Answer :
To solve the equation \(2x + 3y = 5\) for \(x\), we need to check each of the given options to see which one satisfies the equation.
1. Option 1: \(x = -3y + \frac{5}{2}\)
Substitute \(x = -3y + \frac{5}{2}\) into the original equation:
[tex]\[ 2(-3y + \frac{5}{2}) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2(-3y) + 2 \cdot \frac{5}{2} + 3y = 5 \][/tex]
This becomes:
[tex]\[ -6y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ -3y + 5 = 5 \][/tex]
Subtract 5 from both sides:
[tex]\[ -3y = 0 \][/tex]
Divide by -3:
[tex]\[ y = 0 \][/tex]
Substituting \(y = 0\) back, we see that it works only when \(y = 0\).
2. Option 2: \(x = \frac{-3}{2}y + 5\)
Substitute \(x = \frac{-3}{2}y + 5\) into the original equation:
[tex]\[ 2\left(\frac{-3}{2}y + 5\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2 \cdot \frac{-3}{2}y + 2 \cdot 5 + 3y = 5 \][/tex]
This becomes:
[tex]\[ -3y + 10 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 10 \neq 5 \][/tex]
So, this option is incorrect.
3. Option 3: \(x = \frac{-3y + 5}{2}\)
Substitute \(x = \frac{-3y + 5}{2}\) into the original equation:
[tex]\[ 2\left(\frac{-3y + 5}{2}\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{2(-3y + 5)}{2} + 3y = 5 \][/tex]
Which simplifies to:
[tex]\[ -3y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 5 = 5 \][/tex]
This option satisfies the original equation, so it is correct.
4. Option 4: \(x = \frac{3y + 5}{2}\)
Substitute \(x = \frac{3y + 5}{2}\) into the original equation:
[tex]\[ 2\left(\frac{3y + 5}{2}\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{2(3y + 5)}{2} + 3y = 5 \][/tex]
Which simplifies to:
[tex]\[ 3y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 6y + 5 = 5 \][/tex]
Subtract 5 from both sides:
[tex]\[ 6y = 0 \][/tex]
Divide by 6:
[tex]\[ y = 0 \][/tex]
Substituting \(y = 0\) back, we see that it works only when \(y = 0\).
Therefore, after checking all the options, the correct solution for the equation \(2x + 3y = 5\) is:
[tex]\[ \boxed{x = \frac{-3y + 5}{2}} \][/tex]
1. Option 1: \(x = -3y + \frac{5}{2}\)
Substitute \(x = -3y + \frac{5}{2}\) into the original equation:
[tex]\[ 2(-3y + \frac{5}{2}) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2(-3y) + 2 \cdot \frac{5}{2} + 3y = 5 \][/tex]
This becomes:
[tex]\[ -6y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ -3y + 5 = 5 \][/tex]
Subtract 5 from both sides:
[tex]\[ -3y = 0 \][/tex]
Divide by -3:
[tex]\[ y = 0 \][/tex]
Substituting \(y = 0\) back, we see that it works only when \(y = 0\).
2. Option 2: \(x = \frac{-3}{2}y + 5\)
Substitute \(x = \frac{-3}{2}y + 5\) into the original equation:
[tex]\[ 2\left(\frac{-3}{2}y + 5\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2 \cdot \frac{-3}{2}y + 2 \cdot 5 + 3y = 5 \][/tex]
This becomes:
[tex]\[ -3y + 10 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 10 \neq 5 \][/tex]
So, this option is incorrect.
3. Option 3: \(x = \frac{-3y + 5}{2}\)
Substitute \(x = \frac{-3y + 5}{2}\) into the original equation:
[tex]\[ 2\left(\frac{-3y + 5}{2}\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{2(-3y + 5)}{2} + 3y = 5 \][/tex]
Which simplifies to:
[tex]\[ -3y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 5 = 5 \][/tex]
This option satisfies the original equation, so it is correct.
4. Option 4: \(x = \frac{3y + 5}{2}\)
Substitute \(x = \frac{3y + 5}{2}\) into the original equation:
[tex]\[ 2\left(\frac{3y + 5}{2}\right) + 3y = 5 \][/tex]
Simplify inside the parentheses:
[tex]\[ \frac{2(3y + 5)}{2} + 3y = 5 \][/tex]
Which simplifies to:
[tex]\[ 3y + 5 + 3y = 5 \][/tex]
Combine like terms:
[tex]\[ 6y + 5 = 5 \][/tex]
Subtract 5 from both sides:
[tex]\[ 6y = 0 \][/tex]
Divide by 6:
[tex]\[ y = 0 \][/tex]
Substituting \(y = 0\) back, we see that it works only when \(y = 0\).
Therefore, after checking all the options, the correct solution for the equation \(2x + 3y = 5\) is:
[tex]\[ \boxed{x = \frac{-3y + 5}{2}} \][/tex]
