SOLVING SYSTEM OF EQUATIONS USING SUBSTITUTION

Example 1 :

Find all solutions to the given system of equations.

x² − 2y² = 3

x + 2y = 1

Solution :

x² − 2y² = 3   -------(1)

x + 2y = 1   -------(2)

x  =  1 - 2y

Substitute x = 1 - 2y in (1)

(1 - 2y)² - 2y²  =  3

1 + 4y² - 4y - 4y²  =  3

-4y  =  3 - 1

-4y  =  2

y  =  2/(-4)  =  -1/2

x  =  1 - 2(-1/2)

  =  1 + 1

x  =  2

So, the solution is

(2, -1/2)

Example 2 :

Find all solutions to the given system of equations.

(1/x) − (1/y)  =  2

4x + y = 3 

Solution :

(1/x) − (1/y)  =  2  ----(1)

4x + y = 3  ----(2)

y  =  3 - 4x

Substitute y = 3 - 4x in (1)

 (1/x) - (1/(3 - 4x))  =  2

(3 - 4x - x)/(x(3-4x))  =  2

(3 - 5x)/(3x - 4x²)  =  2

3 - 5x  =  2(3x - 4x²)

3 - 5x  =  6x - 8x²

8x² - 6x - 5x + 3  =  0

8x² - 11x + 3  =  0

(8x - 3) (x - 1)  =  0

x  =  3/8 and x  =  1

So, the solutions are

(3/8, 3/2) and (1, -1)

Example 3 :

Find all solutions to the given system of equations.

x² + 3y² = 5

x − 3y = 2

Solution :

x² + 3y² = 5  ----(1)

x − 3y = 2  ----(2)

x  =  2 + 3y

Substitute x = 2 + 3y in (1)

(2 + 3y)² + 3y²  =  5

4 + 9y² + 2(2)(3y) + 3y²  =  5

12y² + 12y + 4 - 5  =  0

12y² + 12y - 1  =  0

a  =  12, b  =  12 and c  =  -1

y  =  -b ± √(b² - 4ac) / 2a

y  =  -12 ± √(12² - 4(12) (-1) / 2(12)

y  =  -12 ± √(144 + 48) / 24

y  =  (-12 ± √192) / 24

y  =  (-12 ± 8√3) / 24

y  =  4(-3 ± 2√3) / 24

y  =  (-3 ± 2√3) / 6

y  =  (-3 + 2√3)/6   (or)  y  =  (-3 - 2√3)/6

So, the solutions are

((1 + 2√3)/2, (-3 + 2√3)/6) and ((1 - 2√3)/2, (-3 - 2√3)/6)

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