SOLVING LINEAR EQUATIONS IN TWO VARIABLES BY SUBSTITUTION

Solve the following pairs of linear equations by substitution.

Example 1 :

x + y = 14 and x - y = 4

Solution :

x + y = 14 ----(1)

x - y = 4 ----(2)

In (1), solve for y in terms of x.

y = 14 - x

Substitute y = 14 - x.

x - (14 - x) = 4

x - 14 + x = 4

2x - 14 = 4

Add 14 to both sides.

2x = 18

Divide both sides by 2.

x = 9

Substitute x = 9 in y = 14 - x.

y = 14 - 9

  y = 5

So, the solution is (x, y) = (9, 5).

Example 2 :

s - t = 3 and s/3 + t/2 = 6

Solution :

s - t = 3 ----(1)

s/3 + t/2 = 6 ----(2)

In (1), solve for s in terms of t.

s = 3 + t

Substitute s = 3 + t in (2).

(3 + t)/3 + t/2 = 6

Least common multiple of (3, 2) is 6. Multiply both sides of the equation by 6 to get rid of the denominators 3 and 2.

6[(3 + t)/3 + t/2] = 6(6)

6[(3 + t)/3] + 6(t/2) = 36

2(3 + t) + 3t = 36

6 + 2t + 3t = 36

6 + 5t = 36

Subtract 6 from both sides.

5t = 30

Divide both sides by 5.

t = 6

Substitute t = 6 in s = 3 + t.

s = 3 + 6

s = 9

So, solution is (s, t) = (9, 6).

Example 3 :

3x - y = 3 and 9x - 3y = 9

Solution :

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

9x - 3y = 9 ----(2)

In (1), solve for y in terms of x.

3x - y = 3

-y = -3x + 3

y = 3x - 3

Substitute y = 3x - 3 in (2).

9x - 3(3x - 3) = 9

9x - 9x + 9 = 9

9 = 9

In the last step, the variable is no more. and also the result 9 = 9 is a true statement. So, the given pair of linear equations has infinitely many solutions.

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