = a, 1/y = b**

** (1/2)(1/x) + (1/3)(1/y) = 2**

**(a/2) + (b/3) = 2**

**(3a + 2b)/6 = 2**

**3a + 2b = 12 -----(1)**

**(1/3x) + (1/2y) = 13/6**

**(a/3) + (b/2) = 13/6**

**(2a + 3b)/6 = 13/6**

**2a + 3b = 13**** -----(2)**

3a + 2b = 12 -------- (1)

2a + 3b = 13 -------- (2)

(1) ⋅ 3 => 9 a + 6 b = 36

(2) ⋅ 2 => 4 a + 6 b = 26

(-) (-) (-)

--------------------

5 a = 10 ==> a = 2

By applying the value of a in (1), we get

3(2) + 3b = 13

6 + 3 b = 13

3 b = 13 - 6

3 b = 7

b = 7/3

1/x = a 1/x = 2 x = 1/2 |
1/y = b 1/y = 7/3 y = 3/7 |

**Example 2 :**

Solve the following pairs of equations by reducing them to a pair of linear equations

(2/√x) + (3/√y) = 2 and (4/√x) – (9/√y) = -1

**Solution :**

**Let 1/x = a, 1/y = b**

1/√x = a 1/√y = b

2 a + 3 b = 2 ------(1)

4 a – 9 b = -1 ------(2)

(1) ⋅ 3 = > 6 a + 9 b = 6

4 a – 9 b = - 1

-------------------

10 a = 5 ==> a = 1/2

By applying the value of a in (1), we get

2(1/2) + 3 b = 2

1 + 3 b = 2

3 b = 2 – 1

3 b = 1

b = 1/3

From the values of a and b, we may solve for x and y.

1/√x = 1/2 2 = √x x = 2 x = 4 |
1/√y = 1/3 3 = √y y = 3² y = 9 |

**Example 3 :**

Solve the following pairs of equations by reducing them to a pair of linear equations

(4/x) + 3 y = 14 and (3/x) – 4 y = 23

**Solution :**

Let 1/x = a and y = b

4a + 3b = 14 ------(1)

3a – 4b = 23 ------(2)

(1) ⋅ 4 => 16 a + 12 b = 56

(2) ⋅ 3 = > 9 a – 12 b = 69

-----------------

25a = 125 ===> a = 5

By applying the value of a in (1), we get

4(5) + 3 b = 14

3 b = 14 – 20

3 b = -6

b = -2

1/x = a 1/x = 5 x = 1/5 |
y = b y = -2 |

After having gone through the stuff and examples, we hope that the students would have understood, how to solve linear equations using elimination method.

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