Get detailed and accurate responses to your questions on IDNLearn.com. Get timely and accurate answers to your questions from our dedicated community of experts who are here to help you.
Sagot :
To solve for the slopes of the given pair of equations and determine the relationship between the lines, let's follow these steps:
### Step 1: Identify the equations
We have the following pair of linear equations:
[tex]\[ \begin{array}{l} 5x + 4y = 1 \\ 20x + 16y = 3 \end{array} \][/tex]
### Step 2: Put each equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] )
#### For the first equation [tex]\( 5x + 4y = 1 \)[/tex]:
1. Solve for [tex]\( y \)[/tex]:
[tex]\[ 4y = -5x + 1 \][/tex]
2. Divide everything by 4:
[tex]\[ y = -\frac{5}{4}x + \frac{1}{4} \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of the first equation is:
[tex]\[ m_1 = -\frac{5}{4} \][/tex]
#### For the second equation [tex]\( 20x + 16y = 3 \)[/tex]:
1. Solve for [tex]\( y \)[/tex]:
[tex]\[ 16y = -20x + 3 \][/tex]
2. Divide everything by 16:
[tex]\[ y = -\frac{5}{4}x + \frac{3}{16} \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of the second equation is:
[tex]\[ m_2 = -\frac{5}{4} \][/tex]
### Step 3: Determine the relationship between the lines
To check whether the lines are parallel, perpendicular, or neither:
- Lines are parallel if their slopes are equal.
- Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
- If neither condition is met, then the lines are neither parallel nor perpendicular.
Comparing the slopes:
[tex]\[ m_1 = -\frac{5}{4}, \quad m_2 = -\frac{5}{4} \][/tex]
Since [tex]\( m_1 = m_2 \)[/tex], the lines are parallel.
### Summary
- The slope of the first equation [tex]\( 5x + 4y = 1 \)[/tex] is [tex]\(-\frac{5}{4}\)[/tex].
- The slope of the second equation [tex]\( 20x + 16y = 3 \)[/tex] is [tex]\(-\frac{5}{4}\)[/tex].
- The lines are parallel.
So, the correct answers are:
- The slope of [tex]\( 5x + 4y = 1 \)[/tex]: [tex]\(-\frac{5}{4}\)[/tex]
- The slope of [tex]\( 20x + 16y = 3 \)[/tex]: [tex]\(-\frac{5}{4}\)[/tex]
- The lines are parallel.
### Step 1: Identify the equations
We have the following pair of linear equations:
[tex]\[ \begin{array}{l} 5x + 4y = 1 \\ 20x + 16y = 3 \end{array} \][/tex]
### Step 2: Put each equation in slope-intercept form ( [tex]\( y = mx + b \)[/tex] )
#### For the first equation [tex]\( 5x + 4y = 1 \)[/tex]:
1. Solve for [tex]\( y \)[/tex]:
[tex]\[ 4y = -5x + 1 \][/tex]
2. Divide everything by 4:
[tex]\[ y = -\frac{5}{4}x + \frac{1}{4} \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of the first equation is:
[tex]\[ m_1 = -\frac{5}{4} \][/tex]
#### For the second equation [tex]\( 20x + 16y = 3 \)[/tex]:
1. Solve for [tex]\( y \)[/tex]:
[tex]\[ 16y = -20x + 3 \][/tex]
2. Divide everything by 16:
[tex]\[ y = -\frac{5}{4}x + \frac{3}{16} \][/tex]
Therefore, the slope ([tex]\(m\)[/tex]) of the second equation is:
[tex]\[ m_2 = -\frac{5}{4} \][/tex]
### Step 3: Determine the relationship between the lines
To check whether the lines are parallel, perpendicular, or neither:
- Lines are parallel if their slopes are equal.
- Lines are perpendicular if the product of their slopes is [tex]\(-1\)[/tex].
- If neither condition is met, then the lines are neither parallel nor perpendicular.
Comparing the slopes:
[tex]\[ m_1 = -\frac{5}{4}, \quad m_2 = -\frac{5}{4} \][/tex]
Since [tex]\( m_1 = m_2 \)[/tex], the lines are parallel.
### Summary
- The slope of the first equation [tex]\( 5x + 4y = 1 \)[/tex] is [tex]\(-\frac{5}{4}\)[/tex].
- The slope of the second equation [tex]\( 20x + 16y = 3 \)[/tex] is [tex]\(-\frac{5}{4}\)[/tex].
- The lines are parallel.
So, the correct answers are:
- The slope of [tex]\( 5x + 4y = 1 \)[/tex]: [tex]\(-\frac{5}{4}\)[/tex]
- The slope of [tex]\( 20x + 16y = 3 \)[/tex]: [tex]\(-\frac{5}{4}\)[/tex]
- The lines are parallel.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.