Substitution method

Solving simultaneous equations by substitution method.

Writing one variable in terms of the other variables.

Find y in terms of x.

4x - y = 10

Solution

4x - y = 10

Subtract 4x form both sides of the equation.

4x - 4x -y = 10 - 4x

-y = 10 - 4x

Multiply both sides of the equation by -1

-y x - 1 + (10 - 4x) -1

y = -10 + 4x

In y = 4x - 10, y is expressed in terms of x.

b) 3x + 7y = 29

Solution

3x + 7y = 29

Subtract 3x form both sides of the equation.

3x - 3x + 7y = 29 - 3x

7y = 29 - 3x

Dividig both sides of the equation by 7.

7y = 29 - 3x

7 7

y = 29 - 3x

Find x in terms of y.

Subtracting 2y form both sides of the equation.

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

x = 5 - 2y

c) 3x- 6y = 10

adding 6y on both sides of the equation

3x - 6y +6y = 10 + 6y

3x = 10 + 6y

Dividing both sides of the equation by 3.

3x = 10 + 6y

3 3

x = 10 + 6y

The processof writing a variable in terms of the other is similar

to the process of solving linear equation with one unkown.

Solve the simultaneous equation by substitution method.

4x - 3y = 1

x - 4 = 2y

Let 4x - 3y= 1 be equation (i) and

x - 4 = 2y be equation (ii)

4x - 3y = 1 -------(i)

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

Using equation (ii) write x in terms of y.

x - 4 + 4 = 2y + 4

x = 2y + 4 Let this be equation (iii)

Substitute 2y + 4 for x in equation i)

4(2y + 4) - 3y = 1

8y + 16 - 3y = 1

8y - 3y + 16 = 1

5y + 16 = 1

5y + 16 - 16 = 1 – 16

5y = -15

5y 5

y = -3

Substituting -3 for y in either equations (i) or (ii) to get the value of x.

Using equation (i)

4x - 3y = 1

4x - 3(-3) = 1

4x -(-9) = 1

4x + 9 =1

4x + 9 - 9 = 1 -9

4x = -8

4 4

x = -2

Using Equation (ii)

x - 4 = 2y

x - 4 = 2(-3)

x - 4 = -6

x - 4 + 4 = -6 + 4

x = -2

If an equation is used to write one variabe in terms of the other, it cannot be used again for substitution.

Solve the simultaneous linear equations by substitution.

2t + 3s = 8

6t + 5 s = 16

Let 2t + 3s = 8 be equation (i) and

6t + 5s = 16 be equation (ii)

Using equation (i) write in terms of s

2t + 3s - 3s = 8 - 3s

2t = 8-3s

2 2

t = 8-3s

Let this equation be equation (iii)

Using equation (i) write t in terms of s.

Substitute 8-3s for t in equation (ii)

6(8-3s) + 5s = 16)2

Multiply both sides of the equation by 2.

6(8 - 3s) + 10S = 32

48 - 18S + 10S = 32

48 - 48 - 8S = 32 – 48

-8S = -16

-8 -8

S = 2.

Substituting 2 for s in equation either equation i, ii or iii to get the value of t in equation (iii)

t = 8 - 3(2)

t = 8 - 6

t = 2

t = 1

Solve the simultaneous linear equations by substitution method.

5x + 2y = 10

3y + 7x = 29

Let 5x + 2y be equation (i) and

3y + 7x = 29

5x + 2y = 10 -----------------(i)

3y + 7x = 29-----------------(i

Using Equation (i) write x in terms of y.

5x + 2y = 10

5x + 2y - 2y + 10 - 2y

5x = 10 - 2y

Divide both sides by 5

5x = 10 - 2y

5 5

x = 10 - 2y

Substituting x = 10 - 2y

for x in equation (ii)

3y + 7(10 - 2y)

= 29 5

Multiply both sides by 5.

(3y + 7(10 - 2y)

29) 5 5

15y + 7(10 - 2y) = 145

15y + 70 - 14y = 145

15y - 14y + 70 = 145

y + 70 = 145

y + 70 - 70 = 145 – 70

y = 75

Substitute 75 for y in equation (iii)

x = 10 - 2(75)

x = 10 - 150

x = -140

x = -28