## SOLVING SYSTEM OF EQUATIONS USING SUBSTITUTION

Solving System of Equations Using Substitution :

In this section, you will learn how to solve system of equations using substitution method.

## Solving Systems of Equations Using Substitution - Procedure

(a) Use one of the equations in the system of equations to solve for one of the variables in terms of the other variables.

(b) Substitute the expression obtained in the previous step into the other equations, resulting in a new system of equations with one less variable and one less equation.

(c) Repeat the first two steps until you can solve the remaining system.

(d) Then substitute the values you have found into the previously obtained equations to get the complete solutions.

## Solving System of Equations Using Substitution - Examples

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 - 4y =  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

 If x  =  3/8y  =  3 - 4(3/8)y  =  3 - (3/2)y  =  (6 - 3)/2y  =  3/2 If x  =  1y  =  3 - 4(1)y  =  3 - 4y  =  -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

 y  =  (-3 + 2√3)/6 x  =  2 + 3yx  =  2 + 3(-3 + 2√3)/6 x  =  2 + (-3 + 2√3)/2  =  (4 - 3 + 2√3)/2  =  (1 + 2√3)/2 y  =  (-3 - 2√3)/6 x  =  2 + 3yx  =  2 + 3(-3 - 2√3)/6 x  =  2 + (-3 - 2√3)/2  =  (4 - 3 - 2√3)/2  =  (1 - 2√3)/2

So, the solutions are

((1 + 2√3)/2, (-3 + 2√3)/6) and ((1 - 2√3)/2, (-3 - 2√3)/6) After having gone through the stuff given above, we hope that the students would have understood how to solve system of equations using substitution.

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