Until now, we have solved linear equations with two variables. Let's see how we can solve two linear equations having two variables x and y.

Solve x + y = 5

x - y = 1

If an equation contains two variables, there are a number of ordered pairs that satisfy the given equation.

To solve an equation having two variables, we need two independent equations, i.e., to determine the ordered pairs that satisfy both equations.

Each ordered pair that satisfies a pair of equations is called the solution of that pair of equations.

Method of elimination using addition or subtraction

**Example 1**

Solve x + 2y = 5

x - 2y = 1

In this method, we eliminate one variable by addition or subtraction. The equation consists of the other variable after elimination.

**Step 1: **

x + 2y = 5

x - 2y = 1

The coefficient is the numerical value assigned to the variables. In this case, the coefficient of y is 2 in both the equations but with different signs.

**Step 2: **

If we add the two equations, the variable y will be eliminated.

x + 2y = 5 ------------- (1)

x - 2y = 1 ------------(2)

Add equations (1) and (2)

to get 2x = 6

x = 6/2 = 3

**Step 3: **

Substituting the value of x, i.e., 3, in equation (1) or (2)

x + 2y = 5

3 + 2y = 5

2y = 5 - 3 = 2

2y = 2

y = 1

**Step 4:**

If x = 3 then y = 1

Solution set = { (3,1) }

**Verification:**

We have to verify whether or not the solution set (3,1) satisfies both equations.

Substitute x = 3 and y = 1 into both equations

x + 2y = 5 x - 2y = 1

3 + 2 = 5 ∗3 - 2 = 1

The solution (3, 1) is correct.

**Example 2**

Solve x + 3y = 7, 3x + 4y = 11

In this example, we cannot eliminate x and y by addition or subtraction because the coefficients of x and y are different.

If every term in an equation is multiplied with a real number, the value of the equation remains unaltered.

Therefore, to make the coefficients of either x or y the same, we should multiply each equation with a number so that the value of the coefficient in both equations becomes the same.

**Solution:**

Step 1:21

x + 3y = 7 ------------ (1)

3x + 4y = 11 ------------ (2)

Multiply all the terms of the first equation with 3 to make the coefficients of x equal in both the equations.

x + 3y = 7 ----------- (1)

3x + 4y = 11 ---------- (2)

Equation (1) gives

3x + 9y = 21 ------------(3)

Subtracting the equation (3) from (2), we get

- 5y = - 10

y = -10/ -5 =2

y = 2

Substitute y = 2 in any one of the given equations to find x.

x + 3y = 7

x + 3 (2) = 7

x + 6 = 7

Therefore x = 1 and y = 2

Solution set = { (1,2) }

**Verification:**

Substitute x = 1 and y = 2 in the given equations

x + 3y = 7 3x + 4y = 11

1 + 3 (2) = 7 3 (1) + 4 (2) = 11

The solution (1,2) satisfies both the equations.

Therefore, the solution is correct.

In this method, we express one variable (say x) in terms of the other variable (say y) in one of the equations and solve the resulting equation for the value of x.

**Example 1**

Solve 3x + 2y = 8 ---------------(1)

2x + 3y = 7 ----------------(2)

**Solution:**

**Step 1: **

Find the value of x in terms of y

2x + 3y = 7

2x = 7 - 3y

x = (7-3y)/2 ------------------(3)

**Step 2: **

The equation is a simultaneous linear equation. So the values of x and y are the same in both equations.

Substitute the values of x in equation (1)

3[ 7-3y / 2 ] + 2y = 8

**Step 3:**

Multiplying on both sides by 2

3(7 - 3y) + 2 ∗ 2y = 2 ∗8

21 - 9y + 4y = 16

-5y = 16 - 21

-5y = -5

y = -5/-5 =1

**Step 4: **

Substituting the value of y in equation (3) we get the value of x

x = 7-3 ∗ 1 / 2 = 7-3/ 2 = 4/2 = 2

x = 2; y = 1

Therefore, the solution set = {(2,1)}

**Example 2: **

Solve 1/x + 1/y = 9 -------------- (1)

3/x +2/y = 22 -------------- (2)

**Solution:**

Let 1/x = a and 1/y = b

The given equations can be written as

a + b = 9 -------------- (3)

3a + 2b = 22 ----------------(4)

Multiply the equation (3) by 2

2a + 2b = 18 --------------- (5)

Subtracting (5) from (4)

2a + 2b = 18 ----------------(5)

+3a + 2b = 22 ------------------(4)

Subtract (4) from (5), we get

-a = -4

a = 4

Substitute a = 4 in equation (3)

4 + b = 9

b = 9 - 4 = 5

But a = 1/x

therefore, 1/x = 4

and x =1/4

b = 1/y

therefore, 1/y = 5

and y = 1/5.

Solution set = {(1/4, 1/5)}

Verify whether or not { 1/4, 1/5 } is a solution set.

**Example 3**

Solve (x + y)/2 - (x - y )/3 = 6 -------------- (1)

(x - 3y)/2 + (x + 2y)/6 = 1 -------------- (2)

**Solution:**

Write both equations in the form ax + by + c = 0

(x + y)/2 - (x - y)/3 - 6 = 0

The L.C.M. of 2 and 3 is 6

Multiply the equation by 6

(x + y)/2 ∗ 6 - (x - y)/3 ∗ 6 - 6 ∗ 6 = 0

3( x + y ) - 2 ( x - y ) - 36 = 0

x + 5y - 36 = 0 ---------------- (3)

(x - 3y)/2 + (x + 2y)/6 = 1

The L.C.M. of 2 and 6 is 6

Multiply the equation by 6

(x - 3y)/2 ∗ 6 + (x + 2y)/6 ∗ 6 - 1∗ 6 = 0

3( x - 3y ) + x + 2y - 6 = 0

3x - 9y + x + 2y - 6 = 0

4x - 7y - 6 = 0 ---------------- (4)

Solve equations (3) and (4) for the values of x and y

Solution set = { (12,6) }

If the equation ax + by + c = 0 is multiplied by a constant k, the resulting equation is k ax + k by + kc = 0.

The equation ax + by + c = 0 and k ax + k by + kc = 0 are known as dependent equations.

**Example 1**

Solve 8x - 5y = 3 ---------------(1)

16x - 10y = 6 ------------------ (2)

In the above example, the first equation is multiplied by 2 to get the second equation.

Therefore, 8x - 5y = 8 and 16x - 10y = 6 are dependent equations.

8x - 5y = 3----------------(1) 16x - 10y = 6 -------------- (2)

8x = 3 + 5y 16x = 6 + 10y

x = (3+5y)/8 -------------- (3) x = (6+ 10y)/16 -------------(4)

Solving the equation for the value of y

(3+ 5y) / 8 = (6 + 10y) / 16

Cross-multiplying

48 + 80y = 48 + 80y

80y = 80y

The equation is true for any value of y.

For every value of y, we can determine the corresponding value of x from equation (1) or (2). The ordered pair that satisfies equation (1) will also satisfy equation (2).

The solution set for the above equations is infinite.

**Find the solution sets of the following systems of equations**

- x + y = 6

x - y = 2

**Answer:**x + y = 6 ________ (1)

x - y = 2 ________ (2)

Adding (1) & (2) 2x = 8

Therefore x =8/2= 4

Substituting x = 4 in (1) we get 4 + y = 6

Therefore y = 6 - 4 = 2

Therefore solution set = { (4, 2) } - x + y = 7

x - y = 5

**Answer:**x + y = 7 ________ (1)

x - y = 5 ________ (2)

Adding (1) & (2) 2x = 12

Therefore x = 12/2= 6

Substituting x = 6 in (1) we get

6 + y = 7

Therefore y = 7 -6 = 1

Therefore solution set = { (6, 1) } - 4x - y = 5

4x + 4y = 20**Answer:**4x - y = 5 ________ (1)

4x + 4y = 20 ________ (2)

Subtract 1 & 2

-5y = -15

y = 3

Substituting y = 3 in equation 1

4x - 3 = 5

4x = 8

x =8/4

x =2

Therefore solution set = {( 2, 3)} - a - b = 5

a + b = 9

**Answer:**a - b = 5 ________ (1)

a + b = 9 ________ (2)

Adding (1) & (2) 2a = 14

Therefore a = 14/2= 7

Substituting a = 7 in (1) we get

7 - b = 5

Therefore b = 7 - 5 = 2

Therefore solution set = { (7, 2) } - 2x + 2y = 10

3x - 2y = 5**Answer:**2x + 2y = 10 ________ (1)

3x - 2y = 5 ________ (2)

Adding (1) & (2) 5x = 15

Therefore x = 15/5= 3

Substituting x = 3 in (1) we get

2 ∗3 + 2y = 10

Therefore 2y = 4, y = 2

Therefore solution set = { (3, 2) }

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