Systems of Linear Equations

8.E.1Solve by graphing01/04

Solve a system by graphing.

The system

{y = x + 2 y = - 2 x + 8

Step01of 04

The solution to a system of two linear equations is the point $(x, y)$ where the two lines INTERSECT. Plot both, find the crossing.

8.E.1 — Solve a system by graphing.

The system

{y = x + 2 y = - 2 x + 8

Step 01 of 04

The solution to a system of two linear equations is the point $(x, y)$ where the two lines INTERSECT. Plot both, find the crossing.

Step 02 of 04

Plot each line using slope-intercept (y-intercept + slope).

$y = x + 2$	→starts at (0, 2), slope 1 (up 1 right 1)
$y = - 2 x + 8$	→starts at (0, 8), slope -2 (down 2 right 1)

Draw both lines on the same axes.

Step 03 of 04

Read the intersection. The lines cross at $(2, 4)$ . That point satisfies BOTH equations:

4 = 2 + 2 ✓ 4 = - 2 (2) + 8 ✓

Step 04 of 04

Limitations of graphing.

Solutions with fractions	→hard to read precisely
Very steep / shallow lines	→crossing point hard to identify
Solutions far off-grid	→requires scaling
Useful for...	→visualizing, checking algebraic answers, problems with clean integer solutions

Key insight

Solution = intersection point. Graph both lines, read coordinates of crossing. Best for clean integer solutions; switch to substitution or elimination when the algebra gets messy.

8.E.2 — Solve a system by substitution.

The system

{y = 2 x - 1 3 x + y = 14

Step 01 of 04

Substitution is best when one equation already isolates a variable (or is one easy step from doing so). Here, $y$ is already isolated in the first equation.

Step 02 of 04

Step 1 — substitute into the OTHER equation. Replace $y$ in eq. 2 with $(2 x - 1)$ .

3 x + (2 x - 1) = 14

5 x - 1 = 14 \Rightarrow x = 3

Step 03 of 04

Step 2 — back-substitute. Plug $x = 3$ back into eq. 1 (the simpler one).

y = 2 (3) - 1 = 5

Solution: $(3, 5)$ .

Step 04 of 04

Always check. Plug into BOTH originals.

5 = 2 (3) - 1 ✓ 3 (3) + 5 = 14 ✓

Key insight

Substitution shines when one variable is already isolated (or has coefficient 1). Replace it in the OTHER equation. Solve, then back-substitute for the second variable. Always check.

8.E.3 — Solve a system by elimination.

The system

{2 x + 3 y = 13 4 x - 3 y = 5

Step 01 of 04

Elimination adds (or subtracts) the two equations to make one variable disappear. Best when coefficients of one variable are already opposites (or easy to scale to opposites).

Here, $+ 3 y$ and $- 3 y$ are already opposites — adding the equations cancels $y$ for free.

Step 02 of 04

Step 1 — add the equations.

(2 x + 3 y) + (4 x - 3 y) = 13 + 5

6 x = 18 \Rightarrow x = 3

Step 03 of 04

Step 2 — back-substitute. Plug into either original.

2 (3) + 3 y = 13 \Rightarrow 3 y = 7 \Rightarrow y = 7/3

Solution: $(3, 7/3)$ . Check both: $2 (3) + 3 (7/3) = 6 + 7 = 13$ $✓$ ; $4 (3) - 3 (7/3) = 12 - 7 = 5$ $✓$ .

Step 04 of 04

When coefficients aren't opposites, scale first.

{2 x + 3 y = 12 x - y = 1

Multiply eq. 2 by $- 2$ so the x-terms cancel:

- 2 x + 2 y = - 2

Now add to eq. 1: $5 y = 10 \Rightarrow y = 2$ .

Key insight

Elimination cancels a variable by adding (or subtracting) the equations. If coefficients aren't opposites, multiply one or both equations to make them so. Best for systems where neither variable is already isolated.

8.E.4 — Systems word problems.

The problem

A movie theater sells adult tickets for $12 and child tickets for $7. On Saturday they sold 200 tickets total for $2,000. How many adult tickets and how many child tickets?

Step 01 of 04

Step 1 — name the unknowns.

s = adult tickets sold L = child tickets sold

Step 02 of 04

Step 2 — write one equation per fact.

{s + L = 200 12 s + 7 L = 2000 (count) (revenue)

Step 03 of 04

Step 3 — solve. Eq. 1 is easy to solve for $s$ : $s = 200 - L$ . Substitute into eq. 2:

12 (200 - L) + 7 L = 2000

2400 - 12 L + 7 L = 2000 \Rightarrow - 5 L = - 400 \Rightarrow L = 80

Then $s = 200 - 80 = 120$ .

Step 04 of 04

Step 4 — interpret and check. 120 adult tickets and 80 child tickets.

Count: 120 + 80 = 200 ✓

Revenue: 12 (120) + 7 (80) = 1440 + 560 = 2000 ✓

Key insight

Define variables in words. Translate each fact into an equation. Solve the system. ALWAYS check both quantities (count + revenue, here) — passing the math but failing the count usually means a setup error.

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8.E.1 – 8.E.4 · CCSS 8.EE.C.8 · Full curriculum →