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$Solve for y. Express your answer as a proper or improper fraction in simplest terms. -(1)/(2)=-(1)/(4)-(4)/(7)y Answer: y=$

Solve for $y$ . Express your answer as a proper or improper fraction in simplest terms. $\newline$ $-\frac{1}{2}=-\frac{1}{4}-\frac{4}{7} y$ $\newline$ Answer: $y=$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Solve for

y

. Express your answer as a proper or improper fraction in simplest terms.

\newline

-\frac{1}{2}=-\frac{1}{4}-\frac{4}{7} y

\newline

Answer:

y=

Isolate term containing y: First, we want to isolate the term containing $y$ on one side of the equation. To do this, we can start by adding $(1/4)$ to both sides of the equation to move it away from the right side. $-\frac{1}{2} + \frac{1}{4} = -\frac{1}{4} + \frac{1}{4} - \frac{4}{7}y$ Simplify the right side by combining like terms.
Simplify by adding fractions: Now we simplify the left side by finding a common denominator for the fractions $-\frac{1}{2}$ and $\frac{1}{4}$ . The common denominator is $4$ . $\newline$ $-\frac{2}{4} + \frac{1}{4} = 0 - \frac{4}{7}y$ $\newline$ Combine the fractions on the left side.
Combine fractions on left side: After combining the fractions on the left side, we get: $\newline$ $-\frac{1}{4} = - \frac{4}{7}y$ $\newline$ Now we want to solve for $y$ , so we need to get rid of the coefficient $-\frac{4}{7}$ that is multiplied by $y$ . To do this, we can multiply both sides of the equation by the reciprocal of $-\frac{4}{7}$ , which is $-\frac{7}{4}$ . $\newline$ $-\frac{7}{4} \times -\frac{1}{4} = -\frac{7}{4} \times - \left(\frac{4}{7}y\right)$
Multiply by reciprocal: Multiplying both sides by $-\frac{7}{4}$ , we get: $\newline$ $\left(\frac{7}{4}\right) \times \left(\frac{1}{4}\right) = y$ $\newline$ Now we multiply the numerators and the denominators separately.
Multiply numerators and denominators: After multiplying, we get: $\newline$ $(7 \times 1) / (4 \times 4) = y$ $\newline$ $7 / 16 = y$ $\newline$ This is the value of $y$ in its simplest form.