/*! 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 7 Construct a sketch of each syste... [FREE SOLUTION] | 91Ó°ÊÓ

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Construct a sketch of each system by hand and then estimate the solution(s) to the system (if any). $$ \begin{array}{ll} \text { a. } x+2 y=1 & \text { b. } x+y=9 \\ x+4 y=3 & 2 x-3 y=-2 \end{array} $$

Short Answer

Expert verified
System (a) solution: \((-1, 1)\), System (b) solution: \((5, 4)\)

Step by step solution

01

Title - Write down the system of equations

Consider the two systems of equations given:System (a): 1. \( x + 2y = 1 \)2. \( x + 4y = 3 \)System (b): 1. \( x + y = 9 \)2. \( 2x - 3y = -2 \)
02

Title - Solve system (a) using substitution or elimination

Using elimination method for system (a):First equation: \( x + 2y = 1 \)Second equation: \( x + 4y = 3 \)Subtract the first equation from the second to eliminate x:\((x + 4y) - (x + 2y) = 3 - 1 \)\( 2y = 2 \)\( y = 1 \)Substitute \( y = 1 \) back into the first equation:\( x + 2(1) = 1 \)\( x + 2 = 1 \)\( x = -1 \)
03

Title - Solve system (b) using substitution or elimination

Using substitution method for system (b):First equation: \( x + y = 9 \)Second equation: \( 2x - 3y = -2 \)Express \( x \) from the first equation:\( x = 9 - y \)Substitute \( x = 9 - y \) into the second equation:\( 2(9 - y) - 3y = -2 \)\( 18 - 2y - 3y = -2 \)\( 18 - 5y = -2 \)\( -5y = -20 \)\( y = 4 \)Substitute \( y = 4 \) back into \( x = 9 - y \):\( x = 9 - 4 \)\( x = 5 \)
04

Title - Sketch the graphs for each system

Sketch the lines for each equation in both systems.System (a):1. First equation: \( x + 2y = 1 \), y-intercept (0, 0.5) and x-intercept (1, 0)2. Second equation: \( x + 4y = 3 \), y-intercept (0, 0.75) and x-intercept (3, 0)System (b):1. First equation: \( x + y = 9 \), y-intercept (0, 9) and x-intercept (9, 0)2. Second equation: \( 2x - 3y = -2 \), y-intercept (0, 2/3) and x-intercept (-1, 0)
05

Title - Estimate the solutions

From the sketches:For system (a), the lines intersect at (-1, 1).For system (b), the lines intersect at (5, 4).

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Key Concepts

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

Elimination Method
The elimination method is a way to solve systems of equations by removing one variable. This is achieved by adding or subtracting the equations so that one of the variables cancels out.
Let's consider system (a):
1. First equation: \( x + 2y = 1 \)
2. Second equation: \( x + 4y = 3 \)
By subtracting the first equation from the second, we eliminate \( x \):
\( (x + 4y) - (x + 2y) = 3 - 1 \)
This simplifies to \( 2y = 2 \), so \( y = 1 \).
Now we substitute \( y = 1 \) back into the first equation to find \( x \):
\( x + 2(1) = 1 \),
which simplifies to \( x = -1 \).
Therefore, the solution to system (a) is \( (x, y) = (-1, 1) \).
Substitution Method
The substitution method involves solving one of the equations in terms of one variable and then substituting that expression into the other equation.
Let's consider system (b):
1. First equation: \( x + y = 9 \)
2. Second equation: \( 2x - 3y = -2 \)
We solve the first equation for \( x \):
\( x = 9 - y \).
Next, we substitute this expression for \( x \) into the second equation:
\( 2(9 - y) - 3y = -2 \).
This simplifies to \( 18 - 2y - 3y = -2 \), then to \( 18 - 5y = -2 \).
Solving for \( y \):
\( -5y = -20 \), thus \( y = 4 \).
We then substitute \( y = 4 \) back into the expression for \( x \):
\( x = 9 - 4 \),
which simplifies to \( x = 5 \).
Therefore, the solution to system (b) is \( (x, y) = (5, 4) \).
Graphical Solution
The graphical solution method involves plotting both equations of a system on a coordinate plane and identifying their intersection point(s).
For system (a), we plot:
1. \( x + 2y = 1 \) with intercepts at (0, 0.5) and (1, 0).
2. \( x + 4y = 3 \) with intercepts at (0, 0.75) and (3, 0).
The intersection of these lines, which is the solution for system (a), occurs at \( (-1, 1) \).
For system (b), we plot:
1. \( x + y = 9 \) with intercepts at (0, 9) and (9, 0).
2. \( 2x - 3y = -2 \) with intercepts at (0, 2/3) and (-1, 0).
The intersection of these lines, which is the solution for system (b), occurs at \( (5, 4) \).
By graphing the equations, we visually confirm the solutions found through algebraic methods.

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