Problem 18
Question
Solve each equation. Check each solution. $$ \frac{2}{y}+\frac{1}{2}=\frac{5}{2 y} $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \( y = 1 \)
1Step 1: Clear the fractions
The first step to solving this fraction-based equation is to eliminate the fractions. A common method of doing this is to multiply every term by the denominator of every fraction in the equation, which is known as the 'least common multiple' (LCM). Here, the LCM is \( 2y \). So, multiply every term in the equation as follows:\( 2y * \frac{2}{y} + 2y * \frac{1}{2} = 2y * \frac{5}{2y} \)That yields: \( 4 + y = 5 \)
2Step 2: Solve for \( y \)
Now, solve for \( y \) by bringing the \( y \)-term from right side of the equation to left. We subtract \( y \) from both sides to get:\( 4 + y - y = 5 - y \)That simplifies to: \( 4 = 5 - y \)
3Step 3: Final calculation for \( y \)
Finally, solve for \( y \) by subtracting 5 from both sides of the equation and multiply the result by -1 to get \( y \) positive:\( 4 - 5 = 5 - y - 5 \)This simplifies to:\( -1 = - y \) Multiplying by -1 gives:\( y = 1 \)
4Step 4: Checking the solution
Substitute \( y = 1 \) back into the original equation to check if the left side equals the right side:\(\frac{2}{1}+\frac{1}{2} = 2 + 0.5 = 2.5\) and \(\frac{5}{2*1} = 2.5 \)Since both sides of the equation are equal, \( y = 1 \) is the correct solution.
Other exercises in this chapter
Problem 17
Sketch the asymptotes and the graph of each equation. \(y=\frac{1}{x-3}+4\)
View solution Problem 18
A standard number cube is tossed. Find each probability. \(P(3 \text { or odd })\)
View solution Problem 18
Divide. State any restrictions on the variables. $$ \frac{y^{2}-5 y+6}{y^{3}} \div \frac{y^{2}+3 y-10}{4 y^{2}} $$
View solution Problem 18
Describe the vertical asymptotes and holes for the graph of each rational function. $$ y=\frac{6 x^{2}+x-2}{3 x^{2}+17 x+10} $$
View solution