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To determine which equations have one solution, infinitely many solutions, or no solution, we will analyze each equation step by step.
1. Equation 1: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
Combine like terms:
[tex]\[ \left(\frac{1}{2} + 3.2\right)y = 20 \implies \left(\frac{1}{2} + \frac{32}{10}\right) y = 20 \implies \left(\frac{1}{2} + \frac{16}{5}\right)y = 20 \implies \left(\frac{5 + 32}{10}\right)y = 20 \implies \frac{37}{10}y = 20 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{20 \times 10}{37} \implies y = \frac{200}{37} \][/tex]
Thus, this equation has One Solution.
2. Equation 2: [tex]\(\frac{15}{2} + 2z - \frac{1}{4} = 4z + \frac{29}{4} - 2z\)[/tex]
Combine like terms:
[tex]\[ \frac{15}{2} - \frac{1}{4} + 2z = 2z + \frac{29}{4} \][/tex]
Simplify the constants:
[tex]\[ \frac{30}{4} - \frac{1}{4} + 2z = 2z + \frac{29}{4} \implies \frac{29}{4} + 2z = 2z + \frac{29}{4} \][/tex]
Since the terms on both sides are identical, the equation always holds true, regardless of [tex]\( z \)[/tex]. Thus, this equation has Infinitely Many Solutions.
3. Equation 3: [tex]\(3z + 2.5 = 3.2 + 3z\)[/tex]
Subtract [tex]\( 3z \)[/tex] from both sides:
[tex]\[ 2.5 = 3.2 \][/tex]
This is a contradiction, so the equation has No Solution.
4. Equation 4: [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Combine like terms:
[tex]\[ 1.1 + 2 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \implies 3.1 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
Since the terms on both sides are identical, the equation always holds true, regardless of [tex]\( x \)[/tex]. Thus, this equation has Infinitely Many Solutions.
5. Equation 5: [tex]\(4.5r = 3.2 + 4.5r\)[/tex]
Subtract [tex]\( 4.5r \)[/tex] from both sides:
[tex]\[ 0 = 3.2 \][/tex]
This is a contradiction, so the equation has No Solution.
6. Equation 6: [tex]\(2x + 4 = 3x + \frac{1}{2}\)[/tex]
Subtract [tex]\( 2x \)[/tex] and [tex]\(\frac{1}{2}\)[/tex] from both sides:
[tex]\[ 4 - \frac{1}{2} = x \implies \frac{8}{2} - \frac{1}{2} = x \implies \frac{7}{2} = x \][/tex]
Thus, this equation has One Solution.
To summarize, the classification of the solutions is as follows:
- No Solution: [tex]\(3z + 2.5 = 3.2 + 3z\)[/tex] and [tex]\(4.5r = 3.2 + 4.5r\)[/tex]
- One Solution: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex] and [tex]\(2x + 4 = 3x + \frac{1}{2}\)[/tex]
- Infinitely Many Solutions: [tex]\(\frac{15}{2} + 2z - \frac{1}{4} = 4z + \frac{29}{4} - 2z\)[/tex] and [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
1. Equation 1: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex]
Combine like terms:
[tex]\[ \left(\frac{1}{2} + 3.2\right)y = 20 \implies \left(\frac{1}{2} + \frac{32}{10}\right) y = 20 \implies \left(\frac{1}{2} + \frac{16}{5}\right)y = 20 \implies \left(\frac{5 + 32}{10}\right)y = 20 \implies \frac{37}{10}y = 20 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{20 \times 10}{37} \implies y = \frac{200}{37} \][/tex]
Thus, this equation has One Solution.
2. Equation 2: [tex]\(\frac{15}{2} + 2z - \frac{1}{4} = 4z + \frac{29}{4} - 2z\)[/tex]
Combine like terms:
[tex]\[ \frac{15}{2} - \frac{1}{4} + 2z = 2z + \frac{29}{4} \][/tex]
Simplify the constants:
[tex]\[ \frac{30}{4} - \frac{1}{4} + 2z = 2z + \frac{29}{4} \implies \frac{29}{4} + 2z = 2z + \frac{29}{4} \][/tex]
Since the terms on both sides are identical, the equation always holds true, regardless of [tex]\( z \)[/tex]. Thus, this equation has Infinitely Many Solutions.
3. Equation 3: [tex]\(3z + 2.5 = 3.2 + 3z\)[/tex]
Subtract [tex]\( 3z \)[/tex] from both sides:
[tex]\[ 2.5 = 3.2 \][/tex]
This is a contradiction, so the equation has No Solution.
4. Equation 4: [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
Combine like terms:
[tex]\[ 1.1 + 2 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \implies 3.1 + \frac{3}{4} x = 3.1 + \frac{3}{4} x \][/tex]
Since the terms on both sides are identical, the equation always holds true, regardless of [tex]\( x \)[/tex]. Thus, this equation has Infinitely Many Solutions.
5. Equation 5: [tex]\(4.5r = 3.2 + 4.5r\)[/tex]
Subtract [tex]\( 4.5r \)[/tex] from both sides:
[tex]\[ 0 = 3.2 \][/tex]
This is a contradiction, so the equation has No Solution.
6. Equation 6: [tex]\(2x + 4 = 3x + \frac{1}{2}\)[/tex]
Subtract [tex]\( 2x \)[/tex] and [tex]\(\frac{1}{2}\)[/tex] from both sides:
[tex]\[ 4 - \frac{1}{2} = x \implies \frac{8}{2} - \frac{1}{2} = x \implies \frac{7}{2} = x \][/tex]
Thus, this equation has One Solution.
To summarize, the classification of the solutions is as follows:
- No Solution: [tex]\(3z + 2.5 = 3.2 + 3z\)[/tex] and [tex]\(4.5r = 3.2 + 4.5r\)[/tex]
- One Solution: [tex]\(\frac{1}{2} y + 3.2 y = 20\)[/tex] and [tex]\(2x + 4 = 3x + \frac{1}{2}\)[/tex]
- Infinitely Many Solutions: [tex]\(\frac{15}{2} + 2z - \frac{1}{4} = 4z + \frac{29}{4} - 2z\)[/tex] and [tex]\(1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x\)[/tex]
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