Algebra II - Solving Equations

Solve for .

Explanation

In order to solve this equation, we need to isolate the variable, , on the left side of the equals sign. We will do this by performing the same operations on both sides of the equation.

Subtract from both sides of the equation.

Solve.

$\frac{2}{3}x+\frac{2}{3}=-\frac{2}{3}$

$x=-\frac{3}{2}$

$x=\frac{3}{2}$

The answer is not present

Explanation

$\frac{2}{3}x+\frac{2}{3}=-\frac{2}{3}$

Isolate the term with x:

$\frac{2}{3}x=-\frac{2}{3}-\frac{2}{3}$

Simplify:

$\frac{2}{3}x=-\frac{4}{3}$

Isolate x entirely:

$\frac{3}{2}*\frac{2}{3}x=-\frac{4}{3}*\frac{3}{2}$

$\mathbf{x=-2}$

A large tub has two faucets. The hot water faucet, if turned all the way up, can fill the tub in 16 minutes; the cold water faucet, in 12 minutes. Which of the following choices comes closest to the amount of time it takes for both faucets working together to fill the tub?

$7 \textup{ min}$

$8 \textup{ min}$

$5 \textup{ min}$

$6 \textup{ min}$

$4\textup{ min}$

Explanation

Work problems can be solved by looking at them as rate problems.

The hot faucet can fill up the tub at a rate of 16 minutes per tub, or $\frac{1}{16}$ tub per minute. The cold faucet, similarly, can fill up the tub at a rate of 12 minutes per tub, or $\frac{1}{12}$ tub per minute.

Suppose the tub fills up in minutes. Then, at the end of this time, the hot faucet has filled up $\frac{1}{16} X$ tub, and the cold faucet has filled up $\frac{1}{12} X$ tub, for a total of one tub. We can set up this equation and solve for :

$\frac{1}{16} X + \frac{1}{12} X = 1$

$48 \cdot \left (\frac{1}{16} X + \frac{1}{12} X \right )= 48 \cdot 1$

$48 \cdot \frac{1}{16} X + 48 \cdot \frac{1}{12} X = 48$

$7X\div 7 = 48 \div 7$

$X \approx 6.86$

Of the given choices, 7 minutes comes closest.

A large water tank has a water pipe that can be used to fill the tank in forty-five minutes. It has a drain that can empty the tank in one hour and twenty minutes.

One day, someone left the drain open when filling the tank. The tank was completely full by the time someone realized the error. Which of the following comes closest to the amount of time it took to fill the tank?

$1\textup{ hr }45\textup{ min}$

$1\textup{ hr }30\textup{ min}$

$2\textup{ hr }15\textup{ min}$

$2\textup{ hr }30\textup{ min}$

$2\textup{ hr }$

Explanation

Work problems can be solved by looking at them as rate problems. Therefore, we can look at this problem in terms of tanks per minute, rather than minutes per tank. Let be the number of minutes it took to fill the tank.

The pipe filled the tank at a rate of 45 minutes per tank, or $\frac{1}{45}$ tank per minute; over a period of minutes, it filled $\frac{1}{45} X$ tank.

The drain emptied the tank at a rate of 80 minutes per tank, so we can see this as a drain of $\frac{1}{80}$ tank per minute. We can look at draining as "filling negative tanks" - $-\frac{1}{80}$ tank per minute; over a period of minutes, it "filled" $-\frac{1}{80} X$ tank.

Since their work adds up to one tank filled, We can set up, and solve for in, the equation:

$\frac{1}{45} X -\frac{1}{80} X = 1$

$\left (\frac{1}{45} -\frac{1}{80} \right )X = 1$

Using decimal approximations:

$\left (0.02222 - 0.01250 \right )X = 1$

$X = 1 \div 0.00972 \approx 103$ minutes, or 1 hour 43 minutes.

Of the given choices, 1 hour 45 minutes is closest.

Solve for :

Explanation

To solve for x we want to isolate x on one side of the equation and all other numbers on the other side. To do this we start with adding 5 to both sides.