Equations have one solution, infinitely many solutions, or no solution.

Determine the type of solution for each equation:

[tex]\[
\begin{array}{lll}
\frac{1}{2} y + 3.2 y = 20 & \frac{15}{2} + 2 z - \frac{1}{4} = 4 z + \frac{29}{4} - 2 z & 3 z + 2.5 = 3.2 + 3 z \\
1.1 + \frac{3}{4} x + 2 = 3.1 + \frac{3}{4} x & 4.5 r = 3.2 + 4.5 r & 2 x + 4 = 3 x + \frac{1}{2}
\end{array}
\][/tex]

[tex]\[
\begin{tabular}{|l|l|}
\hline
No Solution & \\
\hline
One Solution & \\
\hline
Infinitely Many Solutions & \\
\hline
\end{tabular}
\][/tex]



Answer :

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]