Linear inequalities in one or two variables

An inequality compares two amounts with $\lt$, $\gt$, $\leq$ or $\geq$.

Read the last line first. It might say:

• “Which inequality represents this situation?”
• “What is the greatest number of shirts the club can order without exceeding its budget?”
• “Which point $(x, y)$ is a solution to the given system of inequalities in the $xy$-plane?”
• “For which of the following tables are all the values of $x$ and their corresponding values of $y$ solutions to the given inequality?”
• “The shaded region shown represents the solutions to which inequality?”
• “Which statement is the best interpretation of $3.80b$ in this context?”

Inequality choices often share every number and differ only in the sign or in which price goes with which letter, so look at what changes.

Leave Desmos closed when the question only asks which sign to use or what a number means.

A club is ordering printed T-shirts. The shop charges an $18 setup fee plus $4.75 per shirt, and the club can spend no more than $250. What is the greatest number of shirts the club can order?

1. Name the letter and what it counts.

   $$s=\text{number of shirts}$$

2. Build the total cost. The $18 is paid once, and $4.75 is paid for each shirt.

   $$18+4.75s$$

3. “No more than $250” lets the total equal 250 but not go over, so use $\leq$.

   $$18+4.75s\leq250$$

4. Subtract 18 from both sides. The sign stays the same.

   $$4.75s\leq232$$

5. Divide both sides by 4.75. It's positive, so the sign still stays.

   $$s\leq\frac{232}{4.75}\approx48.84$$

6. Round down. The money runs out before shirt 49.

   $$s=48$$

7. Check 48 in the original cost. $246 fits in $250.

   $$18+4.75(48)=246$$

8. Check one more. 49 shirts go over, so 48 is the answer.

   $$18+4.75(49)=250.75$$

On a grid-inA question where you type your own answer instead of picking one of four.Read the lesson, you'd enter 48.

Here's one to try.

“At least” and “at most” include the number itself. “More than” and “less than” leave it out, and an inequality that leaves it out is a strict inequalityAn inequality that uses less than or greater than without “or equal to,” so the boundary value itself doesn’t count., written with $\lt$ or $\gt$.

The T-shirt order has an $18 setup fee plus $4.75 per shirt, with no more than $250 to spend.

With two items, each price goes with its own letter. Say a school pays $1.60 for each notebook and $3.80 for each binder, and it can spend up to $500. With $n$ notebooks and $b$ binders, that's $1.60n+3.80b\leq500$.

Now try one with four choices.

A 6% sales tax turns a price $p$ into $p+0.06p$, which is $1.06p$.

Say the T-shirt shop ($18 setup plus $4.75 a shirt) adds 6% tax to the whole order, and the club's $250 has to cover it. Write the cost before tax first, $18+4.75s$. Then multiply all of it by 1.06.

Some limits are on what's left. A 500-gallon tank has to keep at least 80 gallons after $g$ gallons are pumped out. What's left is $500-g$.

1. Start.

   $$500-g\geq80$$

2. Add $g$ to both sides.

   $$500\geq80+g$$

3. Subtract 80 from both sides. At most 420 gallons can be pumped out.

   $$420\geq g$$

Twice or half works like the tax. If the water left had to be at least twice what's pumped out, write the amount being doubled first, then multiply: $500-g\geq2g$.

A summer job wants you to work at least 12 but no more than 20 hours a week. Write each limit by itself first, $h\geq12$ and $h\leq20$. Then put them together with the smaller number on the left.

“Between” with no “inclusive” leaves out both ends, so “between 12 and 20 hours” is $12\lt h\lt20$. With “inclusive,” use $\leq$ on both sides.

To solve a range, do the same thing to all three parts.

1. Start.

   $$2\leq x+3\leq10$$

2. Subtract 3 from all three parts.

   $$-1\leq x\leq7$$

A triangle's third side has two limits too. It's longer than the difference of the other two sides and shorter than their sum, so with sides 5 and 9, the third side $x$ fits $4\lt x\lt14$.

A school pays $1.60 for each of its $n$ notebooks and $3.80 for each of its $b$ binders, and it can spend up to $500.

A question can also ask about one term, like $3.80b$. That's $3.80 times the number of binders, so it's what the binders cost by themselves.

Solve an inequality with the moves from linear equations in one variable. Adding, subtracting, and multiplying or dividing by a positive number leave the sign alone.

You can skip the flip by moving the letter to the side where its coefficientThe number multiplying a letter, like the 5 in 5x.Read the lesson comes out positive.

1. Start.

   $$-4x+7\geq19$$

2. Add $4x$ to both sides. Now the $x$ term is positive.

   $$7\geq19+4x$$

3. Subtract 19 from both sides.

   $$-12\geq4x$$

4. Divide both sides by 4, a positive number.

   $$-3\geq x$$

5. Read it from the side of $x$: $x$ is at most $-3$.

   $$x\leq-3$$

A count of shirts or cans has to be a whole number, so the last step is often rounding. For the greatest count under a limit, round down, as with the 48 shirts. For the least count that meets a need, round up.

Say 162 gallons of water go into 5-gallon cans, and the question asks for the least number of cans.

1. The cans have to hold at least 162 gallons.

   $$5c\geq162$$

2. Divide both sides by 5.

   $$c\geq32.4$$

3. Round up. 32 cans hold only 160 gallons.

   $$c=33$$

A count can't be negative either. For the supplies, $(-5,50)$ makes $1.60n+3.80b\leq500$ true, but no school buys $-5$ notebooks.

A point is a solution when its numbers make the inequality true. A tank starts with $a$ gallons and fills at $m$ gallons a minute. After 6 minutes it has to hold at least 480 gallons, and after 15 minutes it can hold no more than 1,200 gallons.

A table works only if every row does. Here's one table for $y\gt6x-20$.

1. Row 1 passes.

   $$6(3)-20=-2,\ \ 0\gt-2$$

2. Row 2 passes.

   $$6(4)-20=4,\ \ 5\gt4$$

3. Row 3 fails. 10 isn't greater than 10.

   $$6(5)-20=10,\ \ 10\gt10 \text{ is false}$$

Some questions ask for the greatest possible value of one count. Give the other count the smallest value the story allows, then solve. For the supplies, $1.60n+3.80b\leq500$, here's the greatest number of binders.

1. Fewer notebooks leave more money for binders, so use $n=0$.

   $$1.60(0)+3.80b\leq500$$

2. Divide both sides by 3.80.

   $$b\leq\frac{500}{3.80}\approx131.58$$

3. Round down.

   $$b=131$$

Zero works only when the story allows it. If the school has to buy at least 20 notebooks, use $n=20$.

1. The fewest notebooks allowed is 20.

   $$1.60(20)+3.80b\leq500$$

2. $1.60(20)=32$. Subtract it from both sides.

   $$3.80b\leq468$$

3. Divide both sides by 3.80.

   $$b\leq\frac{468}{3.80}\approx123.16$$

4. Round down.

   $$b=123$$

If the question fixes one count, as in “if the school buys 20 notebooks,” put that number in the same way.

Check what one grid square is worth first. Here the horizontal axisOne of the two lines a graph is measured against, one running across and one running up. counts by 3s and the vertical axis by 1s, so the solid line crosses at $(0,5)$, the y-interceptThe output when the input is 0, which is where a graph crosses the vertical axis.Read the lesson, and $(15,0)$, the x-interceptThe value of x where a graph crosses the horizontal axis, so the output there is 0.Read the lesson. Find the slopeHow steep a line is, measured by how much it goes up for each unit it moves to the right.Read the lesson and the line from those two points, as in the linear functions lesson.

1. Find the slope: change in $y$ over change in $x$.

   $$m=\frac{0-5}{15-0}=-\frac13$$

2. The line meets the vertical axis at $(0,5)$, so the starting value is 5.

   $$y=-\frac13x+5$$

3. Solid line, shading below.

   $$y\leq-\frac13x+5$$

When the inequality isn't solved for $y$, the sign alone doesn't tell you which side is shaded.

If the line runs through the origin, as $y=2x$ does, test another point, such as $(0,3)$.

Say the shaded region below shows every solution of $kx+9y\leq45$, where $k$ is a constantA number that doesn’t change, like the 7 in 5x + 7, or a letter such as p that stands for one fixed value.Read the lesson, and the question asks for $k$.

The horizontal axis counts by 3s, so the line crosses the axes at $(0,5)$ and $(15,0)$.

1. Points on the line make the two sides equal, so use an equals sign.

   $$kx+9y=45$$

2. Put in $(15,0)$. That's the crossing where the $k$ term isn't zero.

   $$k(15)+9(0)=45$$

3. $9(0)=0$.

   $$15k=45$$

4. Divide both sides by 15.

   $$k=3$$

5. Check the shading. The origin gives $0\leq45$, which is true, and the origin is shaded.

   $$3(0)+9(0)\leq45$$

Now try one on a different grid.

+−100%Reset⤢

A taxi company limits its combined staffing cost for daytime and overnight shifts to $3,780. Daytime staffing costs $67.50 per hour.

In the graph, $x$ and $y$ represent daytime and overnight shift hours, respectively. The shaded region represents the solutions to $67.5x + ky \leq 3{,}780$, where $k$ is a constant.

What is the value of $k$?

1. A

   54

2. B

   67.5

3. C

   63

4. D

   70

Show the answer

• A

  54

  Correct answer

  The boundary line passes through $(0,70)$, so $70$ overnight hours use the full $3{,}780$-dollar budget when there are no daytime hours. Substituting gives $70k=3{,}780$, so $k=54$.

• B

  67.5

  Dividing $3{,}780$ by $56$ uses the horizontal interceptWhere a graph crosses an axis.Read the lesson in place of the vertical intercept. The $56$ represents daytime hours, so this calculation gives the daytime hourly cost.

• C

  63

  This comes from reading the vertical interceptWhere a graph crosses an axis.Read the lesson as $60$, one grid interval below $70$, and calculating $3{,}780\div60=63$. Each vertical grid interval represents $10$ hours.

• D

  70

  This treats the vertical interceptWhere a graph crosses an axis.Read the lesson, $70$, as the hourly cost. That intercept represents overnight hours at the full budget; you need to substitute it into the boundary equationTwo expressions set equal to each other, like 5x + 7 = 22.Read the lesson to find $k$.

Solution

On the boundary line, the staffing cost equals the budget, so $67.5x+ky=3{,}780$. The line crosses the vertical axisOne of the two lines a graph is measured against, one running across and one running up. at $(0,70)$, where there are $0$ daytime hours and $70$ overnight hours.

Substitute these coordinates: $67.5(0)+k(70)=3{,}780$. This gives $70k=3{,}780$, so $k=\frac{3{,}780}{70}=54$. The overnight staffing cost is $54$ dollars per hour.

Report a problem

Before each one, find the limit words and decide whether the boundary counts. Then check your answer in the original words.