Systems of Equations

A1.D.1Substitution01/05

Solve a system by substitution.

The system we'll solve

{x + y = 7 2 x - y = 2

Step01of 05

Substitution shines when one equation already isolates a variable, or is one quick step from doing so. Equation 1 already has y with coefficient 1 — solve it for y:

y = 7 - x

A1.D.1 — Solve a system by substitution.

The system we'll solve

{x + y = 7 2 x - y = 2

Step 01 of 05

Substitution shines when one equation already isolates a variable, or is one quick step from doing so. Equation 1 already has y with coefficient 1 — solve it for y:

y = 7 - x

Step 02 of 05

Substitute that into the OTHER equation (not the one you just rearranged). Replace y with $(7 - x)$ in eq 2:

2 x - (7 - x) = 2

Step 03 of 05

Simplify and solve the one-variable equation.

2 x - 7 + x 3 x x = 2 = 9 = 3

Step 04 of 05

Back-substitute the value into either original. Eq 1 is easier:

y = 7 - 3 = 4

Step 05 of 05

Always check. Plug $(x, y) = (3, 4)$ into both originals.

x + y 2 x - y = 7 = 2 \Rightarrow 3 + 4 = 7 ✓ \Rightarrow 2 (3) - 4 = 2 ✓

Key insight

The solution is

(x, y) = (3, 4)

. Substitution wins when one variable's coefficient is already

1

- 1

A1.D.2 — Solve a system by elimination.

The system we'll solve

{x + y = 7 2 x - y = 2

Step 01 of 04

Elimination adds or subtracts the equations so one variable cancels. Look for matching opposite coefficients first — here the y-terms are already $+ y$ and $- y$ . Adding the equations will cancel y for free.

Step 02 of 04

Add vertically. Add the left sides; add the right sides. The y-terms vanish.

x + y + 2 x - y 3 x = = = 729

Step 03 of 04

Solve and back-substitute.

3 x = 9 ⟹ x = 3

3 + y = 7 ⟹ y = 4

Same answer as substitution: $(x, y) = (3, 4)$ . Two methods, one destination.

Step 04 of 04

What if coefficients DON'T line up? Scale one (or both) equations first. To eliminate x in:

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

Multiply the second equation by $- 2$ so the x-terms cancel:

2 x + 3 y + (- 2 x + 2 y) 5 y y = = = = 12 - 2 102

A1.D.3 — How many solutions does the system have?

Step 01 of 04

Every linear equation is a line on a graph. The system's solution is wherever the lines cross. Three things can happen.

Case 1 — one solution. Different slopes → lines cross at exactly one point. Most systems fall here.

{y = 2 x + 1 y = - x + 4 ⟹ (1, 3)

Step 02 of 04

Case 2 — no solution. Lines are parallel: same slope, different y-intercept. They never meet.

{y = 2 x + 1 y = 2 x + 5 ⟹ no solution

When you try to solve algebraically, you hit a contradiction like $0 = 4$ . That's the signal.

Step 03 of 04

Case 3 — infinitely many solutions. Same line, written two ways. Every point on the line satisfies both equations.

{y = 2 x + 1 2 y = 4 x + 2 ⟹ \infty solutions

Solving algebraically gives $0 = 0$ — always true, doesn't pin down x or y.

Step 04 of 04

Quick test without graphing. Rearrange both equations to $y = m x + b$ form and compare slope and intercept.

Different $m$	→one solution
Same $m$ , different $b$	→no solution
Same $m$ , same $b$	→infinitely many

Key insight

On multiple choice, you almost never need to actually solve for the point — just identify the case from the slopes and intercepts.

A1.D.4 — Systems of linear inequalities.

Step 01 of 04

A system of linear inequalities swaps $=$ for one of $\leq, \geq, <, >$ . The solution isn't a single point — it's an entire region where every inequality is true.

Step one: graph each line as if it were an equation. For $y \leq 2 x + 1$ , first graph $y = 2 x + 1$ .

Step 02 of 04

Solid or dashed line?

$\leq$ or $\geq$	→SOLID line — boundary points count
$<$ or $>$	→DASHED line — boundary points don't count

Step 03 of 04

Pick a test point not on the line and plug it in. Often $(0, 0)$ is easiest if the line doesn't pass through the origin.

(0, 0) in y \leq 2 x + 1 : 0 \leq 1 ✓

True → shade the side containing $(0, 0)$ . False → shade the other side.

Step 04 of 04

Repeat for every inequality, then look at the OVERLAP of all shaded regions. That overlap is the system's solution set.

Try this

On multiple-choice tests you can usually skip the graphing entirely. Pick a candidate point and plug it into every inequality. The point that satisfies all of them is in the solution region.

A1.D.5 — Word problems — set up the system, then solve.

The problem

A coffee shop sells small drinks for $3 and large for $5. On Monday they sold 80 drinks for $310 total. How many small? How many large?

Step 01 of 05

The hard step is translating the story into two equations. Name the unknowns clearly first.

s = small drinks sold L = large drinks sold

Step 02 of 05

Write one equation per fact in the problem.

{s + L = 80 3 s + 5 L = 310 (count) (revenue)

Step 03 of 05

Pick a method. Equation 1 is easy to solve for either variable, so substitution is clean. Solve eq 1 for s:

s = 80 - L

Step 04 of 05

Substitute into eq 2 and solve.

3 (80 - L) + 5 L 240 + 2 L 2 L L = 310 = 310 = 70 = 35

Step 05 of 05

Back-substitute and check.

s = 80 - 35 = 45

Count: Revenue: 45 + 35 = 80 ✓ 3 (45) + 5 (35) = 310 ✓

Key insight

Answer: 45 small, 35 large. The math is identical to the symbolic systems from skills 1 and 2 — the extra step is the translation.

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