How do I know when to make the boundary line dashed instead of solid?

Look only at the inequality symbol. If it is , points exactly on the line do not satisfy the statement, so the line is dashed. If it is ≤ or ≥, points on the line do satisfy it, so the line is solid. In context, "at most 40 dollars" allows spending exactly 40, so that boundary is solid; "under 40 dollars" does not, so it is dashed.

Which side do I shade if I can't remember the rule?

Use a test point. Pick any point not on the boundary line — (0,0) is easiest whenever the line misses the origin — and substitute its coordinates into the inequality. If the result is a true statement, shade the side containing that point. If it is false, shade the other side. This works for every linear inequality, in any form, and takes about ten seconds.

What if my two equations from a word problem give no solution or infinitely many?

No solution means the conditions in the problem contradict each other — for example, prices and totals that cannot both be met. Infinitely many means the two sentences said the same thing in different words, so you did not actually find two independent relationships. Reread the problem and look for a second, genuinely different piece of information.

Do I have to graph a system of inequalities, or can I just solve it algebraically?

You cannot solve it to a single point, because the answer is a region rather than one ordered pair. You can find the corner points algebraically by solving the boundary lines as systems of equations, and that is useful, but describing the full solution set requires the graph and its shading.

ALG1-5.4

Systems Word Problems & Systems of Inequalities

Learn to define two variables, build a system from a word problem, pick the fastest solving method, and graph systems of linear inequalities to find the overlapping region.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Systems Word Problems & Systems of Inequalities, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Systems Word Problems & Systems of Inequalities free

What this lesson covers

You already know three ways to solve a system: graphing, substitution, and elimination. The hard part in real problems is not the algebra — it is getting from a paragraph of English to two correct equations. This lesson gives you a reliable routine: name your variables precisely, find the two separate relationships hiding in the words, and then choose the method that will cost you the least work.

Then the lesson widens the idea. When a situation says "at most," "no more than," or "at least," you get inequalities instead of equations, and the answer is not a single point but a whole region of the plane. You will learn to graph two or more linear inequalities on the same axes and read off the overlap — the set of every combination that satisfies all the conditions at once.

From Sentences to Two Equations

Every system word problem contains two different relationships between the same two unknowns. Your job is to find both.

Start by writing a variable definition sentence, with units, not just a letter. "Let

a

= number of adult tickets sold" is a definition. "Let

a

= adults" is vague and leads to mixed-up equations later. If a quantity is a rate or a price, say so: "Let

c

= cost of one shirt, in dollars."

Next, hunt for the two relationships. Most Algebra 1 problems fall into a few families.

Type	Relationship 1 (count)	Relationship 2 (value)
Tickets or coins	number of items	total money
Mixture	total amount of mix	total amount of the ingredient
Two purchases	first receipt	second receipt
Comparison	"twice as many as"	total

The classic structure is one equation that adds the counts and one that adds the values. If 250 tickets are sold in all, that is

a + s = 250

. If those same tickets bring in 2,400 dollars at 12 dollars and 8 dollars each, that is

12a + 8s = 2400

.

Where students go wrong: writing

a + s = 2400

, mixing counts and dollars in one equation. Check units on every term. In

12a

, the 12 is dollars per ticket and

a

is tickets, so the product is dollars — it belongs with the 2,400. A term measured in tickets can never be added to a term measured in dollars.

Choosing an Efficient Method

All three methods give the same answer, so "efficient" means fewest steps and least chance of an arithmetic slip. Look at the form the equations are already in.

Situation	Best method	Why
A variable is already alone, like $y = 3x + 5$	Substitution	Plug it straight in; no rearranging
A coefficient is 1 or $-1$	Substitution	Solving for that variable creates no fractions
Both in $Ax + By = C$ form	Elimination	Line them up and add or subtract
Matching or opposite coefficients	Elimination	One step and a variable is gone
You need a rough answer or a picture	Graphing	Shows the intersection visually

For the ticket system

a + s = 250

and

12a + 8s = 2400

, elimination is fast: multiply the first equation by

-8

to get

-8a - 8s = -2000

, add, and you are left with

4a = 400

.

Substitution also works here because

a + s = 250

rearranges cleanly to

s = 250 - a

. Either is fine. What is not efficient is graphing a system whose solution is

(100, 150)

— you would need enormous axes to see it.

After solving, always answer the actual question in a sentence. Some problems ask for the value of one variable, some ask for a difference or a total. And always check both original equations, not just the easy one, because a slip in the second equation is invisible if you only verify the first.

Graphing a System of Linear Inequalities

A linear inequality such as

y \le 2x + 1

has infinitely many solutions: every point on one side of the boundary line, plus possibly the line itself. Graphing a system of inequalities means graphing each one and keeping only what they share.

The routine for each inequality:

Graph the boundary line by treating the inequality symbol as an equals sign. Make it solid for

\le

\ge

(points on the line satisfy the statement) and dashed for

<

>

(points on the line do not).

Shade the correct half-plane. The safest way is a test point. Pick

(0,0)

if the line does not pass through the origin, substitute, and see whether the statement is true. If it is true, shade the side containing

(0,0)

; if false, shade the other side. Shortcut: once the inequality is solved for

y

, "greater than" shades above and "less than" shades below.

The solution to the system is the overlap — the region covered by every shading at once. Many students shade heavily and then cannot see the intersection. Use light shading, arrows along each boundary, or different diagonal hatching directions, and then outline the overlap.

Two checks that catch most errors: a point in the overlap must make both inequalities true, and a point in a single-shaded strip must fail at least one. If the shaded regions never overlap, as with

y > x + 4

and

y < x - 1

, the system genuinely has no solution — parallel boundaries with the shading pointing away from each other.

Constraints, Feasible Regions, and Real Contexts

Real situations attach limits to variables, and those limits become inequalities. A booster club with 40 dollars to spend on snacks priced at 2 dollars and 5 dollars gives

2x + 5y \le 40

. "At least 12 items total" gives

x + y \ge 12

. Because you cannot buy a negative number of snacks, you also get

x \ge 0

and

y \ge 0

, which restricts everything to the first quadrant.

Translate the phrases carefully.

Phrase	Symbol
at most, no more than, maximum of	$\le$
at least, no less than, minimum of	$\ge$
fewer than, under	$<$
more than, over	$>$

The overlapping region is often called the feasible region: every point in it is a combination that satisfies all the constraints. In context, though, not every point in the region is usable. If

x

and

y

count physical objects, only whole-number points make sense, so the shaded region shows possibilities while the actual answers are the lattice points inside it. This is the difference between a mathematical solution and a viable solution, and it is a favorite thing for teachers to ask about in a written explanation.

A good habit: after graphing, name one specific point in the overlap and interpret it in words. "The point

(10, 4)

means buying 10 of the cheaper snacks and 4 of the pricier ones, which costs 40 dollars and gives 14 items — within budget and above the 12-item minimum." That sentence proves you understand the picture, not just how to draw it.

Key terms

System of linear equations.: Two or more linear equations considered together; a solution is an ordered pair that makes every equation true at once.
Variable definition.: A written statement naming exactly what each letter represents, including units, such as "let $s$ = number of student tickets sold."
Linear inequality.: A statement like $y \ge 2x - 3$ whose graph is a half-plane bounded by a line, solid when the symbol includes equality and dashed when it does not.
Boundary line.: The line obtained by replacing the inequality symbol with an equals sign; it separates the plane into the two half-planes.
Half-plane.: One of the two regions a line divides the coordinate plane into; the shaded half-plane holds the solutions of the inequality.
Overlapping solution region.: The set of points shaded by every inequality in the system, so each point satisfies all conditions simultaneously.
Feasible region.: The overlapping region of a system of constraint inequalities in a real-world context.
Viable solution.: A point in the solution region that also makes sense in context, for example a whole, non-negative number of objects.

Worked example

A school play sells 250 tickets and collects 2,400 dollars. Adult tickets cost 12 dollars each and student tickets cost 8 dollars each. How many of each type were sold?

Define the variables. Let

a

= number of adult tickets sold and

s

= number of student tickets sold.

Find the two relationships. The count relationship uses the total number of tickets:

a + s = 250

. The value relationship uses money, where each adult ticket contributes 12 dollars and each student ticket 8 dollars:

12a + 8s = 2400

.

Choose a method. Both equations are in standard form and the first has coefficients of 1, so elimination is quick. Multiply the first equation by

-8

-8a - 8s = -2000

Add this to

12a + 8s = 2400

4a = 400

a = 100

.

Back-substitute into

a + s = 250

100 + s = 250

, so

s = 150

.

Check both original equations. Counts:

100 + 150 = 250

. Money:

12(100) + 8(150) = 1200 + 1200 = 2400

. Both are true.

Answer in a sentence: the play sold 100 adult tickets and 150 student tickets.

Practice questions

A gym charges a one-time fee of 30 dollars plus 15 dollars per month. A rival gym charges no joining fee but 20 dollars per month. If

m

is the number of months and

c

is total cost in dollars, which system models the two plans?

$c = 30m + 15$ and $c = 20m$
$c = 15m + 30$ and $c = 20m$
$c = 30 + 15 + m$ and $c = 20 + m$
$c = 45m$ and $c = 20m$

Answer: $c = 15m + 30$ and $c = 20m$

The 15 dollars is a rate, so it multiplies the number of months:

15m

. The 30 dollar fee is charged once, so it is a constant added on, giving

c = 15m + 30

. The rival plan has no constant term, only the rate:

c = 20m

. The first choice reverses the roles of the rate and the one-time fee — a common slip. Setting the two expressions equal gives

15m + 30 = 20m

, so

m = 6

: the plans cost the same after 6 months.

Graph the system

y > 2x - 3

and

x + y \le 4

. Decide whether

(0,0)

and

(3,1)

are solutions, and explain how the graph shows your answers.

Answer: $(0,0)$ is a solution; $(3,1)$ is not.

Test

(0,0)

in the first inequality:

0 > 2(0) - 3

becomes

0 > -3

, true. In the second:

0 + 0 = 0 \le 4

, true. Both hold, so

(0,0)

lies in the overlap. Test

(3,1)

1 > 2(3) - 3

becomes

1 > 3

, false, so it fails immediately and no further checking is needed. On the graph, the line

y = 2x - 3

is dashed with shading above it, and the line

x + y = 4

is solid with shading below it. The point

(0,0)

sits in the wedge where the two shadings overlap, while

(3,1)

lies below the dashed line in the single-shaded strip.

A chemist mixes a 20 percent acid solution with a 50 percent acid solution to make 60 milliliters of a 30 percent acid solution. Write and solve a system for the amount of each solution used.

Answer: 40 milliliters of the 20 percent solution and 20 milliliters of the 50 percent solution.

Let

x

= milliliters of 20 percent solution and

y

= milliliters of 50 percent solution. The volume relationship is

x + y = 60

. The acid relationship counts only the pure acid:

0.20x + 0.50y = 0.30(60) = 18

. Substitution is efficient since

y = 60 - x

. Then

0.20x + 0.50(60 - x) = 18

, so

0.20x + 30 - 0.50x = 18

, giving

-0.30x = -12

and

x = 40

. So

y = 20

. Check the acid:

0.20(40) + 0.50(20) = 8 + 10 = 18

milliliters of acid, and

18/60 = 0.30

. Notice the resulting concentration, 30 percent, sits between 20 and 50 percent but closer to 20, which matches using more of the weaker solution.

FAQ

How do I know when to make the boundary line dashed instead of solid?: Look only at the inequality symbol. If it is $<$ or $>$ , points exactly on the line do not satisfy the statement, so the line is dashed. If it is $\le$ or $\ge$ , points on the line do satisfy it, so the line is solid. In context, "at most 40 dollars" allows spending exactly 40, so that boundary is solid; "under 40 dollars" does not, so it is dashed.
Which side do I shade if I can't remember the rule?: Use a test point. Pick any point not on the boundary line — $(0,0)$ is easiest whenever the line misses the origin — and substitute its coordinates into the inequality. If the result is a true statement, shade the side containing that point. If it is false, shade the other side. This works for every linear inequality, in any form, and takes about ten seconds.
What if my two equations from a word problem give no solution or infinitely many?: No solution means the conditions in the problem contradict each other — for example, prices and totals that cannot both be met. Infinitely many means the two sentences said the same thing in different words, so you did not actually find two independent relationships. Reread the problem and look for a second, genuinely different piece of information.
Do I have to graph a system of inequalities, or can I just solve it algebraically?: You cannot solve it to a single point, because the answer is a region rather than one ordered pair. You can find the corner points algebraically by solving the boundary lines as systems of equations, and that is useful, but describing the full solution set requires the graph and its shading.

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