Page 184 - Math Course 2 (Book 1)

Systems of Equations: Substitution

Mo. 6

Lesson 2
Let’s Begin

KEY CONCEPTS:
1. Solve systems of equations algebraically Solve Using Substitution
by using substitution.
2. Solve real-world problems involving
systems of equations. Examples

Use substitution to solve the system of equations.
x = 4y
MO. 6 - L2a 4x – y = 75

Solving Systems of Equations Since x = 4y, substitute 4y for x in the second
by Substitution equation.

4x – y = 75 Second equation
Vocabulary A-Z 4(4y) – y = 75 x = 4y

Let us learn some vocabulary
16y – y = 75 Simplify.

15y = 75 Combine like terms.
substitution 15y = 75
15 15 Divide each side by 15.
The exact solution of a system of equations can
be found by using algebraic methods. One such y = 5 Simplify.
method is called substitution.

3x + y = 8 and y = x – 4
so Use x = 4y to f nd the value of x.
3x + (x – 4) = 8
x = 4y First equation
3x + y = 8 and y = x – 4
x = 4(5) y = 5

x x x 1 1 1 x = 20 Simplify.
3x
= 1 1 1
–1 –1
y x 1 1 Answer The solution is (20, 5)
–1 –1

Use substitution to solve the system of equations. Find the value of x by substituting 12 – 4x for y in the
4x + y = 12 second equation.
–2x – 3y = 14
–2x – 3y = 14 Second equation
Solve the f rst equation for y since the coef cient
of y is 1.
–2x – 3(12 – 4x) = 14 y = 12 – 4x
4x + y = 12 First equation
–2x – 36 + 12x = 14 Distributive Property
4x + y – 4x = 12 – 4x Subtract 4x from each side.
10x – 36 = 14 Combine like terms.
y = 12 – 4x Simplify.
10x – 36 + 36 = 14 + 36 Add 36 to each side.

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