Section $2$ . $1$ $\newline$ Question $10$ , $2$ . $1$ . $71$ $\newline$ Solve the equation using the multiplication property $\newline$ $\frac{2}{15} = -\frac{y}{5}$ $\newline$ The solution set is

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Grade 7

Add and subtract rational numbers

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Q. Section

2

1

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Question

10

2

1

71

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Solve the equation using the multiplication property

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\frac{2}{15} = -\frac{y}{5}

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The solution set is

Isolate variable y: First, we need to isolate the variable $y$ by getting rid of the fraction on the right side of the equation. We can do this by multiplying both sides of the equation by $5$ , which is the denominator of the fraction on the right side. $\newline$ Calculation: $(\frac{2}{15}) \cdot 5 = -(\frac{y}{5}) \cdot 5$
Multiply by $5$ : After multiplying both sides by $5$ , the $5$ on the right side cancels out with the denominator of the fraction, leaving us with $y$ alone on the right side. $\newline$ Calculation: $(\frac{2}{15}) \times 5 = -y$
Perform left side multiplication: Now we perform the multiplication on the left side of the equation. $\newline$ Calculation: $(2 \times 5) / 15 = -y$
Simplify left side: Simplify the multiplication on the left side. $\newline$ Calculation: $10 / 15 = -y$
Simplify fraction: We can simplify the fraction $\frac{10}{15}$ by dividing both the numerator and the denominator by their greatest common divisor, which is $5$ . $\newline$ Calculation: $\left(\frac{10}{5}\right) / \left(\frac{15}{5}\right) = -y$
Final fraction: After simplifying the fraction, we get: $\newline$ Calculation: $\frac{2}{3} = -y$
Get rid of negative sign: To solve for $y$ , we need to get rid of the negative sign. We can do this by multiplying both sides of the equation by $-1$ . $\newline$ Calculation: $-1 \times (\frac{2}{3}) = -1 \times (-y)$
Final value for y: After multiplying both sides by $-1$ , we get the final value for $y$ . $\newline$ Calculation: $-\frac{2}{3} = y$