/*! 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 20 (a) Graph the pair of equations,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 20

(a) Graph the pair of equations, and by zooming in on the intersection point, estimate the solution of the system (each value to the nearest one-tenth ). (b) Use the substitution method to determine the solution. Check that your answer is consistent with the graphical estimate in part (a). $$\left\\{\begin{array}{l} \sqrt{2} x-\sqrt{3} y=\sqrt{3} \\ \sqrt{3} x-\sqrt{8} y=\sqrt{2} \end{array}\right.$$

Short Answer

Expert verified
The solution is approximately \( (x, y) = (1.4, 0.6) \).

Step by step solution

01

Graphing the Equations

To graph the equations, rewrite them in slope-intercept form. For the first equation, \( \sqrt{2}x - \sqrt{3}y = \sqrt{3} \), solve for \( y \): \[ y = \frac{\sqrt{2}}{\sqrt{3}}x - 1 \]For the second equation, \( \sqrt{3}x - \sqrt{8}y = \sqrt{2} \), solve for \( y \):\[ y = \frac{\sqrt{3}}{\sqrt{8}}x - \frac{\sqrt{2}}{\sqrt{8}} \]Plot both lines on a graph and locate their intersection point to estimate the solution.
02

Estimating the Intersection

By zooming in on the graph, observe where the two lines intersect. The approximate intersection point is estimated as \( (x, y) \approx (1.4, 0.6) \).
03

Substitution Method: Solving for One Variable

Use the substitution method by solving the first equation for \( x \):\[ x = \frac{\sqrt{3} + \sqrt{3}y}{\sqrt{2}} \]Substitute this expression for \( x \) in the second equation to find \( y \).
04

Solving for the Second Variable

Substitute the expression for \( x \) into the second equation \( \sqrt{3}x - \sqrt{8}y = \sqrt{2} \):\[ \sqrt{3}\left(\frac{\sqrt{3} + \sqrt{3}y}{\sqrt{2}}\right) - \sqrt{8}y = \sqrt{2} \]Simplify and solve for \( y \).
05

Calculate and Simplify Solution for y

Expand and simplify the resulting equation:\[ \frac{3}{\sqrt{2}} + \frac{3\sqrt{3}y}{\sqrt{2}} - \sqrt{8}y = \sqrt{2} \]Combine like terms and solve for \( y \). After calculations, we find \( y \approx 0.6 \).
06

Solving for x Using Found y

Substitute \( y = 0.6 \) back into the expression for \( x \):\[ x = \frac{\sqrt{3} + \sqrt{3}(0.6)}{\sqrt{2}} \] Calculate to find \( x \approx 1.4 \).
07

Verification with Graphing

Check if the calculated solution \( (1.4, 0.6) \) is consistent with the estimated value from the graph. It matches closely, confirming the accuracy of both methods.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Graphing Linear Equations
Graphing linear equations is a crucial skill in solving systems of equations because it visually shows how two equations interact on a coordinate plane. Each linear equation represents a line. To graph these equations, they must be expressed in a convenient form, often the slope-intercept form, which we will discuss later. The goal is to plot each line accurately using the equation's properties, such as its slope and y-intercept. Once both lines are drawn, observing where they intersect can visually show the solution to the system. In our exercise, we plotted the given equations after converting them into slope-intercept form and estimated the intersection point on the graph.
Substitution Method
The substitution method is a reliable algebraic technique used to find the exact solution of a system of equations. It involves expressing one variable in terms of the other from one equation, then replacing that expression in the second equation. This method effectively simplifies the system to a single equation with one variable, making it easier to solve. For instance, in our scenario, we solved the first equation for \( x \), then substituted this expression into the second equation, allowing us to solve for \( y \). Once \( y \) was found, we substituted back to find \( x \). This method is particularly useful when one equation is easily solvable for a single variable.
Intersection Point
The intersection point of two lines derived from linear equations represents the solution to the system, where both equations are simultaneously satisfied. In graphical representation, the intersection point's coordinates
  • tell us the values of \( x \) and \( y \) that solve both equations,
  • are easily estimated by observing where the lines meet on the graph
For our exercise, after plotting the equations, we zoomed in on the graph to closely estimate the intersection point at approximately \( (1.4, 0.6) \). This visual estimation serves as a preliminary check against the exact solutions obtained through algebraic methods.
Solution Estimation
Solution estimation in the context of graphing involves approximating the coordinates of the intersection point. It provides a quick qualitative insight into the solutions of the system. By looking closely at the intersection, especially by zooming in, we can make a fairly accurate guess of where the solution lies. Estimating solutions graphically is a helpful verification step when an exact algebraic solution is also obtained. In our case, the graphical estimation of the solution as \( (1.4, 0.6) \) was cross-checked with results from the substitution method, providing confidence in the accuracy of our solution.
Slope-Intercept Form
The slope-intercept form is a popular way to express linear equations. It is given by the formula \( y = mx + b \), where \( m \) is the slope of the line, and \( b \) is the y-intercept.
  • Slope \( m \) tells us how steep the line is and whether it slopes upwards or downwards.
  • The y-intercept, \( b \), indicates where the line crosses the y-axis.
By converting equations to this form, we easily find these features, which simplifies the graphing process. In our exercise, both given equations were converted into slope-intercept form to plot them and find their intersection. This form is critical not just for graphing but also for understanding the line's behavior in the coordinate system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as follows: $$ A=\left(\begin{array}{rr} 2 & 3 \\ -1 & 4 \end{array}\right) \quad B=\left(\begin{array}{rr} 1 & -1 \\ 3 & 0 \end{array}\right) \quad C=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$\begin{aligned} &D=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right) \quad E=\left(\begin{array}{rr} 2 & 1 \\ 8 & -1 \\ 6 & 5 \end{array}\right)\\\ &F=\left(\begin{array}{rr} 5 & -1 \\ -4 & 0 \\ 2 & 3 \end{array}\right) \quad G=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned}$$ In each exercise, carry out the indicated matrix operations if they are defined. If an operation is not defined, say so. $$C^{2}$$

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{aligned} 4 u+3 v-2 w &=14 \\ u+2 v-3 w &=6 \\ 2 u-v+4 w &=2 \end{aligned}\right.$$

Find all right triangles for which the perimeter is 24 units and the area is 24 square units.

In a two-digit number, the sum of the digits is \(14 .\) Twice the tens digit exceeds the units digit by one. Find the number.

The trace of a \(2 \times 2\) matrix \(\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) is defined by \(\operatorname{tr}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=a+d\) (a) If \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right),\) verify that \(\operatorname{tr}(A+B)=\operatorname{tr} A+\operatorname{tr} B\) (b) If \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(B=\left(\begin{array}{ll}e & f \\ g & h\end{array}\right),\) show that \(\operatorname{tr}(A+B)=\operatorname{tr} A+\operatorname{tr} B\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.