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Sagot :
Let's analyze each of the given equations step-by-step to classify them according to the type of solution they have. We will determine if an equation has no solution, one solution, or infinitely many solutions.
### Equation 1: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
First, let's combine like terms:
[tex]\[ \left(\frac{1}{2} + 3.2\right) y = 20 \][/tex]
This simplifies to:
[tex]\[ 3.7 y = 20 \][/tex]
To solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{20}{3.7} \][/tex]
This equation has one solution.
### Equation 2: [tex]\(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 \)[/tex]
Simplify both sides of the equation:
[tex]\[ \frac{15}{2} + 2 z - \frac{1}{4} = 4 z + 5 - 2 z + 3 z + 2.5 \][/tex]
By simplifying the constants and combining the terms involving [tex]\( z \)[/tex]:
[tex]\[ \frac{30}{4} + 8z/4 - 1/4 = 4z + 5 - 2z + 3z + 2.5 \][/tex]
[tex]\[ 7.25 + 2z = 5z + 7.5 \][/tex]
Rearrange to solve for [tex]\( z \)[/tex]:
[tex]\[ 7.25 - 7.5 = 5z - 2z \][/tex]
[tex]\[ -0.25 = 3z \][/tex]
[tex]\[ z = -\frac{0.25}{3} \][/tex]
This equation has one solution.
### Equation 3: [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Simplify and combine like terms on both sides:
[tex]\[ \left(1.1 + 2\right) + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
[tex]\[ 3.1 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
This holds true for all [tex]\( x \)[/tex].
This equation has infinitely many solutions.
### Equation 4: [tex]\(4.5 r = 3.2 + 4.5 r\)[/tex]
To solve this equation, isolate the term involving [tex]\( r \)[/tex] on one side:
[tex]\[ 4.5 r - 4.5 r = 3.2 \][/tex]
[tex]\[ 0 = 3.2 \][/tex]
This is a contradiction.
This equation has no solution.
### Equation 5: [tex]\(2 x + 4 = 3 x + \frac{1}{2}\)[/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 2 x + 4 = 3 x + \frac{1}{2} \][/tex]
[tex]\[ 4 - \frac{1}{2} = 3 x - 2 x \][/tex]
[tex]\[ \frac{7.5}{4} = x \][/tex]
This equation has one solution.
### Classification Summary:
- No Solution: [tex]\(4.5 r = 3.2 + 4.5 r\)[/tex]
- One Solution:
- [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
- [tex]\(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 = 3.2 + 3 z \)[/tex]
- [tex]\(2 x + 4 = 3 x + \frac{1}{2} \)[/tex]
- Infinitely Many Solutions:
- [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Hence, the completed table is:
[tex]\[ \begin{tabular}{|l|l|} \hline No Solution & \(4.5 r = 3.2 + 4.5 r\) \\ \hline One Solution & \begin{tabular}{l} \(\frac{1}{2} y + 3.2 y = 20\) \\ \(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 = 3.2 + 3 z\) \\ \(2 x + 4 = 3 x + \frac{1}{2}\) \\ \end{tabular} \\ \hline Infinitely Many Solutions & \(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\) \\ \hline \end{tabular} \][/tex]
### Equation 1: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
First, let's combine like terms:
[tex]\[ \left(\frac{1}{2} + 3.2\right) y = 20 \][/tex]
This simplifies to:
[tex]\[ 3.7 y = 20 \][/tex]
To solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{20}{3.7} \][/tex]
This equation has one solution.
### Equation 2: [tex]\(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 \)[/tex]
Simplify both sides of the equation:
[tex]\[ \frac{15}{2} + 2 z - \frac{1}{4} = 4 z + 5 - 2 z + 3 z + 2.5 \][/tex]
By simplifying the constants and combining the terms involving [tex]\( z \)[/tex]:
[tex]\[ \frac{30}{4} + 8z/4 - 1/4 = 4z + 5 - 2z + 3z + 2.5 \][/tex]
[tex]\[ 7.25 + 2z = 5z + 7.5 \][/tex]
Rearrange to solve for [tex]\( z \)[/tex]:
[tex]\[ 7.25 - 7.5 = 5z - 2z \][/tex]
[tex]\[ -0.25 = 3z \][/tex]
[tex]\[ z = -\frac{0.25}{3} \][/tex]
This equation has one solution.
### Equation 3: [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Simplify and combine like terms on both sides:
[tex]\[ \left(1.1 + 2\right) + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
[tex]\[ 3.1 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
This holds true for all [tex]\( x \)[/tex].
This equation has infinitely many solutions.
### Equation 4: [tex]\(4.5 r = 3.2 + 4.5 r\)[/tex]
To solve this equation, isolate the term involving [tex]\( r \)[/tex] on one side:
[tex]\[ 4.5 r - 4.5 r = 3.2 \][/tex]
[tex]\[ 0 = 3.2 \][/tex]
This is a contradiction.
This equation has no solution.
### Equation 5: [tex]\(2 x + 4 = 3 x + \frac{1}{2}\)[/tex]
Rearrange the equation to solve for [tex]\( x \)[/tex]:
[tex]\[ 2 x + 4 = 3 x + \frac{1}{2} \][/tex]
[tex]\[ 4 - \frac{1}{2} = 3 x - 2 x \][/tex]
[tex]\[ \frac{7.5}{4} = x \][/tex]
This equation has one solution.
### Classification Summary:
- No Solution: [tex]\(4.5 r = 3.2 + 4.5 r\)[/tex]
- One Solution:
- [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
- [tex]\(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 = 3.2 + 3 z \)[/tex]
- [tex]\(2 x + 4 = 3 x + \frac{1}{2} \)[/tex]
- Infinitely Many Solutions:
- [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Hence, the completed table is:
[tex]\[ \begin{tabular}{|l|l|} \hline No Solution & \(4.5 r = 3.2 + 4.5 r\) \\ \hline One Solution & \begin{tabular}{l} \(\frac{1}{2} y + 3.2 y = 20\) \\ \(\frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{20}{4} - 2 z + 3 z + 2.5 = 3.2 + 3 z\) \\ \(2 x + 4 = 3 x + \frac{1}{2}\) \\ \end{tabular} \\ \hline Infinitely Many Solutions & \(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\) \\ \hline \end{tabular} \][/tex]
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