Tier 5: Systems of Equations

Level 5.1 – Substitution Method

Solve systems by substituting one equation into the other.

\( y = 2x + 1 \), \( 3x + y = 16 \)

Substitute: \( 3x + (2x + 1) = 16 \Rightarrow 5x = 15 \Rightarrow x = 3 \)
\( y = 2(3) + 1 = 7 \) → Solution: \( (3, 7) \)

\( y = x - 4 \), \( x + y = 10 \)

Substitute into second: \( x + (x - 4) = 10 \Rightarrow 2x = 14 \Rightarrow x = 7 \)
\( y = 3 \)

\( x = 2y \), \( 3x + y = 14 \)

\( 3(2y) + y = 14 \Rightarrow 7y = 14 \Rightarrow y = 2 \), \( x = 4 \)

\( y = -x + 5 \), \( 2x + y = 1 \)

Substitute: \( 2x + (-x + 5) = 1 \Rightarrow x = -4 \), \( y = 9 \)

Solve systems by adding or subtracting equations to eliminate one variable.

\( x + y = 5 \), \( x - y = 1 \)
Add equations: \( 2x = 6 \Rightarrow x = 3 \), \( y = 2 \)

\( 2x + 3y = 12 \), \( -2x + y = 0 \)
Add: \( 4y = 12 \Rightarrow y = 3 \), \( x = 1.5 \)

\( 3x + 2y = 16 \), \( 6x + 4y = 32 \)
Multiply first by 2 and subtract → Infinite solutions

\( 4x + 2y = 8 \), \( 2x - 2y = 0 \)
Add: \( 6x = 8 \Rightarrow x = \frac{4}{3}, y = \frac{2}{3} \)

Translate word problems into equations and solve using substitution or elimination.

Tom and Tai have 24 coins. Tom has 4 more than Tai. How many each?

\( a + b = 24 \), \( a = b + 4 \) → \( b = 10, a = 14 \)

Tickets cost 10 Baht for adults, 5 Baht for students. 5 tickets total 35 Baht. Find numbers.

\( a + s = 5 \), \( 10a + 5s = 35 \) → \( s = 3, a = 2 \)

A rectangle's perimeter is 50, length is 3 more than width. Find dimensions.

\( 2l + 2w = 50 \), \( l = w + 3 \) → \( w = 11, l = 14 \)

Two numbers sum to 30, and their difference is 4. What are they?

\( x + y = 30 \), \( x - y = 4 \) → \( x = 17, y = 13 \)