The equation x+53=15 can be solved by performing two operations on both sides.

1 answer:

3 0

Hey!

You're right! This equation can definitely be solved by performing two operations on both sides. Let me show you how it's done.

First, let's write out our equation.

Original Equation :
x + 53 = 15

Now that that's done, we'll be performing two operations on both sides. The operation we'll want to do is subtracting both sides of the equation by 53. We do this to get x on its own.

Original Equation :
x + 53 = 15

New Equation {Added Subtract 53 to Both Sides} :
x + 53 - 53 = 15 - 53

Now we have to solve the equation. Let's do the left side first.

Left Side of the Equation :
x + 53 - 53

Left Side of the Equation {Solved} :
x

Now, we'll solve the right side of the equation.

Right Side of the Equation :
15 - 53

Right Side of the Equation {Solved} :
-38

Now we can put both solutions to both sides of the equation together.

New Equation :
x = -38

Since this cannot be simplified any farther, this is our final answer. And that's it!

So, now we know that in the equation x + 53 = 15, x = -38

Hope this helps!

- Lindsey Frazier ♥

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Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .