/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 (a) Mark on the coordinate plane... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Mark on the coordinate plane all points \((x, y)\) satisfying the inequality $$ |x-y|<3 $$ (b) Mark on the coordinate plane all points \((x, y)\) satisfying the inequality $$ |x-y+5|<3 $$ (c) Mark on the coordinate plane all points \((x, y)\) satisfying the inequality $$ |x-y|<|x+y| $$

Short Answer

Expert verified
(a) Points between lines \(x-y=3\) and \(x-y=-3\). (b) Points between \(x-y=-2\) and \(x-y=-8\). (c) Points above the line \(y=0\).

Step by step solution

01

Understanding the First Inequality

The first inequality to solve is \(|x-y| < 3\). This means the absolute difference between \(x\) and \(y\) is less than 3. This can be rewritten as two separate inequalities: 1. \(x - y < 3\)2. \(x - y > -3\)Each inequality represents a region between two lines on the coordinate plane: - The line \(x - y = 3\)- The line \(x - y = -3\)The solution region includes all the points between these two lines, but not on these lines.
02

Marking Region for the First Inequality

On a coordinate plane, plot the lines \(x - y = 3\) and \(x - y = -3\). Both lines will have a slope of 1, with the first line intercepting the y-axis at -3 and the second at 3. The solution region is the strip between these lines but excluding the lines themselves. This area can be shaded to indicate the solution set \(|x-y|<3\).
03

Understanding the Second Inequality

The second inequality \(|x-y+5| < 3\) implies the expression inside the absolute value must range between -3 and 3. Rewrite this as an inequality set: 1. \(x - y + 5 < 3\)2. \(x - y + 5 > -3\)Simplifying these gives:- \(x - y < -2\)- \(x - y > -8\)These inequalities form a region between two parallel lines on the coordinate plane.
04

Marking Region for the Second Inequality

Plot the lines \(x - y = -2\) and \(x - y = -8\) on the coordinate plane, both having a slope of 1, intercepting the y-axis at 2 and 8 respectively. The region of interest is between these lines but excluding the lines. Shade this space to indicate the region satisfying \(|x-y+5|<3\).
05

Understanding the Third Inequality

For \(|x-y| < |x+y|\), the solution involves comparing the absolute values of two expressions. Consider splitting it into two cases:**Case 1:** Both expressions positive or both negative:1. \(x - y < x + y\)Simplifies to: - \(-y < y\), which transforms to any point where \(y > 0\).**Case 2:** Different signs:2. \(|x - y| < -(x + y)\)Combining cases, for symmetric boundary consideration:- All points above \(y = 0\) satisfy this inequality.
06

Marking Region for the Third Inequality

On the coordinate plane, the line \(y = 0\) (the x-axis) can be drawn. The solution to \(|x-y| < |x+y|\) consists of shading the entire region above the x-axis, indicating that all points where \(y > 0\) satisfy this inequality.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value
Absolute value is a concept that measures the distance of a number from zero on a number line, regardless of direction. It's often denoted as \( |a| \), where \( a \) is any real number. The key property of absolute value is that it turns any negative number into a positive version of itself.
- **Positive Numbers**: For \( a > 0 \), \( |a| = a \). - **Negative Numbers**: For \( a < 0 \), \( |a| = -a \). - **Zero**: For \( a = 0 \), \( |a| = 0 \).
Let's apply this to inequalities. Consider the expression \( |x-y| < 3 \). This states that the absolute difference between \( x \) and \( y \) is strictly less than 3. We write it as two simultaneous inequalities: \( x-y < 3 \) and \( x-y > -3 \). In coordinate geometry, this represents a band or strip of points between two parallel lines. The understanding of absolute value helps in defining such regions on the coordinate plane. It guides you to determine how far points can be from each other while still satisfying the given inequality.
Graphing Inequalities
Graphing inequalities on the coordinate plane involves shading the region that satisfies an inequality. We start graphing points but extend that to regions, giving a visual interpretation of all possible solutions.
To handle inequalities like \( |x-y+5| < 3 \), we begin by expressing the absolute value as a compound inequality: \( -3 < x-y+5 < 3 \). This splits into two linear inequalities: \( x-y < -2 \) and \( x-y > -8 \). Here, graphing involves plotting the boundary lines \( x-y=-2 \) and \( x-y=-8 \), which are not included in the solution (hence dashed lines).
- **Shade the Correct Region**: Once the lines are drawn, shade the area between them to illustrate the solution. - **Boundary Lines**: Always determine if lines are included in the solution using dashed or solid lines.When graphing, it helps to test a point not on the boundary to see if it lies in the solution region. Graphing inequalities turns abstract expressions into tangible visual realities, facilitating deeper understanding of their solutions.
Coordinate Geometry
Coordinate geometry, or analytic geometry, combines algebra and geometry. It allows us to study geometry using a coordinate plane, giving a precise method for plotting points and lines.
The basic idea involves plotting points with coordinates \( (x, y) \). Lines on the plane can be expressed as equations, like \( x-y=3 \), which rearranges to describe a line with a certain slope and intercept. For example, lines such as \( x-y=3 \) or \( x-y=-3 \) have the same slope but different intercepts. This creates parallel lines.
- **Intersections and Regions**: When solving inequalities, these equations define boundaries. The region of solutions is often between or outside these boundary lines.- **Slopes and Intercepts**: Understanding the slope (rise over run) and intercepts (where the line crosses axes) is crucial for plotting accurate lines.In problems like \( |x-y| < |x+y| \), coordinate geometry helps to visualize and understand which regions satisfy the condition. You're visually separating the plane into regions that meet the inequality's criteria, significantly easing the solution process.

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Most popular questions from this chapter

(Farey series) When the fully cancelled fractions in [0,1] with denominator \(\leqslant n\) are arranged in increasing order, the result is called the Farey series (or Farey sequence) of order \(n\). \(\begin{array}{ll}\text { Order 1: } & \frac{0}{1}<\frac{1}{1} \\ \text { Order } 2: & \frac{0}{1}<\frac{1}{2}<\frac{1}{1} \\ \text { Order 3: } & \frac{0}{1}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{1}{1} \\ \text { Order 4: } & \frac{0}{1}<\frac{1}{4}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<\frac{1}{1}\end{array}\) (a) Write down the full Farey series (or sequence) of order 7 . (b)(i) Imagine the points \(0.1,0.2,0.3, \ldots, 0.9\) dividing the interval [0,1] into ten subintervals of length \(\frac{1}{10}\). Now insert the eight points corresponding to $$ \frac{1}{9}, \frac{2}{9}, \frac{3}{9}, \ldots, \frac{8}{9} $$ Into which of the ten subintervals do they fall? (ii) Imagine the \(n\) points $$ \frac{1}{n+1}, \frac{2}{n+1}, \frac{3}{n+1}, \ldots, \frac{n}{n+1} $$ dividing the interval [0,1] into \(n+1\) subintervals of length \(\frac{1}{n+1}\). Now insert the \(n-1\) points $$ \frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \ldots, \frac{n-1}{n} $$ Into which of the \(n+1\) subintervals do they fall? (iii) In passing from the Farey series of order \(n\) to the Farey series of order \(n+1,\) we insert fractions of the form \(\frac{k}{n+1}\) between certain pairs of adjacent fractions in the Farey series of order \(n\). If \(\frac{a}{b}<\frac{c}{d}\) are adjacent fractions in the Farey series of order \(n\), prove that, when adding fractions for the Farey series of order \(n+1,\) at most one fraction is inserted between \(\frac{a}{b}\) and \(\frac{c}{d}\). (c) Note: It is worth struggling to prove the two results in part (c). But do not be surprised if they prove to be elusive \(-\) in which case, be prepared to simply use the result in part (c)(ii) to solve part (d). (i) In the Farey series of order \(n\) the first two fractions are \(\frac{0}{1}<\frac{1}{n},\) and the last two fractions are \(\frac{n-1}{n}<\frac{1}{1} .\) Prove that every other adjacent pair of fractions \(\frac{a}{b}<\frac{c}{d}\) in the Farey series of order \(n\) satisfies \(b d>n\) (ii) Let \(\frac{a}{b}<\frac{c}{d}\) be adjacent fractions in the Farey series of order \(n\). Prove (by induction on \(n)\) that \(b c-a d=1\). (d) Prove that if $$ \frac{a}{b}<\frac{c}{d}<\frac{e}{f} $$ are three successive terms in any Farey series, then $$ \frac{c}{d}=\frac{a+e}{b+f} $$

Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation $$ x^{2}+b x+c=0 $$ (a)(i) Write \(\alpha^{2}+\beta^{2}\) in terms of \(b\) and \(c\) only. (ii) Write \(\alpha^{2} \beta+\beta^{2} \alpha\) in terms of \(b\) and \(c\) only. (iii) Write \(\alpha^{3}+\beta^{3}-3 \alpha \beta\) in terms of \(b\) and \(c\) only. (b)(i) Write \(\alpha-\beta\) in terms of \(b\) and \(c\) only. (ii) Write \(\alpha^{2} \beta-\beta^{2} \alpha\) in terms of \(b\) and \(c\) only. (iii) Write \(\alpha^{3}-\beta^{3}\) in terms of \(b\) and \(c\) only.

(a)(i) For any positive real numbers \(a, b,\) prove that $$ \sqrt{a}+\sqrt{b}=\sqrt{a+b+\sqrt{4 a b}} $$ (ii) Simplify \(\sqrt{5+\sqrt{24}}\). (b) (i) Find a similar formula for \(\sqrt{a}-\sqrt{b}\). (ii) Simplify \(\sqrt{5-\sqrt{16}}\) and \(\sqrt{6-\sqrt{20}}\).

Solve the following systems of simultaneous equations. (a)(i) \(x+y=1, \quad y+z=2, \quad x+z=3\) (ii) \(u v=2, \quad v w=4, \quad u w=8\) (b) (i) \(x+y=2, \quad y+z=3, \quad x+z=4\) (ii) \(u v=6, \quad v w=10, \quad u w=15\) (iii) \(u v=6, \quad v w=10, \quad u w=30\) (iv) \(u v=4, \quad v w=8, \quad u w=16\)

Solve the following quadratic equations: (a) \(x^{2}-3 x+2=0\) (b) \(x^{2}-1=0\) (c) \(x^{2}-2 x+1=0\) (d) \(x^{2}+\sqrt{2} x-1=0\) (e) \(x^{2}+x-\sqrt{2}=0\) (f) \(x^{2}+1=0\) (g) \(x^{2}+\sqrt{2} x+1=0\)

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