Substitution Definition

Substitution is a fundamental mathematical technique used to solve equations and simplify complex expressions by replacing one term or variable with an equivalent expression. This powerful method is widely applied in algebra, particularly for solving systems of equations, and is often used in conjunction with other problem-solving approaches. This article explores the concept of substitution, its applications, and how it compares to other mathematical techniques.

Definition of Substitution

It is a technique in mathematics that involves replacing a term or expression with its equivalent definition. This process is used to simplify, clarify, or transform mathematical statements

Algebraic Substitution Method

Algebraic substitution method is a technique used in mathematics to solve systems of equations. This method involves using one equation to express a variable in terms of another, and then substituting that expression into a second equation.

Steps for Substitution Method

Step 1: Choose one equation and isolate a variable.

Step 2: Substitute the expression for that variable into the other equation(s).

Step 3: Solve the resulting equation for the remaining variable.

Step 4: Use the solution to find the value of the other variable.

This method is particularly useful when dealing with linear equations or when one equation is simpler than the other. Substitution can often simplify complex problems and make them more manageable.

For example, consider the system of equations: x + y = 10 and 2x - y = 4

Using substitution, we could isolate y in the first equation: y = 10 - x

Substitute this expression for y into the second equation: 2x - (10 - x) = 4

Solving this equation will give us the value of x, which we can then use to find y.

Substitution Method by Elimination

Elimination by substitution method is a powerful technique used to solve systems of equations in mathematics. This method combines elements of both substitution and elimination approaches.

Steps for Substitution Method by Elimination

Step 1: Start with a system of two or more equations.

Step 2: Choose one equation and isolate one variable in terms of the others.

Step 3: Substitute this expression into the other equation(s) to eliminate that variable.

Step 4: Solve the resulting equation for one of the remaining variables.

Use this solution to find the values of the other variables by back-substitution.

NOTE: This method is particularly useful when:

• One equation is simpler or easier to rearrange than the others
• Direct elimination would require complex multiplication of equations

The elimination by substitution method can be more efficient than pure substitution or elimination in certain cases. It allows for strategic choices in how to manipulate the equations, potentially simplifying the solving process.

For instance, consider this system:

• 3x + 2y = 12
• 5x - 3y = 7

We could isolate y in the first equation: y = (12 - 3x) / 2

Then substitute this into the second equation to eliminate y: 5x - 3((12 - 3x) / 2) = 7

Solving this equation would give us x, which we could then use to find y.

Difference Between Substitution and Elimination Method

The difference between substitution method and elimination method can be understood from the table given below.

Examples on Substitution

Substitution can be used in different areas. In this section we will see how to use substitution method to solve problems in different scenarios:

Substitution in Geometry

Find the area of a triangle with vertices at (0, 0), (a, 0), and (b, c).

Solution:

Using the formula for the area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃):

Area=1/2 ∣x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)∣

Substitute the given points (0, 0), (a, 0), and (b, c):

Area=1/2 ∣0(0 - c) + a(c - 0) + b(0 - 0)∣

Area=1/2​ ∣ac∣

Area=1/2ac

Solving a Quadratic Equation

4x⁴-12x²+9=0

Solution:

Let u=x²

Now the equation becomes: 4u²-12u+9=0

(2u-3)²=0

2u-3=0

2u=3

u=3/2

Now putting back u=x²

∴ x²=3/2

On solving for x we get,

x=​ ±\sqrt{\frac{3}{2}}

Practice Questions on Substitution

Q1. Given f(x) = 2x² - 3x + 1, find f(-2).

Solution:

f(-2) = 2(-2)² - 3(-2) + 1

= 2(4) + 6 + 1

= 8 + 6 + 1

= 15

Therefore, f(-2) = 15.

Q2. If a = 3 and b = 2, evaluate 2a² - 3b + 1.

Solution:

2a² - 3b + 1 = 2(3)² - 3(2) + 1

= 2(9) - 6 + 1

= 18 - 6 + 1

= 13

Therefore, the expression evaluates to 13.

Q3. Solve for x: 2x + 3 = 11

Solution:

2x + 3 = 11

2x = 11 - 3

2x = 8

x = 8 ÷ 2

x = 4

Therefore, the solution is x = 4.

Q4. Given that sin θ = 0.6, find cos θ using the identity sin²θ + cos²θ = 1.

Solution:

sin²θ + cos²θ = 1

(0.6)² + cos²θ = 1

0.36 + cos²θ = 1

cos²θ = 1 - 0.36

cos²θ = 0.64

cos θ = √0.64

cos θ = 0.8

Therefore, cos θ = 0.8.

Q5. If f(x) = x² + 2 and g(x) = 3x - 1, find f(g(2)).

Solution:

g(2) = 3(2) - 1 = 6 - 1 = 5

f(g(2)) = f(5) = 5² + 2 = 25 + 2 = 27

Therefore, f(g(2)) = 27.

Q6. Solve the system of equations: x + y = 7 y = 2x - 3

Solution:

Substitute y = 2x - 3 into x + y = 7:

x + (2x - 3) = 7

3x - 3 = 7

3x = 10

x = 10/3

Substitute x = 10/3 into y = 2x - 3:

y = 2(10/3) - 3 = 20/3 - 3 = 20/3 - 9/3 = 11/3

Therefore, x = 10/3 and y = 11/3.

Q7. Given h(t) = t² - 4t + 5, find the value of t for which h(t) = 2.

Solution:

t² - 4t + 5 = 2

t² - 4t + 3 = 0

Using the quadratic formula: t = [-b ± √(b² - 4ac)] / 2a

t = [4 ± √(16 - 12)] / 2 = (4 ± 2) / 2

t = 3 or t = 1

Therefore, h(t) = 2 when t = 3 or t = 1.

Q8. If P = 2πr and r = 5, find P.

Solution:

P = 2πr

P = 2π(5)

P = 10π

Therefore, P = 10π.

Q9. Given the function f(x) = √(x + 3), find f(6).

Solution:

f(6) = √(6 + 3)

= √9

= 3

Therefore, f(6) = 3.

Q10. If v = u + at, and u = 10 m/s, a = 2 m/s², t = 5 s, find v.

Solution:

v = u + at

v = 10 + 2(5)

v = 10 + 10

v = 20 m/s

Therefore, v = 20 m/s.