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

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


Sagot :

To solve the equation [tex]\(\frac{x+3}{2} = \frac{3x+5}{5}\)[/tex], one alternative method to cross multiplication is to use the multiplication property of equality. You can multiply both sides of the equation by the same number to clear the denominators.

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

By multiplying both sides by 10, we eliminate the fractions:
[tex]\[ 10 \left(\frac{x+3}{2}\right) = 10 \left(\frac{3x+5}{5}\right) \][/tex]

Simplifying both sides:
[tex]\[ 5(x+3) = 2(3x+5) \][/tex]

Continuing with the distributive property:
[tex]\[ 5x + 15 = 6x + 10 \][/tex]

Using the subtraction property of equality, subtract [tex]\(5x\)[/tex] from both sides to isolate the variable:
[tex]\[ 15 = x + 10 \][/tex]

Subtracting 10 from both sides results in:
[tex]\[ 5 = x \][/tex]

Therefore, the correct method of solving this equation, other than cross multiplication, is using the multiplication property of equality to multiply both sides of the equation by 10.
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