Q. 304 - Intermediate Algebra

Question

Solve the following exercise 

$14y^2-77y=-35$

Step-by-Step Solution

Answer

The solution of the given equation is $y=\frac{1}{2} or y=5$$14y^2-77y=-35$

Step 1. given information

The given equation $14y^2-77y=-35$

Step 2. Simplify the quadratic equation

Simplifying the given equation

$$\begin{gathered} (a - b)(a + b) = 0 \\ a - b = 0 or a + b = 0 \\ a=-b or a=b \\ \mathrm{Simplify} \mathrm{the} \mathrm{quadratic} \mathrm{expression}: \\ 14y^2- 77y = - 35 [\mathrm{Given} \mathrm{the} \mathrm{expression}] \\ 14y^2 - 77y + 35 = 0 [\mathrm{Write} \mathrm{the} \mathrm{quadratic} \mathrm{equation} \mathrm{of} \mathrm{standard} \mathrm{form}] \\ 7(2y^2 - 11y + 5) = 0 [\mathrm{Factor} \mathrm{out} 7 \mathrm{from} \mathrm{expression}] \\ 7(2y^2 - 10y - y + 5) = 0 [\mathrm{Since};-11--10y-y] \\ 7(2y (y - 5) - 1(y - 5\left)\right) = 0 [\mathrm{Factor} \mathrm{out} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{group}] \\ 7(2y - 1) (y - 5) = 0 [\mathrm{Factor} \mathrm{the} \mathrm{common} \mathrm{factor} \mathrm{from} \mathrm{expression}] \\ 7 \simeq 0 or 2y-1-0 or y-5-0 [\mathrm{Use} \mathrm{the} \mathrm{zero} \mathrm{product} \mathrm{property}] \\ or y = 1/2 or y = 5 \end{gathered}$$

Step 3. Checking the equation

$$\begin{gathered} \mathrm{Check} \\ \mathrm{Substitute} \mathrm{them} \mathrm{value} \mathrm{y}=1/2 \mathrm{in} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 14y^2-77y=-35 [Given equation] \\ 14\times (\frac{1}{2}{)}^2-77(\frac{1}{2})=-35 [Put y=\frac{1}{2}] \\ \frac{7}{2}-\frac{77}{2}=-35 \\ -35=-35 \mathrm{True} \\ \mathrm{Substitute} \mathrm{the} \mathrm{value} \mathrm{y}=5 \mathrm{in} \mathrm{the} \mathrm{original} \mathrm{equation}. \\ 14y^2-77y=-35 [\mathrm{Given} \mathrm{equation}\left] \\ 14\right(5)²-77(5)=-35 [Put,y=5] \\ 350-385-35 \\ -35=-35 True \\ The answer is y=1/2,y=5 \end{gathered}$$