/*! 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 69 \(8 x-9 y=6\) \(8 x+6 y=-5\)... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 69

\(8 x-9 y=6\) \(8 x+6 y=-5\)

Short Answer

Expert verified
The solution is \(x = -\frac{3}{40}\) and \(y = -\frac{11}{15}\).

Step by step solution

01

- Define the equations

First, identify the given system of linear equations: 1) \(8x - 9y = 6\) 2) \(8x + 6y = -5\)
02

- Eliminate one variable

Subtract equation 2 from equation 1 to eliminate the variable \(x\). The resulting equation is: \((8x - 9y) - (8x + 6y) = 6 - (-5)\) Simplifying, we get: \(-15y = 11\) Therefore, \(y = \frac{11}{-15}\) or \(y = -\frac{11}{15}\)
03

- Substitute back to find x

Substitute \(y = -\frac{11}{15}\) into one of the original equations. Using \(8x + 6y = -5\), we have: \(8x + 6\left(-\frac{11}{15}\right) = -5\) Simplify and solve for \(x\): \(8x - \frac{66}{15} = -5\) Multiply both sides by 15 to clear the fraction: \(120x - 66 = -75\) Add 66 to both sides: \(120x = -9\) Divide by 120: \(x = -\frac{9}{120}\) Simplify the fraction: \(x = -\frac{3}{40}\)
04

- Verify the solution

Substitute \(x = -\frac{3}{40}\) and \(y = -\frac{11}{15}\) back into the original equations to check the validity. Both equations should hold true.

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Key Concepts

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

Linear Equations
Linear equations are equations of the first degree, which means they have no exponents higher than one. They typically look like this: \[ ax + by = c \].
This is a straight line when graphed on a coordinate plane.
In this exercise, we have two linear equations: \[ 8x - 9y = 6 \]
and \[ 8x + 6y = -5 \]. Our goal is to find the values of the variables 'x' and 'y' that make both equations true at the same time.
By solving, we find the intersection point of these two lines, which gives us the solution.
Elimination Method
The elimination method is a technique to solve systems of linear equations. It involves adding or subtracting the equations to eliminate one of the variables.
This helps in simplifying the system so that we can solve for the other variable.
Here鈥檚 a recap of how we used the elimination method in the exercise:
  • First, we identified the equations: \[ 8x - 9y = 6 \] \[ 8x + 6y = -5 \].
  • Next, we subtracted the second equation from the first to eliminate 'x': \[ (8x - 9y) - (8x + 6y) = 6 - (-5) \].
  • This simplified to: \[ -15y = 11 \].
With 'x' eliminated, we could then solve for 'y'.
Substitution Method
The substitution method involves solving one equation for one variable and then substituting that solution into the other equation.
It's another technique used to solve systems of linear equations, particularly when elimination isn't easy.
In our exercise, after finding \[ y = -\frac{11}{15} \],
we substituted 'y' back into one of the original equations to find 'x':
  • We chose the second equation: \[ 8x + 6(-\frac{11}{15}) = -5 \].
  • This simplified to: \[ 8x - \frac{66}{15} = -5 \].
  • We then cleared the fraction and solved for 'x'.
By substituting back, we found \[ x = -\frac{3}{40} \].
To ensure our solution was correct, we substituted both 'x' and 'y' back into both original equations to verify.

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Most popular questions from this chapter

The table gives the worldwide number \(y\) of monthly active Twitter users in millions for selected years, where \(x=1\) represents \(2014, x=2\) represents \(2015,\) and so on. $$ \begin{aligned} &\begin{array}{|c|c|c|} \hline \text { Year } & x & \begin{array}{c} \text { Twitter Users } \\ \text { (in millions) } \end{array} \\ 2014 & 1 & 271 \\ 2015 & 2 & 304 \\ 2016 & 3 & 313 \\ 2017 & 4 & 326 \\ \hline \end{array}\\\ &\text { Data from Twitter. } \end{aligned} $$

If a line is vertical, what is true of any line that is perpendicular to it?

Graph each linear equation. $$ y-2 x=0 $$

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\(x-5=0\)
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