A student solves the equation [tex]\frac{x+3}{2}=\frac{3x+5}{5}[/tex] using the steps in the table below.

\begin{tabular}{|c|c|}
\hline
Step & Equation \\
\hline
Original equation & [tex]\frac{x+3}{2}=\frac{3x+5}{5}[/tex] \\
\hline
Cross multiplication & [tex]5(x+3)=2(3x+5)[/tex] \\
\hline
Distributive property & [tex]5x+15=2(3x+5)[/tex] \\
\hline
Subtraction property of equality & [tex]5=x[/tex] \\
\hline
\end{tabular}

Which method of solving for the variable could be used instead of cross multiplication?

A. Distributing [tex]x+3[/tex] and then [tex]3x+5[/tex] to both sides of the equation

B. Distributing [tex]x-3[/tex] and then [tex]3x-5[/tex] to both sides of the equation

C. Using the multiplication property of equality to multiply both sides of the equation by 10

D. Using the multiplication property of equality to multiply both sides of the equation by [tex]\frac{1}{10}[/tex]



Answer :

To solve the given equation [tex]\(\frac{x+3}{2}=\frac{3x+5}{5}\)[/tex], let's consider different methods that could also solve the problem.

The original equation is:
[tex]\[ \frac{x+3}{2}=\frac{3x+5}{5} \][/tex]

### Cross Multiplication (Method used in the student's table)
This method involves cross-multiplying to eliminate the fractions:
[tex]\[ 5(x+3)=2(3x+5) \][/tex]

### Other Methods

1. Distributing [tex]\(x+3\)[/tex] and then [tex]\(3x+5\)[/tex] to both sides of the equation:
This method does not make sense mathematically because "distributing" is not an appropriate term in this context. Distributing applies within expressions, not across an equation like this.

2. Distributing [tex]\(x-3\)[/tex] and then [tex]\(3x-5\)[/tex] to both sides of the equation:
Similar to the first point, distributing [tex]\(x-3\)[/tex] and [tex]\(3x-5\)[/tex] also does not make sense and does not follow algebraic rules correctly.

3. Using the multiplication property of equality to multiply both sides of the equation by 10:
To clear the denominators, you can multiply both sides of the equation by the least common multiple (LCM) of 2 and 5, which is 10.
[tex]\[ 10 \left( \frac{x+3}{2} \right) = 10 \left( \frac{3x+5}{5} \right) \][/tex]
Simplifying, we get:
[tex]\[ 5(x+3) = 2(3x+5) \][/tex]
This step is equivalent to what is done using cross-multiplication and will result in the same simplified equation:
[tex]\[ 5x + 15 = 6x + 10 \][/tex]

Using the above steps or cross-multiplication, you would ultimately solve for [tex]\(x\)[/tex] in the same way.

4. Using the multiplication property of equality to multiply both sides of the equation by [tex]\(\frac{1}{10}\)[/tex]:
Multiplying both sides by [tex]\(\frac{1}{10}\)[/tex] will not effectively help to eliminate the denominators. Instead, it will complicate the equation further, and is not a valid approach to solving this particular type of equation.

Given the options, the correct and effective alternative method to cross-multiplication in this scenario is:

Using the multiplication property of equality to multiply both sides of the equation by 10.