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Prep 2 SS · Mathematics
Algebra · Algebra

Inequalities of the First and Second Degree

Before you start, make sure you know: Intervals, rays, and neighborhoods (High School Year 1 (semesters 1 and 2))
Before you start, make sure you know: Quadratic equations (High School Year 1 (semesters 1 and 2))

A high school class is organizing its graduation party: the venue costs $\$9{,}000$ and each guest's meal costs $\$240$. They have $\$30{,}000$ and want at least $\$4{,}200$ left over. How many guests can they invite? There isn't just one answer here, but a whole set of possible values. That is an inequality: a relation with an unknown whose solution is usually an interval. At the end of the linear problems section we solve this case.

First-degree inequalities

Number line showing the solution x greater than negative three, with an open endpoint, obtained by dividing by negative two and flipping the inequality. −2x < 6 ÷(−2) → flips the direction x > −3 −3
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Solve the following first-degree inequality and represent the solution on the real number line using an interval: $5 - 2x \le 11$ Draw a real number line in your…

Solve the following first-degree inequality and represent the solution on the real number line using an interval:

$5 - 2x \le 11$

Draw a real number line in your notebook with a suitable scale, mark the endpoint of the interval and indicate whether it is included or not. Then write the solution using interval notation.

Solve
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Solve the following inequality with brackets and simplify the result: $3(x - 1) + 2 > 5x - 7$

Solve the following inequality with brackets and simplify the result:

$3(x - 1) + 2 > 5x - 7$

Solve
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Solve the following linear inequality, expressing the solution in interval notation and representing it on the number line: $$\frac{2x - 3}{4} - \frac{x + 1}{3} \geq…

Solve the following linear inequality, expressing the solution in interval notation and representing it on the number line:

$$\frac{2x - 3}{4} - \frac{x + 1}{3} \geq \frac{x - 1}{2} - 1$$
Solve

Problems with linear inequalities

To set up a problem, translate the phrases: "at minimum" and "at least" are $\geq$; "at most," "at maximum," and "must not exceed" are $\leq$; "more than" is $>$ and "less than" is $<$.

Two cost lines that cross at 40 extra GB: to the right of that point Plan A is cheaper. x = 40 Extra GB Plan A Plan B
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The area of a rectangle with base $5$ cm and height $(x - 2)$ cm must be less than $30$ cm². What values can the variable $x$ take? Note that the height must always be a…

The area of a rectangle with base $5$ cm and height $(x - 2)$ cm must be less than $30$ cm². What values can the variable $x$ take? Note that the height must always be a positive quantity.

Limit: Area = 30 cm² Area < 30 5 x - 2 x - 2 > 0
Solve
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A group of 1st-year Bachillerato students is organizing a graduation party. They have calculated that the fixed costs of renting the venue and the DJ are $450$ pesos. In…

A group of 1st-year Bachillerato students is organizing a graduation party. They have calculated that the fixed costs of renting the venue and the DJ are $450$ pesos. In addition, the catering costs $12$ pesos per attendee. If they have raised a total of $1500$ pesos through raffle ticket sales and sponsorships, and they want to have at least $210$ pesos left over after the event for an end-of-year trip, write and solve an inequality that determines the maximum number of people, $x$, that they can invite.

In the graph you can see the money raised and the costs as a function of the number of attendees.

Graph of the money raised (1500 €, horizontal line) and of the costs 450 + 12x as a function of the number of attendees 1500 450 10 20 30 40 50 60 70 80 90 100 Number of attendees (x) Pesos ($) Raised ($1500) Costs (450 + 12x) Fixed costs
Solve
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A phone company offers a plan charging a fixed monthly fee of $200$ pesos plus $1$ peso per minute of call. Another company charges a fixed fee of $100$ pesos but $3$…

A phone company offers a plan charging a fixed monthly fee of $200$ pesos plus $1$ peso per minute of call. Another company charges a fixed fee of $100$ pesos but $3$ pesos per minute. After how many minutes of calls per month does the first company become cheaper?

Solve

Second-degree inequalities

Number line divided by the roots 2 and 3 into three intervals with positive, negative, and positive signs. x² − 5x + 6 ≥ 0 2 3 +−+ solution: (−∞, 2] ∪ [3, +∞)
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Determine the solution set of the following quadratic inequality by analyzing the sign of the factors: $x^2 - 9 > 0$ Express the result as intervals.

Determine the solution set of the following quadratic inequality by analyzing the sign of the factors:

$x^2 - 9 > 0$

Express the result as intervals.

Solve
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Check whether the values $x = 0$, $x = 2$ and $x = 5$ are solutions of the quadratic inequality: $x^2 - 5x + 4 \le 0$

Check whether the values $x = 0$, $x = 2$ and $x = 5$ are solutions of the quadratic inequality:

$x^2 - 5x + 4 \le 0$

Solve
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A cell phone case manufacturing company has determined that its monthly profits, in hundreds of pesos, depend on the number of units produced $x$ (in thousands)…

A cell phone case manufacturing company has determined that its monthly profits, in hundreds of pesos, depend on the number of units produced $x$ (in thousands) according to the function $B(x) = -40x^2 + 280x - 240$. Between which production values does the company make a profit (that is, $B(x) > 0$)?

Production x (thousands) Profit B(x) x₁=1 x₂=6 V(3,5, 250) B(x) > 0 Profit zone 200 100 0 −100
Solve

Negative discriminant and double root

If the discriminant $\Delta=b^2-4ac$ is negative, there are no real roots: the parabola doesn't touch the $x$-axis and its sign is the sign of $a$ on the whole line. So, if $a>0$ and $\Delta<0$, the expression $ax^2+bx+c$ is positive for every $x$. If $\Delta=0$, there is a double root and the polynomial is a perfect square.

Parabola with a negative discriminant that stays always above the horizontal axis. always + solution: ℝ (all real numbers) x² − 2x + 5 > 0 (Δ < 0)
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Given the quadratic equation $x^2 - 6x + k = 0$, determine for which values of the real parameter $k$ the equation has no real solutions. Hint: Remember that for no real…

Given the quadratic equation $x^2 - 6x + k = 0$, determine for which values of the real parameter $k$ the equation has no real solutions. Hint: Remember that for no real solutions to exist, the discriminant $\Delta = b^2 - 4ac$ must satisfy a specific inequality.

Analysis of the Equation: x² - 6x + k = 0 Axis of symmetry (x = 3) x y 0 1 2 3 4 5 1 2 3 Δ > 0 (2 solutions) Δ = 0 (1 solution) Δ < 0 (0 real solutions) Vertex V(3, k - 9) Must float! yv > 0 The parameter "k" shifts the parabola Vertex Method X-coordinate of the vertex: xv = -b / 2a = 6 / 2 = 3 Y-coordinate of the vertex: yv = f(3) = k - 9 Condition (zero roots): yv > 0 → k - 9 ? 0 Discriminant Method Formula: Δ = b² - 4ac Substituting: Δ = (−6)² - 4(1)(k) Δ = 36 - 4k Condition (zero solutions): Δ < 0 → 36 - 4k ? 0 Find the values of k: k = ?
Solve
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Determine for which values of the real parameter $k$ the quadratic inequality $x^2 + (k-2)x + k + 1 > 0$ holds for any real value of $x$. Apply this result to explain…

Determine for which values of the real parameter $k$ the quadratic inequality $x^2 + (k-2)x + k + 1 > 0$ holds for any real value of $x$. Apply this result to explain under what conditions an investment fund guaranteeing this minimum yield would never make a loss, regardless of market fluctuation $x$.

Solve

Problems with quadratic inequalities

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In a diving competition, the height $h$ (in meters) reached by an athlete with respect to time $t$ (in seconds) is given by the expression $h(t) = -t^2 + 4t + 1.5$.…

In a diving competition, the height $h$ (in meters) reached by an athlete with respect to time $t$ (in seconds) is given by the expression $h(t) = -t^2 + 4t + 1.5$. Determine the time interval during which the diver is at a height of more than $4.5$ meters.

h = 4.5 m t (s) h (m) 0 1 2 3 4 5 1 2 3 4 5 6 t₁=? t₂=? Time interval: ? h > 4.5 m Diver Problem data: h(t) = -t² + 4t + 1.5 Condición: h > 4.5
Solve
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The area of a rectangular garden must be less than $54$ $m^2$. It is known that the length of the garden is $3$ meters longer than the width. If we call $x$ the measure…

The area of a rectangular garden must be less than $54$ $m^2$. It is known that the length of the garden is $3$ meters longer than the width. If we call $x$ the measure of the width, set up a quadratic inequality and determine the possible values that this measure can take.

x x + 3 Area < 54 m²
Solve
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A rectangular garden is to be designed so that its perimeter is exactly 60 meters. If the area of the garden must be between 125 and 200 square meters, what values can…

A rectangular garden is to be designed so that its perimeter is exactly 60 meters. If the area of the garden must be between 125 and 200 square meters, what values can the length of the longer side of the garden take? Express the result as an interval.

x (longer side) 30 - x Perimeter = 60 m A = x · (30 - x) 125 ≤ A ≤ 200 m²
Solve

Two conditions at once

When a problem requires several inequalities to hold at the same time, you solve each one separately and take the intersection of the solutions: the values that satisfy all of them at once.

Two solutions on the number line and their intersection, the interval from 1 (open) to 3 (closed). x − 1 > 0 and x² − 9 ≤ 0 x > 1 −3 ≤ x ≤ 3 intersection: (1, 3]
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Find the solution of the following system of linear inequalities with one unknown and express it as an interval: $\begin{cases} x + 3 > 1 \\ 2x - 4 \le 2 \end{cases}$…

Find the solution of the following system of linear inequalities with one unknown and express it as an interval:

$\begin{cases} x + 3 > 1 \\ 2x - 4 \le 2 \end{cases}$

You can use this number line to represent the solutions of each inequality and their intersection:

Number line with the integers from −5 to 5−5−4−3−2−1012345
Solve
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Find the set of real numbers that simultaneously satisfy the following system of inequalities: $$\begin{cases} x^2 - 5x + 4 \leq 0 \\ 2x + 1 > 5 \end{cases}$$

Find the set of real numbers that simultaneously satisfy the following system of inequalities:

$$\begin{cases} x^2 - 5x + 4 \leq 0 \\ 2x + 1 > 5 \end{cases}$$
Solve
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In a sociological experiment on the spread of news on a social network, the number of people who know the news (in hundreds) after $t$ hours is modelled by the function…

In a sociological experiment on the spread of news on a social network, the number of people who know the news (in hundreds) after $t$ hours is modelled by the function $N(t) = -t^2 + 12t - 20$. At the same time, a moderation team intervenes by removing posts so that the maximum reach allowed by the platform's rules is given by $L(t) = 3t - 4$, that is, $N(t) < L(t)$ must hold.

Calculate the time interval (values of $t$ in hours) in which the news spreads naturally ($N(t) > 0$) but its reach already exceeds the limit allowed by the platform.

t N N(t) L(t) 2 10 t₁ t₂ V(6,16)
Solve

Rational inequalities

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Solve the following rational inequality, taking special care with values that make the denominator zero: $$\frac{x^2 - 9}{x - 1} \geq 0$$

Solve the following rational inequality, taking special care with values that make the denominator zero:

$$\frac{x^2 - 9}{x - 1} \geq 0$$
Solve
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A consulting services firm estimates that its monthly profits, expressed in thousands of pesos, are given by the rational expression $B(x) = \frac{x^2 - 9}{x - 5}$,…

A consulting services firm estimates that its monthly profits, expressed in thousands of pesos, are given by the rational expression $B(x) = \frac{x^2 - 9}{x - 5}$, where $x$ represents the number of projects managed simultaneously. For operational reasons, the firm can only manage between 0 and 10 projects. Determine the range of projects for which the firm achieves positive profits that are less than or equal to 4 thousand pesos, that is, solve the system:

$0 < \frac{x^2 - 9}{x - 5} \leq 4$

Limit: B(x) = 4 x = 5 (Asymptote) 0 10 4 ? Solution Interval = [ 0, ? ) Simultaneous projects (x) Profit in thousands of $ (y) Profit: B(x) = x² - 9 x - 5 Objective: 0 < B(x) ≤ 4
Solve
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Solve the following quadratic rational inequality, making sure to correctly consider the domain of the expression and the intervals where the quotient changes sign:…

Solve the following quadratic rational inequality, making sure to correctly consider the domain of the expression and the intervals where the quotient changes sign:

$\frac{3}{x-2} \leq \frac{x}{x+3}$

State in detail the values that are excluded from the solution because they make the denominator zero.

Solve

Absolute value in inequalities

$|u|$ is the distance from $u$ to $0$. So, for $a>0$:

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An electronic components manufacturer has found that the unit production cost $C$ depends on the room temperature $T$ (in ºC) according to the expression…

An electronic components manufacturer has found that the unit production cost $C$ depends on the room temperature $T$ (in ºC) according to the expression $C(T) = |T^2 - 40T + 375|$. For the product to be competitive in the market, the unit cost must not exceed 25 pesos.

The figure shows the graph of $C(T)$ and the line $C = 25$.

Solve the inequality $|T^2 - 40T + 375| \leq 25$ and determine the temperature ranges allowed for manufacturing.

Graph of the cost C(T) = |T² − 40T + 375| versus temperature, together with the horizontal line C = 251015202530102030405025C = 25C(T) = |T² − 40T + 375|Temperature T (ºC)Cost C(T) ($)
Solve

Inequalities with two variables

An inequality like $20x+10y\leq2000$ is represented as a half-plane: you draw the line $20x+10y=2000$ (solid if it has $\leq$ or $\geq$, dashed if strict) and test a point, for example $(0,0)$, to find out which side the solution is on. If there are several conditions, the feasible region is the intersection of all the half-planes.

The vertices of the feasible region

When the feasible region is a polygon, its corners are called vertices. Each vertex is the crossing of two boundary lines, so it is found by solving the system of those two equations. If you want to maximize or minimize a linear quantity (profit, cost), the best value is always reached at a vertex. That is why it is enough to evaluate the quantity at each vertex and compare.

Feasible region with four vertices: (0, 0), (5, 0), (5, 3), and (0, 8), bounded by x ≥ 0, y ≥ 0, x ≤ 5, and x + y ≤ 8. (0, 0) (5, 0) (5, 3) (0, 8)
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An NGO distributes food ($x$) and hygiene ($y$) packs. It has a maximum budget of 2000 pesos. Each food pack costs 20 pesos and each hygiene pack costs 10 pesos. For…

An NGO distributes food ($x$) and hygiene ($y$) packs. It has a maximum budget of 2000 pesos. Each food pack costs 20 pesos and each hygiene pack costs 10 pesos. For logistical reasons, they must send at least 40 food packs and the number of hygiene packs cannot exceed twice the number of food packs. Furthermore, for the aid to be balanced, the total sum of packs must be at least 100.

Write the system of linear inequalities in two variables that represents these constraints and graphically represent the feasible region in the plane.

You can do it on paper and send a photo with your answer.

Solve

Summary

See all the exercises in this section